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# IBPS Clerk English Language Questions with Answers Practice online test 8

Description: free IBPS Clerk English Language Questions with Answers Practice test 8 for IBPS Clerk Preliminary and Main online test Prepare bank Clerk banking mock exams adda

1 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

Last April, the Allahabad High Court allowed the farmers' writ pleas and $(1)$ the project. It $(2)$ that farmers who had received $(3)$ for their land should either return the money and take back the land or willingly hand it over for the project. Around 140 farmers did not accept compensation. Those who did are in no condition to repay the amount, causing an $(4)$which the administration is $(5)$to break.

After the $(6)$violence, the $(7)$have been peaceful, but the farmers complain they are being $(8)$ by local $(9)$ and officials to give up their land and $(10)$the protests.
$(6)$
2 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

Last April, the Allahabad High Court allowed the farmers' writ pleas and $(1)$ the project. It $(2)$ that farmers who had received $(3)$ for their land should either return the money and take back the land or willingly hand it over for the project. Around 140 farmers did not accept compensation. Those who did are in no condition to repay the amount, causing an $(4)$which the administration is $(5)$to break.

After the $(6)$violence, the $(7)$have been peaceful, but the farmers complain they are being $(8)$ by local $(9)$ and officials to give up their land and $(10)$the protests.
$(7)$
 gossip condition protests riot turnoil
3 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

Last April, the Allahabad High Court allowed the farmers' writ pleas and $(1)$ the project. It $(2)$ that farmers who had received $(3)$ for their land should either return the money and take back the land or willingly hand it over for the project. Around 140 farmers did not accept compensation. Those who did are in no condition to repay the amount, causing an $(4)$which the administration is $(5)$to break.

After the $(6)$violence, the $(7)$have been peaceful, but the farmers complain they are being $(8)$ by local $(9)$ and officials to give up their land and $(10)$the protests.
$(8)$
 helped rescued intimidated praised protected
4 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

Last April, the Allahabad High Court allowed the farmers' writ pleas and $(1)$ the project. It $(2)$ that farmers who had received $(3)$ for their land should either return the money and take back the land or willingly hand it over for the project. Around 140 farmers did not accept compensation. Those who did are in no condition to repay the amount, causing an $(4)$which the administration is $(5)$to break.

After the $(6)$violence, the $(7)$have been peaceful, but the farmers complain they are being $(8)$ by local $(9)$ and officials to give up their land and $(10)$the protests.
$(9)$
 hawkers workers goons friends person
5 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

Last April, the Allahabad High Court allowed the farmers' writ pleas and $(1)$ the project. It $(2)$ that farmers who had received $(3)$ for their land should either return the money and take back the land or willingly hand it over for the project. Around 140 farmers did not accept compensation. Those who did are in no condition to repay the amount, causing an $(4)$which the administration is $(5)$to break.

After the $(6)$violence, the $(7)$have been peaceful, but the farmers complain they are being $(8)$ by local $(9)$ and officials to give up their land and $(10)$the protests.
$(10)$
 forbid prohibit check enforce discontinue
6 . Read the following passage carefully and answer the questions given below it. Certain words/phrases in the passage are printed in bold to help you locate them while answering some of the questions.

A US intelligence portrait of the world in 2030 predicts that China will be the largest economic power, climate change will create instability by contributing to water and food shortages, and there will be a 'tectonic shift' with the rise of a global middle class.

The National Intelligence Council's (NIC) Global Trends Report, published every five years, paints a future in which US power will greatly diminish but no other individual state rises to supplantit.

'There will not be any hegemonic power. Power will shift to networks and coalitions in a multipolar world,' it says. The report offers a series of potential scenarios for 2030. It says the best outcome would be the one in which 'China and the US collaborate on a range of issues, leading to broader global co-operation.' It says the worst is a world in which 'the US draws inward and globalisation stalls.' 'A collapse or sudden retreatof US power probably wouldresult in an extended period of global anarchy; no leading power would be likely to replace the United States as guarantor of the international order,' it says, working on the assumption that the US is a force for stability - a premise open to challenge in Iraq and elsewhere in the Middle East and beyond.

The NIC report draws a distinctionbetween what it calls 'megatrends' - things that are highly likely to occur - and 'game-changers', which are far less certain. Among the megatrends is growing prosperity across the globe. 'The growth of the global middle class constitutes a tectonic shift: for the first time, a majority of the world's population will not be impoverished, and the middle classes will be the most important social and economic sector in the vast majority of countries around the world,' the report says.

With prosperity spreading across the globe will come shifts in influence and power. 'The diffusion of power among countries will have a dramatic impact by 2030. Asia will have surpassed North America and Europe combined in terms of global power, based upon GDP, population size, military spending, and technological investment. China alone will probably have the largest economy, surpassing that of the United States a few years before 2030,' the report says. In addition to China, India and Brazil, regional players such as Colombia, Indonesia, Nigeria, South Africa and Turkey will become especially important to the global economy.

'Meanwhile, the economies of Europe, Japan, and Russia are likely to continue their slow relative declines,' the report says. 'In a tectonic shift, individuals and small groups will have greater accessto lethal and disruptive technologies (particularly precision-strike capabilities, cyber instruments, and bio-terror weaponry), enabling themto perpetrate large-scale violence - a capability formerly the monopoly of states,' it says.

The megatrends also point to increased instability because of rising demand for water, food and energy compounded by climate change.
As per a US intelligence portrait of the world in 2030, what would be the impact of the diffusion of power among countries?
 Regional players such as Indonesia, Nigeria, Colombia and South Africa will not play any significant role in the global economy Asia will have surpassed North America and Europe combined in terms of global power, population size, military spending and technological investment Europe, Japan and Russia will register a significant economic growth Individuals and small groups will have no access to disruptive technologies. All the above
7 . Read the following passage carefully and answer the questions given below it. Certain words/phrases in the passage are printed in bold to help you locate them while answering some of the questions.

A US intelligence portrait of the world in 2030 predicts that China will be the largest economic power, climate change will create instability by contributing to water and food shortages, and there will be a 'tectonic shift' with the rise of a global middle class.

The National Intelligence Council's (NIC) Global Trends Report, published every five years, paints a future in which US power will greatly diminish but no other individual state rises to supplantit.

'There will not be any hegemonic power. Power will shift to networks and coalitions in a multipolar world,' it says. The report offers a series of potential scenarios for 2030. It says the best outcome would be the one in which 'China and the US collaborate on a range of issues, leading to broader global co-operation.' It says the worst is a world in which 'the US draws inward and globalisation stalls.' 'A collapse or sudden retreatof US power probably wouldresult in an extended period of global anarchy; no leading power would be likely to replace the United States as guarantor of the international order,' it says, working on the assumption that the US is a force for stability - a premise open to challenge in Iraq and elsewhere in the Middle East and beyond.

The NIC report draws a distinctionbetween what it calls 'megatrends' - things that are highly likely to occur - and 'game-changers', which are far less certain. Among the megatrends is growing prosperity across the globe. 'The growth of the global middle class constitutes a tectonic shift: for the first time, a majority of the world's population will not be impoverished, and the middle classes will be the most important social and economic sector in the vast majority of countries around the world,' the report says.

With prosperity spreading across the globe will come shifts in influence and power. 'The diffusion of power among countries will have a dramatic impact by 2030. Asia will have surpassed North America and Europe combined in terms of global power, based upon GDP, population size, military spending, and technological investment. China alone will probably have the largest economy, surpassing that of the United States a few years before 2030,' the report says. In addition to China, India and Brazil, regional players such as Colombia, Indonesia, Nigeria, South Africa and Turkey will become especially important to the global economy.

'Meanwhile, the economies of Europe, Japan, and Russia are likely to continue their slow relative declines,' the report says. 'In a tectonic shift, individuals and small groups will have greater accessto lethal and disruptive technologies (particularly precision-strike capabilities, cyber instruments, and bio-terror weaponry), enabling themto perpetrate large-scale violence - a capability formerly the monopoly of states,' it says.

The megatrends also point to increased instability because of rising demand for water, food and energy compounded by climate change.
Which of the following statements is not true according to the given passage?

(A) According to NIC's Global Trends Report, US power will greatly reduce but no other individual state will rise to replace it.

(B) Power will shift to a multipolar world.

(C) There is no chance of global anarchy even during collapse of US power.
 Only A Only B Only C Only A and B Only B and C
8 . Read the following passage carefully and answer the questions given below it. Certain words/phrases in the passage are printed in bold to help you locate them while answering some of the questions.

A US intelligence portrait of the world in 2030 predicts that China will be the largest economic power, climate change will create instability by contributing to water and food shortages, and there will be a 'tectonic shift' with the rise of a global middle class.

The National Intelligence Council's (NIC) Global Trends Report, published every five years, paints a future in which US power will greatly diminish but no other individual state rises to supplantit.

'There will not be any hegemonic power. Power will shift to networks and coalitions in a multipolar world,' it says. The report offers a series of potential scenarios for 2030. It says the best outcome would be the one in which 'China and the US collaborate on a range of issues, leading to broader global co-operation.' It says the worst is a world in which 'the US draws inward and globalisation stalls.' 'A collapse or sudden retreatof US power probably wouldresult in an extended period of global anarchy; no leading power would be likely to replace the United States as guarantor of the international order,' it says, working on the assumption that the US is a force for stability - a premise open to challenge in Iraq and elsewhere in the Middle East and beyond.

The NIC report draws a distinctionbetween what it calls 'megatrends' - things that are highly likely to occur - and 'game-changers', which are far less certain. Among the megatrends is growing prosperity across the globe. 'The growth of the global middle class constitutes a tectonic shift: for the first time, a majority of the world's population will not be impoverished, and the middle classes will be the most important social and economic sector in the vast majority of countries around the world,' the report says.

With prosperity spreading across the globe will come shifts in influence and power. 'The diffusion of power among countries will have a dramatic impact by 2030. Asia will have surpassed North America and Europe combined in terms of global power, based upon GDP, population size, military spending, and technological investment. China alone will probably have the largest economy, surpassing that of the United States a few years before 2030,' the report says. In addition to China, India and Brazil, regional players such as Colombia, Indonesia, Nigeria, South Africa and Turkey will become especially important to the global economy.

'Meanwhile, the economies of Europe, Japan, and Russia are likely to continue their slow relative declines,' the report says. 'In a tectonic shift, individuals and small groups will have greater accessto lethal and disruptive technologies (particularly precision-strike capabilities, cyber instruments, and bio-terror weaponry), enabling themto perpetrate large-scale violence - a capability formerly the monopoly of states,' it says.

