  Last lecture we have discussed the linear
operators on a vector spaces, operators on
vector spaces.
So, this operator, we have defined as T of
X plus Y is equal to T X plus T Y and T of
alpha X equal to alpha of T X, where the domain
of T is a vector space and range of T lies
in a vector space, lies in Y, which is a vector
space. So, that way we have defined the linear
operator on a vector space, say X T in operator,
from X to Y. ok. Now, if we look the definition
here, we are having the two operation, addition
and scalar multiplication. So, this operator
T has a connection between a scalar multiplication
and addition and addition of the two vectors.
But we do not have any things about the normed.
So, if I replace the vector space X by a normed
space or a Banach space, then, this norm has
no function over here so far. So, we wanted
to extend this concept of the linear operator,
over a Banach space or over a normed space,
which includes another operation, which we
call it as a norm.
So, a operator which includes all these three
operation, its definitions, will be more interesting
and important, comparative to this one. And,
this leads to the concept of bounded linear
operator. The bounded linear operator is defined
as follows. Let X and Y be, X and Y be normed
spaces; normed spaces, here, I am taking the
same norm and can use the another notation,
as this as norm X and this is corresponding
to the norm Y. So, normally, we do not write
it, because it is a understanding that, whenever
the elements of Y is there, that corresponding
norm of Y is used. So, let X and Y be normed
spaces and T is a mapping from domain of T,
which is of course, a subset of X to Y, a
linear operator, a linear operator. Then,
we say the operator T 
is said to be, is said to be a bounded linear
operator, said to be bounded, a linear operator
said to be bounded, if there is a real number
C, of course, C will be positive, such that,
for all X belonging to domain of T, the norm
of T X is less than equal to C times norm
of X.
So, we have defined the bounded linear operator
in this form, means, a linear operator is
said to be bounded, if this extra condition
is also satisfied. So, basically, if we look,
an operator T from domain D T to Y is a bounded
linear operator means that, this should satisfy
this condition alpha X plus beta Y equal to
alpha of T X plus beta of T Y; that is T is
linear; T is linear and apart from this, the
norm of T X should be less than equal to C
times norm of X. Here, this is the norm on
Y and this is the norm on X. So, this gives
the relation between the norms and the operator
also. So, this operator, an operator which
satisfy these conditions, we call it as a
bounded linear operator. Now, the question
arise, what should be the minimum value of
C here, so that, this condition holds. This
can be obtained as a, since norm of T X, Y
norm of X is less than equal to C, this is
true for all X belonging to domain of T D
T.
So, take the supremum of this norm of T X
over norm of X, and X belongs to the domain
of T. If this supremum, that will be the minimum
value of C, which will be satisfy by this.
So, this we denoted by norm of T. So, let,
norm of T is this, which is the minimum value,
minimum value for C. So, if I take the real
is, then, this minimum value of C, we call
it as a norm of the operator, of the bounded
linear operator T. Clear? And, if you use
this thing, again same, so, we get the relation,
the norm of T X replaced by the minimum value
norm of T into norm of X and this is a interesting
equation; equation, which will be, which will
be used very frequently. So, we get from here
that, this norm of this, is like this.
Now, if T is 0, because 0 is also a bounded
operator; so, if T is 0, here, one more thing
which I write, that for all X which are different
from 0, where, because, if X is 0, we cannot
divide by this. So, for all X, which are non-zero,
the minimum value of C will be defined like
this. So, X is not equal to 0, or this will
be T; is, if T is 0, then, the norm of T is
considered to be the 0. So, we do not take
it always, in order to complete the definition,
we take it norm T; when T is 0, it is equal
to 0. Then, we have certain lemma, which will
give you, the another definition of the norm
T. The definition, there is a alternate way
of defining the norm and it is defined as,
let T be a bounded; T be a bounded linear
operator, as defined, bounded linear operator.
Then, the norm of T can also be defined as
supremum of norm T X, where X belongs to the
domain of T and norm of X is equal to 1; norm
of X is equal to 1. This, look the definition
this, here, the definition of the norm which
is assigned by A and the definition of norm,
which is given by B.