The megatrends also point to increased instability because of rising demand for water, food and energy compounded by climate change.
Which of the following is not among the predictions made about the world in 2030 by a US intelligence report?
 There will be a tectonic shift with the rise of a global middle class Due to climate change there will be shortages of food and water China will emerge as the largest economic power India will be a technologically advanced country None of these
9 . Read the following passage carefully and answer the questions given below it. Certain words/phrases in the passage are printed in bold to help you locate them while answering some of the questions.

A US intelligence portrait of the world in 2030 predicts that China will be the largest economic power, climate change will create instability by contributing to water and food shortages, and there will be a 'tectonic shift' with the rise of a global middle class.

The National Intelligence Council's (NIC) Global Trends Report, published every five years, paints a future in which US power will greatly diminish but no other individual state rises to supplantit.

'There will not be any hegemonic power. Power will shift to networks and coalitions in a multipolar world,' it says. The report offers a series of potential scenarios for 2030. It says the best outcome would be the one in which 'China and the US collaborate on a range of issues, leading to broader global co-operation.' It says the worst is a world in which 'the US draws inward and globalisation stalls.' 'A collapse or sudden retreatof US power probably wouldresult in an extended period of global anarchy; no leading power would be likely to replace the United States as guarantor of the international order,' it says, working on the assumption that the US is a force for stability - a premise open to challenge in Iraq and elsewhere in the Middle East and beyond.

The NIC report draws a distinctionbetween what it calls 'megatrends' - things that are highly likely to occur - and 'game-changers', which are far less certain. Among the megatrends is growing prosperity across the globe. 'The growth of the global middle class constitutes a tectonic shift: for the first time, a majority of the world's population will not be impoverished, and the middle classes will be the most important social and economic sector in the vast majority of countries around the world,' the report says.

With prosperity spreading across the globe will come shifts in influence and power. 'The diffusion of power among countries will have a dramatic impact by 2030. Asia will have surpassed North America and Europe combined in terms of global power, based upon GDP, population size, military spending, and technological investment. China alone will probably have the largest economy, surpassing that of the United States a few years before 2030,' the report says. In addition to China, India and Brazil, regional players such as Colombia, Indonesia, Nigeria, South Africa and Turkey will become especially important to the global economy.

'Meanwhile, the economies of Europe, Japan, and Russia are likely to continue their slow relative declines,' the report says. 'In a tectonic shift, individuals and small groups will have greater accessto lethal and disruptive technologies (particularly precision-strike capabilities, cyber instruments, and bio-terror weaponry), enabling themto perpetrate large-scale violence - a capability formerly the monopoly of states,' it says.

The megatrends also point to increased instability because of rising demand for water, food and energy compounded by climate change.
What could be the worst outcome of the US intelligence report about potential scenarios for 2030?
 India would emerge as the most powerful nuclear country in the world China and the US would collaborate on a range of issues leading to a broader global cooperation The United States would shrink inward and the process of globalisation would stall. Japan would merge as the technology giant among the Asian countries None of these
10 . Read the following passage carefully and answer the questions given below it. Certain words/phrases in the passage are printed in bold to help you locate them while answering some of the questions.

A US intelligence portrait of the world in 2030 predicts that China will be the largest economic power, climate change will create instability by contributing to water and food shortages, and there will be a 'tectonic shift' with the rise of a global middle class.

The National Intelligence Council's (NIC) Global Trends Report, published every five years, paints a future in which US power will greatly diminish but no other individual state rises to supplantit.

'There will not be any hegemonic power. Power will shift to networks and coalitions in a multipolar world,' it says. The report offers a series of potential scenarios for 2030. It says the best outcome would be the one in which 'China and the US collaborate on a range of issues, leading to broader global co-operation.' It says the worst is a world in which 'the US draws inward and globalisation stalls.' 'A collapse or sudden retreatof US power probably wouldresult in an extended period of global anarchy; no leading power would be likely to replace the United States as guarantor of the international order,' it says, working on the assumption that the US is a force for stability - a premise open to challenge in Iraq and elsewhere in the Middle East and beyond.

The NIC report draws a distinctionbetween what it calls 'megatrends' - things that are highly likely to occur - and 'game-changers', which are far less certain. Among the megatrends is growing prosperity across the globe. 'The growth of the global middle class constitutes a tectonic shift: for the first time, a majority of the world's population will not be impoverished, and the middle classes will be the most important social and economic sector in the vast majority of countries around the world,' the report says.

With prosperity spreading across the globe will come shifts in influence and power. 'The diffusion of power among countries will have a dramatic impact by 2030. Asia will have surpassed North America and Europe combined in terms of global power, based upon GDP, population size, military spending, and technological investment. China alone will probably have the largest economy, surpassing that of the United States a few years before 2030,' the report says. In addition to China, India and Brazil, regional players such as Colombia, Indonesia, Nigeria, South Africa and Turkey will become especially important to the global economy.

'Meanwhile, the economies of Europe, Japan, and Russia are likely to continue their slow relative declines,' the report says. 'In a tectonic shift, individuals and small groups will have greater accessto lethal and disruptive technologies (particularly precision-strike capabilities, cyber instruments, and bio-terror weaponry), enabling themto perpetrate large-scale violence - a capability formerly the monopoly of states,' it says.

The megatrends also point to increased instability because of rising demand for water, food and energy compounded by climate change.
What would be the impact of megatrends? Give your answer in the context of the given passage.
 There will be a shift of power from upper class to middle class A majority of the world population will not be impoverished The middle classes will be the most important social and economic sector in the vast majority of countries around the world Only 2) and 3) All 1), 2) and 3)


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# Why for number of leaves in a tree (all types of trees) is it true

I have to prove the following claim, given the tree $T=(V,E)$, $|V|\geq3$: $$|V_1| \leq \frac { |V| \times (\Delta (V) - 2) + 2 }{ \Delta (V) - 1 }$$ where $|V_1| -$ number of leaves in a tree, and $\Delta (V) -$ maximum degree of all vertices.

I was sitting on this question for days and have no idea how to prove it.

Can someone help?

Thanks!

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What about induction on $|V|$? Assume $|V|=n+1$ then delete a leaf.. –  Berci Sep 28 '12 at 7:59
@Berci Do you think that induction is correctly solution? I do not understand where may appear a fraction.. –  jofisher Sep 28 '12 at 8:14

Let $V_2=V\setminus V_1$, the set of interior vertices of the tree, and to simplify notation write $\Delta$ for $\Delta(V)$. The inequality can be rearranged to yield

$$(\Delta-2)|V|+2\ge(\Delta-1)|V_1|=(\Delta-2)|V_1|+|V_1|$$ and then

$$|V_1|\le(\Delta-2)(|V|-|V_1|)+2=(\Delta-2)|V_2|+2\;.\tag{1}$$

I’ll prove $(1)$ by induction on $|V|$.

Pick two leaves, $v_0$ and $v_n$, and let $v_0v_1\dots v_n$ be the unique path between them, which I’ll call the spine of the tree. Each of the vertices $v_1,\dots,v_{n-1}$ has degree at most $\Delta$, so it’s an end of at most $\Delta-2$ edges that aren’t in the spine. For each non-spine edge $e$ incident at $v_k$ let $T_k(e)$ be the subtree whose vertices are $v_k$ and all vertices of $T$ whose unique path to $v_k$ included the edge $e$. Let the vertex set of $T_k(e)$ be $V_k(e)$, and let $V_{k,1}(e)$ and $V_{k,2}(e)$ be the sets of leaves and interior vertices, respectively. Finally, let $t_k=\deg(v_k)-2$ be the number of subtrees incident at $v_k$; clearly $t_k\le\Delta-2$.

By the induction hypothesis we have

$$|V_{k,1}(e)|\le(\Delta-2)|V_{k,2}(e)|+2$$

for each of these subtrees.

Each $v_k$ is a leaf of each $T_k(e)$, so

\begin{align*} |V_1|-2+\sum_kt_k&=\sum_{k,e}|V_{k,1}|\\ &\le\sum_{k,e}\left((\Delta-2)|V_{k,2}(e)|+2\right)\\ &=(\Delta-2)\sum_{k,e}|V_{k,2}(e)|+\sum_{k,e}2\\ &=(\Delta-2)\Big(|V_2|-(n-1)\Big)+2\sum_kt_k\;, \end{align*}

and therefore

\begin{align*} |V_1|&\le 2+(\Delta-2)|V_2|-(n-1)(\Delta-2)+\sum_kt_k\\ &\le 2+(\Delta-2)|V_2|\;, \end{align*}

since $$\sum_kt_k\le(n-1)(\Delta-2)\;.$$

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$$2 = 2|V| - 2 |E| = \sum_{v \in V} 2 - \deg(v) \geq |V_1| + (2 - \Delta(V))(|V| - |V_1|).$$

Rearranging as in Brian M. Scott's answer gives the required result.

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Backward Euler method

Can you explain me how does the backward Euler method works? I have seen the formula and try to understand the method, but what I can't understand is why and how to use the Newton-Rapson method.

Do you have a link to a good tutorial? Something that can help me understand it graphically?

UPDATE

According to what Paul said, could you please tell me if what follows is correct?

The Cauchy problem is like this

$$u(t) = f\big ( u(t), t \big ) \quad \text{where} \quad f\big ( u(t), t \big ) = u'\big ( u(t), t \big )$$

then, according to the example given by Paul, I'll have something like:

$$u'(t)=\big [u(t) \big]^3$$

Now, the backward Euler method reads:

$$u_{t+1} = u_t + hf\big ( u_{t+1}, t+1 )~~~~~~~~~~\text{not really sure about this}$$

So for, $t = 0$ and $h = 1/2$ I should have the following

$$u_1 = u_0 + \frac{1}{2}(u_1)^3$$

For this equation I have to solve by means of Newton-Raphson before going to the next time $t = 1$:

$$u_1 - u_0 - \frac{1}{2}(u_1)^3 = 0$$

From here I don't know what to do, how to use Newton's method??

UPDATE

I think I finally got it. From backward Euler's method I have:

$$u_{t+1} = u_t + hf(u_{t+1}, t+1)$$

this can be seen as (just like everybody told me):

$$F(u_{t+1})= u_{t+1} - u_t - hf(u_{t+1}, t+1) = 0$$

where $u_{t+1} = x$ so I have

$$F(x)= x - u_t - hf(x, t+1) = 0$$

which now I see a familiar way to use Newton's (or whatever other method) to find the value of $x$

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I think your last formula is correct. –  Paul Jan 18 '13 at 22:31

You would use backward euler method to solve a differential equation of the form $u_t =f(u,t)$ where $f$ is not necessarily a linear function in u. When f is non-linear, then the backward euler method results in a set of non-linear equations that need to be solved for each time step. Ergo, Newton-raphson can be used to solve it.

For example, take

$$\frac{du}{dt}=u^3(t)$$

Backward euler results in

$$\frac{u_{t+1}-u_{t}}{\Delta t}=u_{t+1}^3$$

Since you can't solve for $u_{t+1}$ directly, you have to estimate it by a numerical scheme. This is where newton-raphson comes into play. Rearrange this equation so that the RHS=0, we get

$$\frac{u_{t+1}-u_{t}}{\Delta t}-u_{t+1}^3=0$$

Now, we can apply newton's method to find the root of this equation $u_{t+1}$, which in turn becomes the solution to the differential equation at the next timestep.

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Ok. Almost got it. First, when you say "of the form $u_t = f(u,t)$", this means $f(u,t) = u'\big ( u(t), t \big )$, I ask just because I don't want to mess up with the notation. Then, $du/dt = u^3(t) = f(u, t)$ right? Finally, when the RHS = 0, from the resulting equation $u_t$, $\Delta t$, are known but not $u_{t+1}$ so how do I evaluate that with Newton-Raphson? –  BRabbit27 Jan 18 '13 at 17:31

Backwards Euler, as you state you know, is time advances using the equation: $U_{t+\Delta t} = U_{t}+ U'(t,U_{t+\Delta t}) \Delta t$.

You can not explicitly evaluate $U'_{t+\Delta t}$ since you don't know $U_{t+\Delta t}$. To make it more clear the equation is:

$X = U(t) + U'(t,X)\Delta t$

Therefore you must solve the equation for X (which is U at the next time step) using some methodology (this is where you use Newton-Raphson in your case).

As far as graphical interpretation, think about your standard Euler. The idea there is that you are taking the current value of $U$ and its derivative at the current time, and assuming that it is nearly linear for some small change in time ($\Delta t$). So you are drawing a straight line from $U(t)$ at a slope of $U'(t)$.

For backwards Euler, all you are doing is using the slope at the end of your line approximation rather than the start of it. As to why you would want to do this, it is a more complicated answer involving the stability of your solution.

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I appreciate your help, but still I can't see where does the Newton-Rapson comes to play... I mean, the equation of backward Euler I got it, and obviously I'm trying to compute the value of $U'(t, U_{t+\Delta t})$ which I can't. Newton-Rapson is used to find the zeros of a function isn't it? Sorry, I can't see how those two things fit... –  BRabbit27 Jan 18 '13 at 16:38
All this subindices confuse me A LOT. Because when I saw the backward Euler combined with Newton-Raphson, there appeared a superindex k, k+1... –  BRabbit27 Jan 18 '13 at 16:41
I made a slight update to hopefully clarify for you. –  Godric Seer Jan 18 '13 at 21:01


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### January 23th, 2005

2000 Mathematics Subject Classification. 46B20, 47A12.Research partially supported by Spanish MCYT project no. BFM2003-01681 and Junta de Andalucıa grant FQM-185 .
<ph f="ecbx">On the intrinsic and the spatial numerical range</ph>
Miguel Martın,1 Javier Merı, and Rafael Paya
mmartins@ugr.es, jmeri@ugr.es, rpaya@ugr.es

• Abstract. For a bounded function $f$  from the unit sphere of a closed subspace $X$  of a Banach space $Y$  , we study when the closed convex hull of its spatial numerical range $W\left(f\right)$  is equal to its intrinsic numerical range $V\left(f\right)$  . We show that for every infinite-dimensional Banach space $X$  there is a superspace $Y$  and a bounded linear operator $T:X⟶Y$  such that $\overline{co}W\left(T\right)\ne V\left(T\right)$  . We also show that, up to renormig, for every non-reflexive Banach space $Y$  , one can find a closed subspace $X$  and a bounded linear operator $T\in L\left(X,Y\right)$  such that $\overline{co}W\left(T\right)\ne V\left(T\right)$  .
Finally, we introduce a sufficient condition for the closed convex hull of the spatial numerical range to be equal to the intrinsic numerical range, which we call the Bishop-Phelps-Bollobas property, and which is weaker than the uniform smoothness and the finite-dimensionality.
We characterize strong subdifferentiability and uniform smoothness in terms of this property.

1 Introduction

Given a Banach space $Y$  over $\mathbb{K}$  ( $=\mathbb{R}$  or $\mathbb{C}$  ), we write ${B}_{Y}$  for the closed unit ball and ${S}_{Y}$  for the unit sphere of $Y$  . The dual space of $Y$  will be denoted by ${Y}^{*}$  . If $Z$  is another Banach space, we write $L\left(Z,Y\right)$  for the Banach space of all bounded linear operators from $Z$  into $Y$  ; if $Z=Y$  we simply write $L\left(Y\right):=L\left(Y,Y\right)$  to denote the the Banach algebra of all bounded linear operators on $Y$  . For an element $u\in {S}_{Y}$  , we write $D\left(Y,u\right):=\left\{{y}^{*}\in {Y}^{*}:\parallel {y}^{*}\parallel ={y}^{*}\left(u\right)=1\right\},$  the ${w}^{*}$  -closed and convex set of all states of $Y$  relative to $u$  . Let us mention two facts, both consequence of the Hahn-Banach Theorem, which will be relevant to our discussion. On one hand, we have
 $\begin{array}{c}{lim}_{\alpha ↓0}\frac{\parallel u+\alpha y\parallel -1}{\alpha }=max\left\{Re{z}^{*}\left(y\right):{z}^{*}\in D\left(Z,u\right)\right\},\end{array}$ (1)
(see [10,Theorem V.9.5for a proof ). On the other hand, if $X$  is a subspace of $Y$  and $u\in X$  , then $D\left(X,u\right)$  coincides with the restriction to $X$  of the elements of $D\left(Y,u\right)$  .
If $Y$  is a Banach space, by a closed subspace of $Y$  we mean a Banach space $X$  and an inclusion operator $J:X⟶Y$  (i.e., $J$  is a linear isometry), and we also say that $Y$  is a superspace of $X$  . When no confusion is possible, we omit $J$  , but all the definitions below depend on the way that $X$  is a subspace of $Y$  . Let us fix $X$  and $Y$  as above. We write $\Pi \left(X,Y\right)$  to denote the subset of ${S}_{X}×{S}_{{Y}^{*}}$  given by $\Pi \left(X,Y\right):=\left\{\left(x,{y}^{*}\right)\in {S}_{X}×{S}_{{Y}^{*}}:{y}^{*}\in D\left(Y,Jx\right)\right\}.$  If $X=Y$  , we just write $\Pi \left(Y\right):=\Pi \left(Y,Y\right)$  . We denote by $B\left({S}_{X},Y\right)$  the Banach space of all bounded functions from ${S}_{X}$  to $Y$  , endowed with the natural supremum norm, and we write ${C}_{\text{u}}\left({S}_{X},Y\right)$  for its closed subspace consisting of all bounded and uniformly continuous functions. For $f\in B\left({S}_{X},Y\right)$  we can define two different numerical ranges, namely, the spatial numerical range defined as $W\left(f\right):=\left\{{y}^{*}\left(f\left(x\right)\right):\left(x,{y}^{*}\right)\in \Pi \left(X,Y\right)\right\},$  and the intrinsic numerical range given by $V\left(f\right):=\left\{\Phi \left(f\right):\Phi \in D\left(B\left({S}_{X},Y\right),J{|}_{{S}_{X}}\right)\right\}.$  The name of intrinsic numerical range comes from the fact that if $f$  belongs to any closed subspace $Z$  of $B\left({S}_{X},Y\right)$  , we can calculate $V\left(f\right)$  using only elements in ${Z}^{*}$  . These numerical ranges appeared in a paper by L. Harris [16for continuous functions. In the particular case when $X=Y$  and $f$  is (the restriction to ${S}_{Y}$  of ) a bounded linear operator, the spatial numerical range was introduced by F. Bauer (field of values subordinate to a norm [1), extending Toeplitz’s numerical range of matrices [25and, concerning applications, it is equivalent to Lumer’s numerical range [18. Also in this case, the intrinsic numerical range appears as the algebra numerical range in the monographs by F. Bonsall and J. Duncan [7, 8; we refer the reader to these books for general information and background. When $f$  is (the restriction to ${S}_{Y}$  of ) a uniformly continuous function from ${B}_{Y}$  to $Y$  which is holomorphic on the interior of ${B}_{Y}$  , both ranges appeared for the first time in [15, where some applications are given. Let us fix a Banach space $Y$  and a closed subspace $X$  . For every $f\in B\left({S}_{X},Y\right)$  , $V\left(f\right)$  is closed and convex, and we have
 $\begin{array}{c}\overline{co}W\left(f\right)\subseteq V\left(f\right),\end{array}$ (2)
where $\overline{co}$  means closed convex hull. (Indeed, for $x\in {S}_{X}$  and ${y}^{*}\in {S}_{{Y}^{*}}$  , the mapping $x\otimes {y}^{*}$  from $B\left({S}_{X},Y\right)$  to $\mathbb{K}$  defined by $\left[x\otimes {y}^{*}\right]\left(g\right):={y}^{*}\left(g\left(x\right)\right)\left(g\in B\left({S}_{X},Y\right)\right)$  is an element of $D\left(B\left({S}_{X},Y\right),J\right)$  .) In the case when $X=Y$  , the inclusion above is known to be an equality whenever $f$  is a uniformly continuous function [16,Theorem 1(see also [7,§9for bounded linear operators, [15for holomorphic functions, and [23for a slightly more general result). On the other hand, the equality $\overline{co}W\left(f\right)=V\left(f\right)$  for arbitrary bounded functions cannot be expected in general. Indeed, this equality holds for every $f\in B\left({S}_{Y},Y\right)$  if and only if $Y$  is uniformly smooth [22. In the general case when $X$  is a proper subspace, two sufficient conditions are given in [16,Theorems2and3for the equality in Eq.  2 , namely, such a equality holds for all $f\in {C}_{\text{u}}\left({S}_{X},Y\right)$  if either $X$  is finite-dimensional or $Y$  is uniformly smooth (see definition below). Let us mention that if $\overline{co}W\left(f\right)=V\left(f\right)$  for a bounded function $f\in B\left({S}_{X},Y\right)$  , then $maxReV\left(f\right)=supReW\left(f\right).$  Therefore, the following formulae, consequence of Eq.  1 , will be useful:
 $\begin{array}{cc}maxReV\left(f\right)& ={lim}_{\alpha ↓0}\frac{\parallel J+\alpha f\parallel -1}{\alpha }={lim}_{\alpha ↓0}{sup}_{x\in {S}_{X}}\frac{\parallel x+\alpha f\left(x\right)\parallel -1}{\alpha },\end{array}$ (3)
 $\begin{array}{cc}supReW\left(f\right)& ={sup}_{x\in {S}_{X}}{lim}_{\alpha ↓0}\frac{\parallel x+\alpha f\left(x\right)\parallel -1}{\alpha }.\end{array}$ (4)
 $\begin{array}{}\end{array}$
To state the main results of the paper, let us recall some definitions and notations.
The norm of a Banach space $Y$  is said to be smooth at $u\in {S}_{Y}$  if $D\left(Y,u\right)$  reduces to a singleton, and it is said to be Frechet-smooth or Frechet differentiable at $u$  whenever there exists
 $\begin{array}{c}{lim}_{\alpha \to 0}\frac{\parallel u+\alpha y\parallel -1}{\alpha }\end{array}$ (5)
uniformly for $y\in {B}_{Y}$  . If this happens for all $u\in {S}_{Y}$  we say that the norm of $Y$  is Frechet differentiable. If, in addition, the limit in  5 is also uniform in $u\in {S}_{X}$  , we say that the norm of $Y$  is uniformly Frechet differentiable at ${S}_{Y}$  or that $Y$  is uniformly smooth. A natural succedanea of Frechet differentiability of the norm when smoothness is not required is the following notion introduced by D. Gregory [13. The norm of $Y$  is strongly subdifferentiable (ssd in short) at $u$  whenever there exists ${lim}_{\alpha ↓0}\frac{\parallel u+\alpha y\parallel -1}{\alpha }$  uniformly for $y\in {B}_{Y}$  . If this happens for all $u\in {S}_{Y}$  , we simply say that the norm of $Y$  is ssd. Thus, the norm of $Y$  is Frechet differentiable at $u$  if and only if it is strongly subdifferentiable at $u$  , and $Y$  is smooth at $u$  . This property has been fully investigated in [11, where we refer the reader for background. It is shown in [11,Theorem 1.2that the norm of $Y$  is ssd at $u$  if and only if $D\left(Y,u\right)$  is strongly exposed by $u$  , i.e., for every $\varepsilon >0$  , there exists $\delta >0$  such that ${y}^{*}\in {B}_{{Y}^{*}},Re{y}^{*}\left(u\right)>1-\delta ⟹d\left({y}^{*},D\left(Y,u\right)\right)<\varepsilon .$  In this paper we study when the equality in Eq.  2 holds. The results of the paper can be divided in two categories.
The first category consists of negative results: we present examples of pairs of Banach spaces $Y$  and closed subspaces $X$  in which the equality in Eq.  2 fails, even for elements of $L\left(X,Y\right)$  . In section  2 we show that for every infinite-dimensional Banach space $X$  , there is a superspace $Y$  and an element $T\in L\left(X,Y\right)$  such that $\overline{co}W\left(T\right)\ne V\left(T\right)$  . In section  3 , we give concrete examples of Banach spaces $Y$  for which there is a closed subspace $X$  and an element $T\in L\left(X,Y\right)$  such that $\overline{co}W\left(T\right)\ne V\left(T\right)$  . Such examples are ${c}_{0}$  , ${\ell }_{2}{\sum }_{\infty }\left({\ell }_{2}{\oplus }_{1}{\ell }_{2}\right)$  , and, up to renorming, every non-reflexive Banach space. We will use the following notation:
a Banach space $Y$  is said to have the FR-property if for every closed subspace $X$  and every $T\in L\left(X,Y\right)$  , the equality $\overline{co}W\left(T\right)=V\left(T\right)$  holds.
The second category is that consisting of positive results. We introduce in section  4 a sufficient condition for the FR-property which covers all the previously known examples and may be interesting by itself. We use the name “Bishop-Phelps-Bollobas property” for it since it is related to the quantitative version of the Bishop-Phelps theorem [4given by B. Bollobas [6. We relate this property to the strong subdifferentiability of the norm and to the uniform smoothness.