This suggests that, norm of an operator T
is basically, obtained by choosing the supremum
value at all the point X, which are on the
unit circle, centered at 0 and radius one.
So, if we take any point here, whose length
is 1, distance from this is 1; the supremum
is taken over all such X, that will be the
norm. So, you need not go for the supremum
for the entire domain D T; simply choose the
point, which has a length 1; those vector
which has a length 1. Let us see the proof
of this.
Proof of this lemma. This is very simple way.
Let us suppose, norm of x is equal to a and
let us set Y equal to 1 by a into x, where
x is not equal to 0. Obviously, from here,
norm of Y will be 1, because norm of Y equal
to norm of x by a, is equal to this. So, norm
of Y is 1. And since T is linear, so, we get
T of norm of T, which is defined as supremum
of norm of T x over norm x, when the x belongs
to the domain of D T. So, but norm x is 1;
so, it is equal to supremum of norm 1 by a
norm of T x where x belongs to the domain
D T and norm of x is equal to a. Then, this
will be x is not equal to 0. Of course, x
is not equal to 0; then, this will be equal
to, supremum is taken over x norm of T x by
a, where the x belongs to D T and x is not
equal to 0; because it is a norm, the property
of the norm, if you remember, the norm of
alpha x is equal to mod alpha; norm of alpha
x is equal to mod alpha into norm x. So, using
this property, we are able to write, this
1 by a inside and we get this one; but x by
a is Y. So, basically, this is the supremum
of norm T y, where the Y belongs to the domain
of D T and norm of Y is equal to 1; and this
proves the results for it.
So, a norm of a bounded linear operator, one
can defined in this way, as the supremum norm
of T X over norm X, X belongs to the domain
of D T and X is not equal to 0; or equivalently,
we can also say, equal to supremum norm of
T X over all X belongs to D T where the norm
X is equal to 1. Clear? So, either this way
or this. Now, the question, whether this really
satisfy the condition of norms. So, if we
look that conditions, this norm satisfy, norm
of T satisfies the conditions of norm; why,
because the first condition is, obviously,
true; norm of T equal to 0, if and only if,
T equal to 0. Because if it is 0, then, according
to this, this supremum has to be 0 for all
X belongs to this, but if norm of T X is 0,
or norm of T Y is 0 for all Y, where Y is,
norm Y is 1, then, it is only possible whenn
10 T is 0.
So, vice versa. So, norm T equal to 0 implies
T is 0 and similarly, if T is 0, this entire
part will be 0 and we get norm T is 0. Second
condition, which we can see, the norm of alpha
X, where alpha is this. So, supremum norm
of T alpha, this is alpha T, sorry alpha T.
So, this is equal to alpha into T X. So, alpha
X; we are taking alpha T or you can write
this, sorry, let me see; norm of alpha T by
definition, it is the supremum of norm alpha
T Y; I am taking the second definition; Y
is D T; norm Y is equal to 1; but alpha can
be taken outside and we get mod alpha norm
of T. Similarly, the third condition, one
can prove that, norm of T 1 plus T 2 is less
than equal to norm of T 1 plus norm of T 2;
because the supremum of the sum, is less than
equal to sum of the supremum. So, this can
be verified. Therefore, all the three conditions
are satisfy and this gives a norm and equivalently,
this gives the norm for it. So, this one.
Now, there are certain examples; is the examples
for the bounded linear functional as well
as unbounded linear functional.
So, first, example of the bounded linear functional.
The identity operator, as we have seen, the
identity operator I is a mapping from X to
X, on a normed space X, such that, on a normed
space , which carriers the image X to I X,
that is equal to X. And, obviously, it is
a linear operator and bounded also; because
the bound of this, we can see, it is 1; norm
of this is 1. It is easily verified. Then,
0 operator; this is also a bounded linear
operator; a 0 from X to X, which carries the
image X to 0, and it has a operator, bounded
with a bound 0. The third operator, which
is a differential operator; we see that, this
operator we have seen, it is already a linear
operator; that is, if we take X to be the
normed space of all polynomials, all polynomials,
on the close interval 0, 1, on the close interval
0, 1, with the norms 
given as norm of X equal to maximum of mod
X T and T belonging to the interval 0, 1.