2 When we fix the subspace

We recall that, when $X$  is finite-dimensional, for every superspace $Y$  and every (uniformly) continuous function $f:{S}_{X}⟶Y$  , the equality $\overline{co}W\left(f\right)=V\left(f\right)$  holds [16,Theorem 2. The aim of this section is to show that this fact characterizes the finite-dimensionality, even if we restrict ourselves to bounded linear operators.
Theorem 2.1. Let $X$  be an infinite-dimensional Banach space. Then, there are a superspace $Y$  and an operator $T\in L\left(X,Y\right)$  such that $\overline{co}W\left(T\right)\ne V\left(T\right)$
We need the following easy lemma.
Lemma 2.2. If $X$  is an infinite-dimensional Banach space, then there exists a norm-one operator $S\in L\left(X,{c}_{0}\right)$  which does not attain its norm.
• Proof. Since $X$  is infinite dimensional, the Josefson-Nissenzweig theorem (see [9,§XII) assures the existence of a sequence $\left\{{x}_{n}^{*}\right\}$  in ${S}_{{X}^{*}}$  ${w}^{*}$  -converging to $0$  . Now, the operator $S:X⟶{c}_{0}$  defined by $\left[Sx\right]\left(n\right)=\frac{n}{n+1}{x}_{n}^{*}\left(x\right)\left(x\in X,n\in \mathbb{N}\right),$  does not attain its norm.
• Proof of Theorem  2.1 . Let $Y=X\oplus {c}_{0}$  endowed with the norm $\parallel \left(x,t\right)\parallel =max\left\{\parallel x\parallel ,\parallel Sx{\parallel }_{\infty }+\parallel t{\parallel }_{\infty }\right\},$  where $S\in L\left(X,{c}_{0}\right)$  is a norm-one operator which does not attain its norm, and let $J:X⟶Y$  be the natural inclusion $Jx=\left(x,0\right)$  for every $x\in X$  . If we define $T\in L\left(X,Y\right)$  by $Tx=\left(0,Sx\right)$  for every $x\in X$  , it is straightforward to check that ${lim}_{\alpha ↓0}{sup}_{x\in {S}_{X}}\frac{\parallel x+\alpha Tx\parallel -1}{\alpha }=1\text{and}{sup}_{x\in {S}_{X}}{lim}_{\alpha ↓0}\frac{\parallel x+\alpha Tx\parallel -1}{\alpha }=0.$  Thus, Eq.  3 and  4 give $V\left(T\right)\ne \overline{co}W\left(T\right)$  , as desired.
Remark 2.3. With a bit more of work, one can show that the superspace $Y$  in the above theorem can be found in such a way that $Y/X$  has dimension $1$  . We divide the proof in two cases, depending on whether $X$  is reflexive or not.
Case 1: Suppose $X$  is not reflexive. Then by the James theorem, there exists ${x}^{*}\in {S}_{{X}^{*}}$  which does not attain its norm. Thus, we can define $Y=X\oplus \mathbb{K}$  endowed with the norm $\parallel \left(x,t\right)\parallel =max\left\{\parallel x\parallel ,|{x}^{*}\left(x\right)|+|t|\right\}\left(x\in X,t\in \mathbb{K}\right),$  which contains $X$  as the subspace $\left\{\left(x,0\right):x\in X\right\}$  . If we take $T\in L\left(X,Y\right)$  defined by $Tx=\left(0,{x}^{*}\left(x\right)\right)$  for every $x\in X$  , it is straightforward to show, by using Eq.  3 and  4 , that $maxReV\left(T\right)=1$  and $supReW\left(T\right)=0$  . Case 2: Suppose $X$  is reflexive. By the Elton-Odell ( $1+\varepsilon$  )-separation theorem, there are ${\varepsilon }_{0}>0$  and a sequence $\left\{{x}_{n}^{*}{\right\}}_{n⩾0}$  of elements of ${S}_{{X}^{*}}$  , satisfying $\parallel {x}_{n}^{*}-{x}_{m}^{*}\parallel ⩾1+{\varepsilon }_{0}\left(n\ne m\right)$  (see [9,§XIV). Since $X$  is reflexive, for each $n\in \mathbb{N}$  there exists ${x}_{n}\in {S}_{X}$  such that $|\left({x}_{n}^{*}-{x}_{0}^{*}\right)\left({x}_{n}\right)|=\parallel {x}_{n}^{*}-{x}_{0}^{*}\parallel ⩾1+{\varepsilon }_{0}.$  Therefore,
 $\begin{array}{c}|{x}_{0}^{*}\left({x}_{n}\right)|⩾|\left({x}_{n}^{*}-{x}_{0}^{*}\right)\left({x}_{n}\right)|-|{x}_{n}^{*}\left({x}_{n}\right)|⩾1+{\varepsilon }_{0}-1={\varepsilon }_{0}.\end{array}$ (6)
On the other hand, for each $n\in \mathbb{N}$  , we take ${y}_{n}^{*}=\frac{{x}_{n}^{*}-{x}_{0}^{*}}{\parallel {x}_{n}^{*}-{x}_{0}^{*}\parallel }\in {S}_{{X}^{*}}$  and we observe that ${y}_{n}^{*}\left({x}_{n}\right)=1$  for every $n\in \mathbb{N}$  . Since $X\mathbb{⊉}{c}_{0}$  , it can be deduced from the proof of the Elton-Odell theorem that $\left\{{x}_{n}^{*}\right\}$  is a basic sequence and so, it converges to zero in the $weak$  topology by the reflexivity of ${X}^{*}$  (see [24,Theorem II.7.2). Using this, and the fact that $\parallel {x}_{n}^{*}-{x}_{0}^{*}\parallel ⩾1+{\varepsilon }_{0}\text{and}\parallel {x}_{0}^{*}\parallel ⩽1,$  we obtain $\overline{lim}{y}_{n}^{*}\left(x\right)<1\left(x\in {B}_{X}\right).$  This clearly implies that the operator $S\in L\left(X,{\ell }_{\infty }\right)$  given by $\left[Sx\right]\left(n\right)=\frac{n}{n+1}{y}_{n}^{*}\left(x\right)\left(x\in X,n\in \mathbb{N}\right)$  does not attain its norm. Now, we take $Y=X\oplus \mathbb{K}$  with the norm given by $\parallel \left(x,t\right)\parallel =max\left\{\parallel x\parallel ,\parallel Sx{\parallel }_{\infty }+|t|\right\}\left(x\in X,t\in \mathbb{K}\right),$  we write $J\in L\left(X,Y\right)$  for the natural inclusion and, we consider the operator $T\in L\left(X,Y\right)$  defined by $Tx=\left(0,{x}_{0}^{*}\left(x\right)\right)$  for all $x\in X$  . Using Eq.  4 and the fact that $S$  does not attain its norm, we obtain $supReW\left(T\right)=0$  . To compute $maxReV\left(T\right)$  , we observe that $\parallel J+\alpha T\parallel ⩾\parallel {x}_{n}+\alpha T{x}_{n}\parallel =\parallel \left({x}_{n},\alpha {x}_{0}^{*}\left({x}_{n}\right)\parallel ⩾\parallel S{x}_{n}\parallel +\alpha |{x}_{0}^{*}\left({x}_{n}\right)|$  so, by using Eq.  6 and the fact that $\parallel S{x}_{n}\parallel ⟶1$  , we get $\parallel J+\alpha T\parallel ⩾1+\alpha {\varepsilon }_{0}\text{for every}\alpha >0\text{.}$  By just using Eq.  3 , we get $maxReV\left(T\right)⩾{\varepsilon }_{0}$  , which finishes the proof.

3 When we fix the superspace

As we commented in the introduction, the following result is a particular case of [16,Theorems2and3.
Proposition 3.1. Finite-dimensional spaces and uniformly smooth spaces have the FR-property.
In the preceding section we have constructed examples ad hoc of Banach spaces $Y$  which do not have the FR-property. The aim of this section is to present some concrete examples of this phenomenon which will also show that some natural extensions of Proposition  3.1 are not possible.
Let us give the first example.
Example 3.2. ${c}_{0}$  does not have the FR-property. Indeed, let $Y={c}_{0}\oplus {\mathbb{K}}^{2}$  endowed with the norm $\parallel \left(x,\lambda ,\mu \right)\parallel =max\left\{\parallel x{\parallel }_{\infty },|\lambda |+|\mu |\right\}\left(x\in {c}_{0},\lambda ,\mu \in \mathbb{K}\right),$  which is isometrically isomorphic to ${c}_{0}$  . We take a norm-one functional ${x}_{0}^{*}$  on ${c}_{0}$  not attaining its norm, we consider the closed subspace $X=\left\{\left(x,{x}_{0}^{*}\left(x\right),0\right):x\in {c}_{0}\right\}$  of $Y$  , and we write $J$  for the natural inclusion of $X$  into $Y$  . If we consider the operator $T:X⟶Y$  given by $T\left(x,{x}_{0}^{*}\left(x\right),0\right)=\left(0,0,{x}_{0}^{*}\left(x\right)\right)\left(x\in {c}_{0}\right),$  by using Eq.  3 , Eq.  4 , and the fact that ${x}_{0}^{*}$  does not attain its norm, it is easy to verify that $maxReV\left(T\right)=1\text{and}supReW\left(T\right)=0,$  which finish the proof.
Since the norm of ${c}_{0}$  is ssd (see [11,corollary 2.6, for instance), the above example shows that Proposition  3.1 cannot be extended to the class of Banach spaces with ssd norm.
On the other hand, using the ideas appearing in the above example, it is easy to prove the following.
Proposition 3.3. Every non-reflexive Banach space admits an equivalent norm failing the FR-property.
• Proof. Let $Z$  be a non-reflexive Banach space. Then, $Z$  is isomorphic to $Y=V{\oplus }_{\infty }\left(\mathbb{K}{\oplus }_{1}\mathbb{K}\right)$  , where $V$  is a $2$  -codimensional closed subspace of $Z$  and, therefore, it is also non-reflexive.
Then, we choose ${v}_{0}^{*}\in {S}_{{V}^{*}}$  which does not attain its norm, we define the closed subspace $X=\left\{\left(v,{v}_{0}^{*}\left(v\right),0\right):v\in V\right\},$  and we consider $J$  the natural inclusion of $X$  in $Y$  . As in the preceding example, the operator $T:X⟶Y$  given by $T\left(v,{v}_{0}^{*}\left(v\right),0\right)=\left(0,0,{v}_{0}^{*}\left(v\right)\right)\left(x\in X\right)$  satisfies $maxReV\left(T\right)=1\text{and}supReW\left(T\right)=0.\Lambda$
In view of Propositions  3.1 and  3.3 , one may wonder if reflexivity implies the FR-property. This is not the case, as the following example shows.
Example 3.4. The superreflexive space $Y={\ell }_{2}{\oplus }_{\infty }\left({\ell }_{2}{\oplus }_{1}{\ell }_{2}\right)$  does not have the FR property. Proof. First of all, it is straightforward to show that the norm-one operator $S:{\ell }_{2}⟶{\ell }_{2}$  defined by $\left[Sx\right]\left(n\right)=\frac{n}{n+1}x\left(n\right)\left(x\in {\ell }_{2},n\in \mathbb{N}\right)$  does not attain its norm. Now, we consider the closed subspace $X=\left\{\left(x,Sx,0\right):x\in {\ell }_{2}\right\}$  with its natural inclusion in $Y$  , and we define the operator $T:X⟶Y$  by $T\left(x,Sx,0\right)=\left(0,0,Sx\right)\left(x\in X\right).$  The proof will be finished if we show that $supReW\left(T\right)=0$  and $maxReV\left(T\right)⩾1$  .
For the first equality, given $x\in {S}_{{\ell }_{2}}$  we may find ${\alpha }_{x}>0$  such that $\left(1+{\alpha }_{x}\right)\parallel Sx\parallel <1$  .
Then, for each $0<\alpha <{\alpha }_{x}$  we have $\parallel \left(x,Sx,0\right)+\alpha T\left(x,Sx,0\right)\parallel =\parallel \left(x,Sx,\alpha Sx\right)\parallel =max\left\{1,\left(1+\alpha \right)\parallel Sx\parallel \right\}=1,$  and therefore ${lim}_{\alpha ↓0}\frac{\parallel \left(x,Sx,0\right)+\alpha T\left(x,Sx,0\right)\parallel -1}{\alpha }=0.$  The arbitrariness of $x\in {S}_{{\ell }_{2}}$  gives $supReW\left(T\right)=0$  . On the other hand, for each $\alpha >0$  , we observe that $\parallel J+\alpha T\parallel ⩾\left(1+\alpha \right)\frac{n}{n+1}\left(n\in \mathbb{N}\right),$  so $\parallel J+\alpha T\parallel ⩾1+\alpha$  , and $maxReV\left(T\right)={lim}_{\alpha ↓0}\frac{\parallel J+\alpha T\parallel -1}{\alpha }⩾1.$