Now, differential operator T is defined on,
T is defined on this X as T X t, is the derivative
of X with respect to t; T x t is the derivative
of X with respect to t, where the prime denotes
the differentiation of X with respect to t.
Now, we can see quickly that, T is linear;
this we have already shown earlier; T is linear.
Now, to show whether T is bounded or unbounded,
let us check. We claim that, this operator
T, differential operator T, is unbounded operator;
is not bounded. It means, we are unable to
get a constant C, such that, norm of T X is
less than equal to C times norm of X. Let
us see, for example, suppose, I take a sequence
of the polynomials, a functions X n t as T
to the power n, where the t belongs to the
interval 0 to 1; this is a polynomial. So,
it belongs to the class and find out the norm
of X n; by definition the norm of X n will
be equal to 1, because the maximum value is
taking. So, norm of this is 1, fine.
When we operate this T of X n t, the operation
will give derivative of X n t, with respect
to T, and that becomes, n t to the power n
minus 1. So, the norm of T X n, because it
is also a polynomial, so, norm is defined
as the maximum of this. So, it will be the,
n into the maximum value of this, T to the
power n minus 1, when T belongs to 0, 1, and
that is nothing, but n, ok. So, what we get
it here is that, norm of T X n divided by
norm X n, that will come out to be the n.
Now, as n increases, the bound C, this is
equivalent to C, the C which is equal to basically
n, increases. It means we are not able to
get, not able to find a constant C, such that,
norm of T x n is less than equal to C times
norm X n, for all n; is not possible; we are
not able to get. So, this will show, this
shows that, T is unbounded; is not a bounded
operator. So, what we see here, this is a
very interesting example. Why in the interest,
because the differential operator is a very
frequently used operator and entire real analysis
and concept is based on this continuous and
differentiation. So, this operator, which
is an unbounded linear operator suggest that,
the theory of the unbounded linear operator
plays a vital role in the development of the
analysis or functional analysis or any branch
of mathematics. We have a lot of application
of an unbounded operators environment, ok.
So, we have also come across about the integral
operator. What is that integral tha, suppose,
T is an operator from C 0, 1 to C 0, 1, where
C 0, 1 is the set of all continuous functions,
continuous functions, defined over the, defined
over the close interval 0, 1; set of all continuous
functions defined over the close interval
with 0 1. So, if we take a point X here a,
the corresponding value Y, is coming to be
in the form of T X, which also continuous,
where the Y, we have defined as integral 0
to 1, K of T tau X tau D tau, where K is known
as the kernel; is a given function, is a given
function, which is called, is called the kernel
of 
the operator T, kernel of T. And, we assume
that, K is continuous on this closed square,
on the closed region or a square, J cross
J, that is on 0, 1 cross 0, 1, 0 1 cross 0
1 in the T tau plane, T tau plane. So, this
definition, this way, when we define the operator
T, which maps the X to Y, where Y is defined
in this way, is known as integral operator.
Now, we claim that, this operator T is linear
and bounded.
Linearity follows immediately, just by replace
X equal to alpha X plus beta Y. So, here it
will change the alpha X tau plus beta X tau,
and then, it can be break up as a sum of the
two integral. So, linearity of the operator
T is guaranteed. To show the boundedness,
let us see, the function is continuous function
and the norm on C 0, 1, let the norm on C
0 is defined as, or is considered as the norm
of X is maximum of mod X t D t, mod X T D
T, where not X is, sorry, and T belongs to
j. Because there are two ways in defining;
one is this way; another is the integral form.
We are choosing the norm on this form. Now,
once you take this, then, T is bounded, obtained
from here. What is the norm of T X? Norm of
T X means, it is equivalent to the norm of
Y, but norm of Y is the maximum of Y, means,
this part; this is, is it not. So, maximum
of mod Y T, where the T belongs to J.