4 A sufficient condition: The Bishop-Phelps-Bollobas property

The aim of this section is to study a sufficient condition for the FR-property which, actually, covers all the examples given previously. The motivation for this property is the quantitative version of the classical Bishop-Phelps’ Theorem [4, 5established by B. Bollobas [6(see [8,§16for the below version).
Theorem 4.1 (Bishop-Phelps-Bollobas). Let $Y$  be a Banach space and $\varepsilon >0$  .
Whenever ${y}_{0}\in {S}_{Y}$  and ${y}_{0}^{*}\in {S}_{{Y}^{*}}$  satisfy that $Re{y}_{0}^{*}\left({y}_{0}\right)>1-\frac{{\varepsilon }^{2}}{4}$  , there exists $\left(y,{y}^{*}\right)\in \Pi \left(Y\right)$  such that $\parallel y-{y}_{0}\parallel <\varepsilon \text{and}\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon .$
This theorem has played an outstanding role in some topics of geometry of Banach spaces (see [12, 20, 21, for instance), specially in the study of ssd norms [11or in the study of spatial numerical range of operators [8,§16and§17. Also, the proof of the fact that $\overline{co}W\left(f\right)=V\left(f\right)$  for every $f\in {C}_{\text{u}}\left({S}_{Y},Y\right)$  given in [16,Theorem 1uses the above result. For bounded linear operators, this equality can be also deduced from [17,Theorem 8, a result whose proof also uses the Bishop-Phelps-Bollobas theorem. Motivated by these facts, we introduce a property which will be sufficient for the FR-property and it may be of independent interest.
Definition 4.2. Let $Y$  be a Banach space and let $X$  be a closed subspace of $Y$  . We say that $\left(X,Y\right)$  is a Bishop-Phelps-Bollobas pair (BPB-pair in short) if for every $\varepsilon >0$  there exists $\delta >0$  such that whenever ${x}_{0}\in {S}_{X}$  , ${y}_{0}^{*}\in {S}_{{Y}^{*}}$  satisfy $Re{y}_{0}^{*}\left({x}_{0}\right)>1-\delta$  , there exists $\left(x,{y}^{*}\right)\in \Pi \left(X,Y\right)$  so that $\parallel {x}_{0}-x\parallel <\varepsilon \text{and}\parallel {y}_{0}^{*}-{y}^{*}\parallel <\varepsilon .$  We say that a Banach space $Y$  has the BPB property if for every closed subspace $X$  of $Y$  , $\left(X,Y\right)$  is a BPB-pair.
The next result shows that the BPB property is sufficient for the FR-property. Actually, it can be proved that the equality in Eq.  2 holds for uniformly continuous functions.
Theorem 4.3. Let $Y$  be a Banach space and $X$  a closed subspace such that $\left(X,Y\right)$  is a BPB-pair. Then, for every $f\in {C}_{\text{u}}\left({S}_{X},Y\right)$  , the equality $\overline{co}W\left(f\right)=V\left(f\right)$  holds.
• Proof. Let $J\in L\left(X,Y\right)$  be the inclusion map. Let $f\in {C}_{\text{u}}\left({S}_{X},Y\right)$  and $\Phi \in D\left({C}_{\text{u}}\left({S}_{X},Y\right),J\right)$  .
By [16,Proposition 1, it suffices to show that  $\begin{array}{c}Re\Phi \left(f\right)⩽supReW\left(f\right).\end{array}$ (7)
For each $n\in \mathbb{N}$  , by using [16,Lemma 1we may find ${x}_{n}\in {S}_{X}$  and ${y}_{n}^{*}\in {S}_{{Y}^{*}}$  such that  $\begin{array}{c}Re\Phi \left(f\right)⩽Re{y}_{n}^{*}\left(f\left({x}_{n}\right)\right)+1/n\end{array}$ (8)
and ${y}_{n}^{*}\left({x}_{n}\right)⟶1$  . Since $\left(X,Y\right)$  is a BPB-pair, it follows that there exists a sequence ${\left\{\left({\stackrel{~}{x}}_{n},{\stackrel{~}{y}}_{n}^{*}\right)\right\}}_{n\in \mathbb{N}}\subseteq \Pi \left(X,Y\right)$  such that $\left\{{x}_{n}-{\stackrel{~}{x}}_{n}{\right\}}_{n\in \mathbb{N}}\mathbb{⟶}0\text{and}\left\{{y}_{n}^{*}-{\stackrel{~}{y}}_{n}^{*}{\right\}}_{n\in \mathbb{N}}\mathbb{⟶}0.$  By Eq.  8 ,
 $\begin{array}{cc}Re\Phi \left(f\right)& ⩽Re{\stackrel{~}{y}}_{n}^{*}\left(f\left({\stackrel{~}{x}}_{n}\right)\right)+Re\left[{y}_{n}^{*}-{\stackrel{~}{y}}_{n}^{*}\right]\left(f\left({\stackrel{~}{x}}_{n}\right)\right)+Re{y}_{n}^{*}\left(f\left({x}_{n}\right)-f\left({\stackrel{~}{x}}_{n}\right)\right)+1/n\end{array}$
 $\begin{array}{cc}& ⩽supReW\left(f\right)+\parallel {y}_{n}^{*}-{\stackrel{~}{y}}_{n}^{*}\parallel \parallel f{\parallel }_{\infty }+\parallel f\left({x}_{n}\right)-f\left({\stackrel{~}{x}}_{n}\right)\parallel +1/n\end{array}$
 $\begin{array}{}\end{array}$
for all $n\in \mathbb{N}$  . Thus, Eq.  7 follows from the above and the uniform continuity of $f$  .
It is worth mentioning that the above proof follows the lines of [16,Theorem 1.
Corollary 4.4. Let $Y$  be a Banach space with the BPB property. Then, $Y$  has the FR-property.
As a consequence of the above corollary and Theorem  2.1 , we get the following.
Corollary 4.5. Let $X$  be an infinite-dimensional Banach space. Then, there is a superspace $Y$  of $X$  such that $\left(X,Y\right)$  is not a BPB-pair.
The above result implies that not every Banach space $Y$  has the BPB property. For instance, the examples given in section  3 of Banach spaces which do not have the FR-property also fail the BPB property.
Example 4.6. The spaces ${c}_{0}$  and ${\ell }_{2}{\oplus }_{\infty }\left({\ell }_{2}{\oplus }_{1}{\ell }_{2}\right)$  fail the BPB property in their canonical norms. Every non-reflexive Banach space admits an equivalent norm failing the BPB property.
On the other hand, if we restrict ourselves to finite-dimensional subspaces, we get a characterization of the ssd norms.
Proposition 4.7. Let $Y$  be a Banach space. Then, the norm of $Y$  is ssd if, and only if, for every finite-dimensional subspace $X\subseteq Y$  , the pair $\left(X,Y\right)$  is BPB.
• Proof. We suppose first that the norm of $Y$  is ssd. Let $X$  be a finite-dimensional subspace of $Y$  and let $\varepsilon >0$  be given. Since the norm of $Y$  is ssd, [11,Theorem 1.2gives us that for each $x\in {S}_{X}$  there exists ${\delta }_{x}>0$  so that ${y}^{*}\in {S}_{{Y}^{*}},Re{y}^{*}\left(x\right)>1-{\delta }_{x}⟹d\left({y}^{*},D\left(Y,x\right)\right)<\varepsilon .$  Therefore, if for each $x\in {S}_{X}$  we define ${A}_{x}=\left\{u\in {S}_{X}:\parallel u-x\parallel   the compactness of ${S}_{X}$  assures the existence of ${x}_{1},...,{x}_{n}\in {S}_{X}$  such that ${S}_{X}{=}^{n}{\bigcup }_{i=1}{A}_{{x}_{i}}.$  Then, $\delta =min\left\{\frac{{\delta }_{{x}_{i}}}{2}:i=1,...,n\right\}$  satisfies the BPB condition. Indeed, let ${x}_{0}\in {S}_{X}$  and ${y}_{0}^{*}\in {S}_{{Y}^{*}}$  be such that $Re{y}_{0}^{*}\left({x}_{0}\right)>1-\delta .$  Since ${x}_{0}\in {S}_{X}$  , there exists $j\in \left\{1,...,n\right\}$  so that ${x}_{0}\in {A}_{{x}_{j}}$  , that is $\parallel {x}_{0}-{x}_{j}\parallel   Therefore, $Re{y}_{0}^{*}\left({x}_{j}\right)>1-{\delta }_{{x}_{j}}$  which implies the existence of ${y}^{*}\in D\left(Y,{x}_{j}\right)$  such that $\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon$  .
To prove the converse, it is enough to fix ${x}_{0}\in {S}_{Y}$  and to show that ${x}_{0}$  strongly exposes $D\left(Y,{x}_{0}\right)$  [11,Theorem 1.2. To do so, let $X$  be the subspace of $Y$  generated by ${x}_{0}$  and, fixed $\varepsilon >0$  , let $\delta >0$  be given by the definition of the BPB for the pair $\left(X,Y\right)$  and $\varepsilon /2$  . Suppose now that ${y}_{0}^{*}\in {S}_{{Y}^{*}}$  is such that $Re{y}_{0}^{*}\left({x}_{0}\right)>1-\delta$  , then there exists $\left(x,{y}^{*}\right)\in \Pi \left(X,Y\right)$  so that $\parallel x-{x}_{0}\parallel <\varepsilon /2\text{and}\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon /2.$  Since $x\in \text{span}\left({x}_{0}\right)$  , there exists a modulus-one $\lambda \in \mathbb{K}$  such that $x=\lambda {x}_{0}$  . Therefore, $|\lambda -1|=\parallel \lambda {x}_{0}-{x}_{0}\parallel =\parallel x-{x}_{0}\parallel <\varepsilon /2,$  and then, $\lambda {y}^{*}\in D\left(Y,{x}_{0}\right)\text{and}\parallel \lambda {y}^{*}-{y}_{0}^{*}\parallel ⩽\parallel \lambda {y}^{*}-{y}^{*}\parallel +\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon /2+\varepsilon /2=\varepsilon ,$  which finishes the proof.
Since the norm of any finite-dimensional Banach space is ssd (see [11,pp. 48), we have the following corollary, which also implies the first part of Proposition  3.1 .
Corollary 4.8. Every finite-dimensional Banach space has the BPB property.
The other class of spaces with the FR-property given in Proposition  3.1 is the one of uniformly smooth spaces. This result can be also deduced from Corollary  4.4 , as the following proposition shows.
Proposition 4.9. Every uniformly smooth space has the BPB property.
• Proof. Let $Y$  be an uniformly smooth space. Then, ${Y}^{*}$  is uniformly convex, so, for every $\varepsilon >0$  , we may find $\delta >0$  (the modulus of convexity of ${Y}^{*}$  ) such that ${x}^{*},{y}^{*}\in {S}_{{Y}^{*}},\parallel {x}^{*}+{y}^{*}\parallel >2-\delta ⟹\parallel {x}^{*}-{y}^{*}\parallel <\varepsilon$  (see [2,Chapter IIfor instance). Let $X$  be a subspace of $Y$  , and let ${x}_{0}\in {S}_{X}$  and ${y}_{0}^{*}\in {S}_{{Y}^{*}}$  be so that $Re{y}_{0}^{*}\left({x}_{0}\right)>1-\delta$  . If we consider ${y}^{*}\in {S}_{{Y}^{*}}$  such that $Re{y}^{*}\left({x}_{0}\right)=1$  , we have $\parallel {y}^{*}+{y}_{0}^{*}\parallel ⩾Re\left({y}^{*}+{y}_{0}^{*}\right)\left({x}_{0}\right)>2-\delta$  and, therefore, $\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon ,$  which finishes the proof.
Observe that, in the above proof, the relation $\varepsilon -\delta$  does not depend on the subspace.
The next result shows that this fact actually characterizes the uniform smoothness.
Proposition 4.10. Let $Y$  be a Banach space with the BPB property in such a way that the relationship between $\varepsilon$  and $\delta$  in Definition  4.2 does not depend on the subspace $X$  . Then, $Y$  is uniformly smooth.
• Proof. In view of [11,Proposition4.1, it is enough to show that the limit ${lim}_{t↓0}\frac{\parallel u+ty\parallel -1}{t}=:\tau \left(u,y\right)$  exists uniformly for $y\in {B}_{Y}$  and $u\in {S}_{Y}$  . Given $\varepsilon >0$  , let $0<\delta <2$  be given by the “uniform” BPB property. Now, for $y\in {B}_{Y}$  , $u\in {S}_{Y}$  and $0  , we consider ${y}_{t}=\frac{u+ty}{\parallel u+ty\parallel }\in {S}_{Y}\text{and}{y}_{t}^{*}\in D\left(Y,{y}_{t}\right).$  It is immediate to check that $Re{y}_{t}^{*}\left(u\right)>1-\delta$  so, if we take $X=\text{span}\left(u\right)$  , the BPB property assures the existence of $\left(x,{z}_{t}^{*}\right)\in \Pi \left(X,Y\right)$  such that $\parallel x-u\parallel <\varepsilon$  and $\parallel {z}_{t}^{*}-{y}_{t}^{*}\parallel <\varepsilon$  . Since $x\in \text{span}\left(u\right)$  , there exists a modulus-one $\lambda \in \mathbb{K}$  such that $x=\lambda u$  . Therefore, $|\lambda -1|=\parallel \lambda u-u\parallel =\parallel x-u\parallel <\varepsilon ,$  and then, $\lambda {z}_{t}^{*}\in D\left(Y,u\right)\text{and}\parallel \lambda {z}_{t}^{*}-{y}_{t}^{*}\parallel ⩽\parallel \lambda {z}_{t}^{*}-{z}_{t}^{*}\parallel +\parallel {z}_{t}^{*}-{y}_{t}^{*}\parallel <\varepsilon +\varepsilon =2\varepsilon ,$  Now, the facts $\frac{\parallel u+ty\parallel -1}{t}=\frac{Re{y}_{t}^{*}\left(u+ty\right)-1}{t}⩽Re{y}_{t}^{*}\left(y\right)$  and $\tau \left(u,y\right)⩾Re\lambda {z}_{t}^{*}\left(y\right)$  (by Eq.  1 ), give $0⩽\frac{\parallel u+ty\parallel -1}{t}-\tau \left(u,y\right)⩽Re{y}_{t}^{*}\left(y\right)-Re\lambda {z}_{t}^{*}\left(y\right)⩽\parallel \lambda {z}_{t}^{*}-{y}_{t}^{*}\parallel <2\varepsilon ,$  and the arbitrariness of $\varepsilon >0$  finishes the proof.
We conclude the paper proving that a pair $\left(X,Y\right)$  is a BPB-pair provided that $X$  is an absolute ideal of $Y$  . Let us introduce the necessary definitions. We refer the reader to [8,§ 21, [19, and references therein for background. A closed subspace $X$  of a Banach space $Y$  is said to be an absolute summand of $Y$  if there exists another closed subspace $Z$  such that $Y=X\oplus Z$  and, for every $x\in X$  and $z\in Z$  , the norm of $x+z$  only depends on $\parallel x\parallel$  and $\parallel z\parallel$  . We also say that $Y$  is an absolute sum of $X$  and $Z$  . This implies that there exists an absolute norm on ${\mathbb{R}}^{2}$  such that $\parallel x+z\parallel =|\left(\parallel x\parallel ,\parallel z\parallel \right){|}_{a}\left(x\in X,z\in Z\right).$  By an absolute norm we mean a norm $|\cdot {|}_{a}$  on ${\mathbb{R}}^{2}$  such that $|\left(1,0\right){|}_{a}=|\left(0,1\right){|}_{a}=1$  and $|\left(a,b\right){|}_{a}=|\left(|a|,|b|\right){|}_{a}$  for every $a,b\in \mathbb{R}$  . Useful results about absolute norms are the following inequality $max\left\{|a|,|b|\right\}⩽|\left(a,b\right){|}_{a}⩽|a|+|b|a,b\in \mathbb{R},$  and the fact that absolute norms are nondecreasing and continuous in each variable. We say that $X$  is an absolute ideal of $Y$  if ${X}^{\perp }$  is an absolute summand of ${Y}^{*}$  , in which case, ${Y}^{*}$  can be identified with ${X}^{*}\oplus {X}^{\perp }$  with a convenient absolute sum. It is clear that absolute summands are absolute ideals, but the converse is not true.
Absolute summands and absolute ideals are generalizations of the well-known M-summands, L-summands, M-ideals, and the more general class of ${L}_{p}$  -summands [3, 14.
Proposition 4.11. Let $Y$  be a Banach space and let $X$  be a closed subspace. If ${X}^{*}$  is an absolute summand of ${Y}^{*}$  , then the pair $\left(X,Y\right)$  is BPB. In particular, this occurs when $X$  is an absolute ideal of $Y$  .
We need the following easy result, which we separate from the proof of the proposition for the sake of clearness.
Lemma 4.12. Let $E$  be $\left({\mathbb{R}}^{2},|\cdot {|}_{a}\right)$  where $|\cdot {|}_{a}$  is an absolute norm. We write ${b}_{0}=max\left\{b⩾0:|\left(1,b\right){|}_{a}=1\right\},$  and we define $A\left(\delta \right)=\left\{\left(a,b\right)\in {B}_{E}:a>1-\delta ,b⩾{b}_{0}\right\}\left(\delta >0\right).$  Then, for every $\varepsilon >0$  there exists $\delta >0$  such that $\text{diam}\left(A\left(\delta \right)\right)<\varepsilon .$
• Proof. Suppose, for the sake of contradiction, that the result does not hold. Then, there exists ${\varepsilon }_{0}>0$  such that for every $n\in \mathbb{N}$  , $\text{diam}\left(A\left(\frac{1}{n}\right)\right)⩾{\varepsilon }_{0}$  . So, we may find $\left({a}_{n},{b}_{n}\right)\in A\left(\frac{1}{n}\right)$  such that $|\left({a}_{n},{b}_{n}\right)-\left(1,{b}_{0}\right){|}_{a}⩾\frac{{\varepsilon }_{0}}{2}$  , and thus  $\begin{array}{c}\frac{{\varepsilon }_{0}}{2}⩽|{a}_{n}-1|+|{b}_{n}-{b}_{0}|\left(n\in \mathbb{N}\right).\end{array}$ (9)
Let $\left\{\left({a}_{{\sigma }_{n}},{b}_{{\sigma }_{n}}\right)\right\}$  be a convergent subsequence of $\left\{\left({a}_{n},{b}_{n}\right)\right\}$  , and let $\left(1,b\right)\in {S}_{E}$  be its limit.
By Eq.  9 and the fact that $\left({a}_{{\sigma }_{n}},{b}_{{\sigma }_{n}}\right)\in A\left(\frac{1}{{\sigma }_{n}}\right)$  , it is immediate to check that $\frac{{\varepsilon }_{0}}{2}⩽|b-{b}_{0}|\text{and}b⩾{b}_{0}.$  So, $b$  is strictly bigger than ${b}_{0}$  , a contradiction.
• Proof of Proposition  4.11 . There exist a subspace $Z$  of ${Y}^{*}$  and an absolute norm $|\cdot {|}_{a}$  on ${\mathbb{R}}^{2}$  so that ${Y}^{*}={X}^{*}\oplus Z$  and $\parallel \left({x}^{*},{z}^{*}\right)\parallel =|\left(\parallel {x}^{*}\parallel ,\parallel {z}^{*}\parallel \right){|}_{a}\left({x}^{*}\in {X}^{*},{z}^{*}\in Z\right).$  For $\varepsilon >0$  fixed, we take ${\delta }_{1}>0$  given by the preceding lemma applied for $\varepsilon /3$  , and we define $\delta :=min\left\{{\delta }_{1},\frac{{\varepsilon }^{2}}{36}\right\}.$  To finish the proof, for ${x}_{0}\in {S}_{X}$  and ${y}_{0}^{*}=\left({x}_{0}^{*},{z}_{0}^{*}\right)\in {S}_{{Y}^{*}}$  satisfying $Re{y}_{0}^{*}\left({x}_{0}\right)=Re{x}_{0}^{*}\left({x}_{0}\right)>1-\delta ,$  we have to find $\left(x,{y}^{*}\right)\in \Pi \left(X,Y\right)$  so that $\parallel {y}^{*}-{y}_{0}^{*}\parallel <\varepsilon \text{and}\parallel x-{x}_{0}\parallel <\varepsilon .$  To this end, since $\parallel {x}_{0}\parallel =1=‖\frac{{x}_{0}^{*}}{\parallel {x}_{0}^{*}\parallel }‖\text{and}Re\frac{{x}_{0}^{*}}{\parallel {x}_{0}^{*}\parallel }\left({x}_{0}\right)⩾Re{x}_{0}^{*}\left({x}_{0}\right)>1-\frac{{\varepsilon }^{2}}{36},$  we can apply the classical Bishop-Phelps-Bollobas Theorem ( 4.1 ) to $\left({x}_{0},\frac{{x}_{0}^{*}}{\parallel {x}_{0}^{*}\parallel }\right)\in X×{X}^{*}$  to get $\left(x,{x}^{*}\right)\in \Pi \left(X\right)$  such that  $\begin{array}{c}‖{x}^{*}-\frac{{x}_{0}^{*}}{\parallel {x}_{0}^{*}\parallel }‖<\frac{\varepsilon }{3}\text{and}\parallel x-{x}_{0}\parallel <\frac{\varepsilon }{3}.\end{array}$ (10)
Now, we distinguish two cases. Suppose first that $\parallel {z}_{0}^{*}\parallel ⩽{b}_{0}$  . Then, we take ${y}^{*}:=\left({x}^{*},{z}_{0}^{*}\right)$  , which satisfies $Re{y}^{*}\left(x\right)=1$  and $\parallel {y}^{*}\parallel =|\left(1,\parallel {z}_{0}^{*}\parallel \right){|}_{a}=1$  . Using Eq.  10 and the definition of $\delta$  , we get $\parallel {y}^{*}-{y}_{0}^{*}\parallel =\parallel {x}^{*}-{x}_{0}^{*}\parallel <\frac{\varepsilon }{3}+\delta <\varepsilon .$  So, the pair $\left(x,{y}^{*}\right)$  satisfies the desired condition.
Suppose otherwise that $\parallel {z}_{0}^{*}\parallel >{b}_{0}$  . In this case, we take ${y}^{*}:=\left({x}^{*},\frac{{b}_{0}}{\parallel {z}_{0}^{*}\parallel }{z}_{0}^{*}\right)$  , which clearly satisfies $Re{y}^{*}\left(x\right)=1=\parallel {y}^{*}\parallel$  . Now, $\left(1,{b}_{0}\right)$  and $\left(\parallel {x}_{0}^{*}\parallel ,\parallel {z}_{0}^{*}\parallel \right)$  belong to $A\left(\delta \right)$  and the diameter of this set is less than $\varepsilon /3$  by Lemma  4.12 , so we have $|\parallel {z}_{0}^{*}\parallel -{b}_{0}|⩽|\left(1,{b}_{0}\right)-\left(\parallel {x}_{0}^{*}\parallel ,\parallel {z}_{0}^{*}\parallel \right){|}_{a}<\frac{\varepsilon }{3}$  and $\parallel {y}^{*}-{y}_{0}^{*}\parallel =|\left(\parallel {x}^{*}-{x}_{0}^{*}\parallel ,\parallel {z}_{0}^{*}\parallel -{b}_{0}\right){|}_{a}⩽\parallel {x}^{*}-{x}_{0}^{*}\parallel +|\parallel {z}_{0}^{*}\parallel -{b}_{0}|<\varepsilon .\Lambda$
By just applying the above proposition and Theorem  4.3 , we get the following.
Corollary 4.13. Let $Y$  be a Banach space and let $X$  be a closed subspace of $Y$  such that ${X}^{*}$  is an absolute summand of ${Y}^{*}$  (in particular, if $X$  is an absolute ideal of $Y$  ). Then, $\overline{co}W\left(f\right)=V\left(f\right)$  for every $f\in {C}_{\text{u}}\left({S}_{X},Y\right)$  .
An interesting particular case is the case of $M$  -embedded and $L$  -embedded spaces. A Banach space $X$  is said to be $M$  -embedded if it is an $M$  -ideal of ${X}^{**}$  , and it is $L$  -embedded if ${X}^{**}=X{\oplus }_{1}Z$  for some closed subspace $Z$  of ${X}^{**}$  .
Corollary 4.14. If $X$  is an $M$  -embedded or an $L$  -embedded space, then $\left(X,{X}^{**}\right)$  is a BPB-pair.
We do not know if the assumption of being $M$  -embedded or $L$  -embedded in the above result is superabundant.
Acknowledgment: The authors would like to express their gratitude to Pradipta Bandyopadhyay, Catherine Finet, Gilles Godefroy, and Gines Lopez for their valuable suggestions and fruitful conversations about the subject of this paper.
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1 Corresponding author. E-mail: mmartins@ugr.es, Fax: +34958243272