But maximum of Y T, this is equal to… So,
norm of T X, norm of Y, which is norm of T
X, is equal to the maximum of t, belongs to
J, modulus of integral 0 to 1, K t tau X tau
D tau; modulus use the modulus. So, it will
be less than equal to maximum of t belongs
to j, integral 0 to 1; take the modulus inside;
mod of K t tau into mod of X tau D tau. Now,
K is a continuous function. Since K is a continuous
function by assumption, on the closed region,
on closed region J cross, on a closed region
J cross J, is it not. On this closed region,
the function is continuous. So, it is bounded
function. Every continuous function in a closed
region is bounded. So, K is bounded; K is
bounded means, we can find a bound for a number,
say m, which is positive and mod also, that
mod of K t tau is less than equal to m for
all t tau, for all t tau belongs to the region
J cross J. So, this is true, bounded.
Hence, from here we can say, the norm of T
X is less than equal to, this will be m and
what will be this maximum of this norm of
X T, is the norm of X; that is all. So, we
get this one. Hence, this shows that, T is
bounded operator, ok. So, integral operator
comes out to be a bounded linear operator.
That is all. Now, yesterday, we have also
taken one example of the matrices, and matrices
we have to consider; in fact, it comes out
to be an operator that, if a is a matrix of
order m cross n, then, it behaves as an operator
from R n to R m, R n to R m, such that, the
image of X goes to Y, where the Y is equal
to a X; and is defined as, if we choose X
to be X i 1, X i 2, X i n, so, if we take
X to be X i 1, X i 2, X i n, of order n cross
1, Y equal to eta 1, eta 2, eta m, of order
say m cross 1, then, a X Y is equal to a X
will give, Y equal to A X will give the eta
i, sigma a n K X i k, K equal to 1 to n and
that will give the eta a n K, so, eta n a
i K.
Here, I am taking i, sorry this is i. So,
it will be written, eta I, as sigma a i K
X i K, K is 1 to n, like this, where the i
will varies from 1 to m, 1 to m. So, this
will be our corresponding map. Now, let us
see the matrix A. We have seen matrix A is
linear. Now, we also claim, A is bounded.
It is not only linear, it is bounded; why?
What is the norm of T X? Norm of T X is square.
This is equal to sigma eta i square, i is
1 to m; because this Y is equal to eta i and
eta i is varying from 1 to m, and it belongs
to… So, Y is equal to eta i of order, eta
i belongs to R m, ok. So, the norm will be
defined in this fashion, raised to the power
half. So, we are getting norm. But eta i is
A. So, we are getting sigma, i is equal to
1 to m, then, sigma K equal to 1 to n, a i
K X i K square X i K, k square, that is all.
Now, apply the Cauchy Schwarz’s inequality.
So, by Cauchy Schwarz’s inequality, Cauchy
Schwarz’s inequality, we get that, this
norm of T X square is less than equal to sigma,
i equal to 1 to m, as usual, then, sigma K
equal to 1 to n, a i K square power half,
sigma K equal to 1 to n X i K square power
half, that is all. Power half and then, whole
square, because this is square. So, we have
applied this Cauchy Schwarz’s inequality
over the product of these two things. So,
sigma of, if these are a i K and X i K, they
are nonnegative numbers, then, we are getting
this sigma a i K, this whole square into power
half, sigma X i K square power half and then,
square will be there.
Now, this shows from here that, this is equal
to… Now, this part is what; this is basically,
is equal to norm of X. So, we are getting,
this is norm of X square into double summation
i equal to 1 to m, K equal to 1 to n, a i
K square, a i K square and square this. Now,
this is nothing, but simply a constant. So,
let it be replaced by another constant, say
C squared into norm of X square. Therefore,
we get from here is, the norm of T X is less
than equal to C times norm of X . So, T is
bounded. So, every matrix of order m cross
n represents a bounded linear operator from
R n to R m, that is all. And, this will finite
case, it is very simple, because we are dealing
basically, the matrices. Now, there is another
advantage of the finite dimensional normed
space. There is so many concepts are valid
for a finite dimensional case. Just like in,
we have seen earlier, if the two norms are
equivalent, then, over a finite dimensional
case, the topology generated by these two
norm will be the same. But this may not be
true, in case of the infinite dimensional
case.