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# Revision history [back]

### Factorizing a polynomial with two variables

Hi, I would like to factorize the following polynomial in $t$, with an integer indeterminate $n$,

$2^n ((n-1)t^n+1)^{n-1} - n^{n-1} t^{(n-1)(n-2)} (t^{n-1} + 1)(t+1)^{n-1}$

I expect it to have a factor of $(t-1)^2$ and hope that after division the polynomial in $t$ would have positive coefficients. Is there any way I can verify this by Sage?

### Factorizing a polynomial with two variables

Hi, I would like to factorize the following polynomial in $t$, with an integer indeterminate $n$,

$2^n$f_n(t) = 2^n ((n-1)t^n+1)^{n-1} - n^{n-1} t^{(n-1)(n-2)} (t^{n-1} + 1)(t+1)^{n-1}$I expect it to have a factor of$(t-1)^2$and hope that after division the polynomial in$t$would have positive coefficients. 1. Is there any way I can verify this by Sage? 2. I have tried to check that$f_n(t)$has a factor of$(t-1)^2$by checking that$f_n(1) = f_n'(1) = 0$. I named my polynomial$f$, and let$h$be f.derivative(t), and try to find$h(t=1)$. Here's what I got, (n - 1)^22^nnn^(n - 2) - (n - 1)2^(n - 1)n^(n - 1) - 2(n - 1)2^(n - 2)n^(n - 1) - 2(n^2 - 3n + 2)2^(n - 1)n^(n - 1) Which turns out to be 0 when I verify by hand. However it seems like Sage is unable to detect the redundance in the expression (e.g.$n \times n^{n-2} = n^{n-1}\$). Did I do the computation in the "wrong way"? If so, what's the proper way to do it?

Thanks!


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# American Institute of Mathematical Sciences

January  2011, 15(1): 61-74. doi: 10.3934/dcdsb.2011.15.61

## Global stability of SIR epidemic models with a wide class of nonlinear incidence rates and distributed delays

 1 Department of Pure and Applied Mathematics, Waseda University, 3-4-1 Ohkubo, Shinjuku-ku, Tokyo, 169-8555 2 Basque Center for Applied Mathematics, Bizkaia Technology Park, Building 500 E-48160 Derio, Spain 3 Department of Mathematics, Waseda University, 3-4-1 Ohkubo, Shinjuku-ku, Tokyo, 169-8555

Received  October 2009 Revised  April 2010 Published  October 2010

In this paper, we establish the global asymptotic stability of equilibria for an SIR model of infectious diseases with distributed time delays governed by a wide class of nonlinear incidence rates. We obtain the global properties of the model by proving the permanence and constructing a suitable Lyapunov functional. Under some suitable assumptions on the nonlinear term in the incidence rate, the global dynamics of the model is completely determined by the basic reproduction number $R_0$ and the distributed delays do not influence the global dynamics of the model.
Citation: Yoichi Enatsu, Yukihiko Nakata, Yoshiaki Muroya. Global stability of SIR epidemic models with a wide class of nonlinear incidence rates and distributed delays. Discrete & Continuous Dynamical Systems - B, 2011, 15 (1) : 61-74. doi: 10.3934/dcdsb.2011.15.61
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## Publications

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Marzouk, M., and A. Abubakr, Automation in Construction, vol. 61, pp. 1-15, 2016.
HadeelElsayedElsherbeiny, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, issue 2, 2017.
HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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Elsherbeiny, H. E., A. A. A. El-Aziz, and N. R. Darwish, Egyptian Computer Science Journal (ECSJ), ISSN: 1110-2586, vol. 41, issue 2: Egyptian Computer Society, pp. 44-52, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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HadeelElsayedElsherbeiny, A. A. Abd El-Aziz, and N. Ramadan, Egyptian Computer Science Journal (ISSN-1110-2586), vol. 41, no. 2: University of Trier database, DBLP, Germany, pp. 9, 2017. Abstract
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Yasser, S., N. Sameh, S. Kassem, Y. Emad, O. Tariq, and I. Fahim, 2020 International Conference on Decision Aid Sciences and Application (DASA): IEEE, pp. 343–349, 2020. Abstract
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Antucheviciene, J., Z. Kala, M. Marzouk, and E. Vaidogas, Mathematical Problems in Engineering, vol. 2015, pp. 1-3, 2015.
Ibrahim, A. A. A. M., A Decision Making Approach for Prioritizing Local Communities’ Development Investments, , vol. 68: Elsevier, pp. 649 - 660, 2012. Abstract
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Abdalla, M., A. Rezayee, D. Cassan, M. V. Ierssel, C. Holdenried, and S. Sadr, DECISION FEEDBACK EQUALIZER AND TRANSCEIVER, , no. 20120201289: Gennum Corp, Rambus Inc, August, 2012. Abstract

A decision feedback equalizer, transceiver, and method are provided, the equalizer having at least one comparator, the at least one comparator comprising a first stage, comprising a main branch having two track switches with a resistive load, an offset cancellation branch, a plurality of tap branches with transistor sizes smaller than the main branch, in which previous decisions of the equalizer are mixed with the tap weights using current-mode switching, and a cross coupled latch branch; and a second stage, comprising a comparator module for making decisions based on the outputs of the first stage and a clock input, and a plurality of flip-flops for storing the output of the comparator module.

Fayed, H., and A. Atiya, Applied Soft Computing, vol. 110, pp. 107628, 2021. asoc-clustering-svm.pdf
, ACS Applied Energy Materials, vol. 5, issue 6, pp. 7661–7673, 2022.
Fahmy, N., A. Egiza, M. Badr, N. Morcos, J. Diclaro, and E. Abd-ElSamie, Deciphering pyrethroid resistance in Cx. pipiens (L): Implications of cytochrome P450; expression profiling and regulatory microRNA, , vol. 52, pp. 101579, 2020/04/01. Abstract

Over the past decades, the extensive use of pyrethroids insecticides for vector control has resulted in the development of insecticide resistance. Cytochrome P450 has been recognized to play a critical role in the metabolic detoxification of insecticides. In the current study, Culex pipiens mosquitoes were collected from Giza Governorate in Egypt and tested for insecticide susceptibility against deltamethrin. First detection of Knockdown resistance gene (Kdr) mutations in field collected mosquitoes was performed. Activities of cytochrome oxidase P450 detoxification enzyme that synchronized with the resistance development, was assessed. Expression profiles of cytochrome P450s and their putative corresponding regulating miRNAs, which was previously reported in Cx. pipiens pallens were evaluated in pyrethroid resistant field-collected Cx. pipiens using RT-qPCR and stem-loop RT-qPCR, respectively. Specific stem-loop reverse transcription primers and forward primers were designed for miRNAs profiling. Our results elucidated the pyrethroid resistance development and revealed its relation to the metabolic and target site modification mechanisms with a first report of L1014F-kdr mutation detection. RT-qPCR results have showed an up-regulation in the expression of the studied P450 transcripts. Negative correlations were found between the expression of P450s and their regulatory miRNAs except for CYP9J35, where positive correlation was found with its corresponding miR-13. Interestingly, our data was the first to detect negative correlation between miR-285 and its putative CYP6Cp1 target gene. These findings highlighted the significance of identifying P450 gene along with regulatory miRNAs as a key mechanism implicated in pyrethroid resistance in field Culex vector population. The elucidation of this mechanism would shed light on the development of insecticide resistance and would help in shaping strategies to combat such vectors.

Jalal, D., M. G. Elzayat, A. A. Diab, H. E. El-Shqanqery, O. Samir, U. Bakry, R. Hassan, M. El anany, L. Shalaby, and A. A. Sayed, mSphere, vol. 6, issue 6: American Society for Microbiology (ASM), 2021. Abstract
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Raafat, R., A. Mohamed, R. Samy, T. ElDeeb, Y. Farouk, M. Elkhouly, and H. A. H. Fahmy, Forty-Second {A}silomar Conference on Signals, Systems, and Computers, {A}silomar, {C}alifornia, {USA}, oct, 2008. Abstract

Decimal floating-point multiplication is important in many commercial applications including financial analysis, banking, tax calculation, currency conversion, insurance, and accounting. This paper presents a fully parallel Decimal64 floating point multiplier compliant to IEEE Std 754-2008 standard for floating point arithmetic. The proposed multiplier possesses novel methods to target low latency. The proposed design is based on previously published fixed point multiplier that uses a novel BCD 4221 recoding for decimal digits to improve the area and latency of the partial product generation and the partial product reduction tree. Several enhancements are introduced to the design; the final carry propagation adder is implemented using a full parallel decimal adder with a Kogge-Stone prefix tree, the sticky bit is generated parallel to the shifter to reduce the critical path delay. The design is extendible to support Decimal128 floating point multiplication. The multiplier is hardware verified for functionality on an FPGA.

Samy, R., H. A. H. Fahmy, R. Raafat, A. Mohamed, T. ElDeeb, and Y. Farouk, Fifty-Third {M}id{W}est {S}ymposium on {C}ircuits and {S}ystems, {(MWSCAS)}, {S}eattle, {W}ashington, {USA}, aug, 2010. Abstract

This paper presents the first hardware implementation of a fully parallel decimal floating-point fused- multiply-add unit performing the operation $\pm (A \times B) \pm C$ on decimal floating-point operands. The proposed design is fully compliant with the IEEE 754-2008 standard and supports the two standard formats decimal64 and decimal128. Furthermore, the proposed design may be controlled to perform the multiplication or the addition/subtraction as standalone operations. Our decimal floating-point FMA may be pipelined so that a complete resultant decimal floating-point is available each clock cycle.

Fahmy, H. A. H., T. ElDeeb, M. Hassan, Y. Farouk, and R. Eissa, The 22nd International Conference on Microelectronics, {ICM} 2010, {E}gypt, dec, 2010. Abstract
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D.Fathy, T. Elsharnouby, and E. AbouAish, International Journal of Business and Emerging Markets, vol. 8, issue 4, pp. 403-425, 2016. deception_in_advertising_revisited_antecedents_and_differences_in_perception_across_consumer_groups2016.pdf
Adel, K., A. A. Elhakeem, and M. Marzouk, Expert Systems with Applications, vol. 194, pp. 116548, 2022.
M.Darwish, H.M.Soliman, and J.Fantin, IEEE Trans. on SMC , vol. 9, pp. 717-20, 1979.
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