So, here also, there are many results, which
are valid for a finite dimensional case, in
general, but may not be true, in a infinite
dimensional case. So, one of them, results
we are telling is that, if a normed space
is a finite dimensional, then, every linear
operator must be bounded; but result is, if
a normed space, if a normed space is finite
dimensional, normed space X is finite dimensional
normed space, then, every linear operator
on X is bounded. Let us see the proof of this.
In case of the finite dimensional normed space,
any linear operator is a bounded operator.
The proof of it. Suppose, we take X to be
a finite dimension. So, let the dimension
of X be n and e 1, e 2, e n, be the basis
element, basis for X . So, any element x belonging
to capital X, can be expressed as sigma alpha
i x i alpha e I, because e 1, e 2, e n are
basis element, i equal to 1 to n. But T is
linear. T is linear, so, the image of X under
T become sigma i equal to 1 to n, alpha i
T of e i, T of e i. Therefore, norm of T X
is less than equal to sigma i equal to 1 to
n, mod of alpha i into norm of T i, T e i.
T e 1, T e 2, T e n, these are the finite
in number and norm we are taking. So, this
is length of the vector T e 1, T e 2, T e
n; replace this by a maximum value. So, let
it be m into sigma i equal to 1 to n, mod
of alpha i, where m denotes the maximum value
of norm T e i, where i e varying from 1 to
n. So, we are getting this. Let it be equation
1.
Now, since X is, can be, is expressed as i
equal to 1 to n, alpha i e i, where e i, e
2, e n these are the linearly independent
set of vectors; because these are the elements
of the basis. So, for a linearly independent
vector, we have seen that, one result that,
if e 1, e 2, e n are linearly independent
vectors, then, one cannot expect a vector
involving the large number of scalars, but
a minimum length; that is one can find a constant
C or epsilon, such that, norm of this is greater
than equal to epsilon times sigma of this.
So, we get from here is, the norm of X which
is equal to norm of sigma, alpha i e i, i
is 1 to n and again apply this, so, there
exist a constant C, such that, this will,
condition holds. By, there, by the earlier
lemma, proved earlier; this is the lemma proved
earlier.
So, from here, we get sigma mod alpha i, i
is 1 to n, is less than equal to 1 by C into
norm X. Let it be 2. Now, if we combine first
and 2, 1 is norm of T X is less than equal
to this; 2 is sigma alpha is less than…
So, 1 and 2 gives that, norm of T X is less
than equal to m by C into norm X; that is,
there is a constant, some constant this, which
is… So, this implies that, T is bounded.
So, every linear operator in a finite dimensional
case, is a bounded linear operator. Now, since
the linear operators, they are basically a
mapping; they are a mapping. Only difference
is that, when we operate from vector space
to a vector space, then, this is called an
operator. So, whether vector space replaced
by the norm or a Banach space, again this
is a mapping. So, in case of the mapping,
we have a concept of continuity. So, similar
concept, we can also define over a bounded
linear operator, over a linear operator or
in general, an operator, when the operator
will be considered to be a continuous operator,
at a certain point or over the entire domain
T.
(Refer Slide time: 40:23)
So, which define the continuity of an operator.
continuity of an operator. Continuous operator.
continuous operator. So, let us suppose, T
is an operator, from D T to Y, not necessary
to be linear, not necessary linear, any arbitrary
operator, that is all, where the D T is contained
in X, X is a vector space and Y is also a
vector space or norm, where X and Y are chosen
to be, say, normed space spaces. We define
the operator T, is said to be continuous 
at a point, at a point 
X naught, belonging to the domain D T, if
for every epsilon greater than 0, there is
a delta greater than 0, such that, the norm
of T X minus T X naught is less than epsilon,
for all X belonging to the domain of T, satisfying
the condition, norm of X minus X naught is
less than delta.
So, it is definition is parallel to our definition
of the continuous function. The only thing,
the mapping is, f is replaced by operator
T. So, an operator which, for which this condition
is satisfied, that for any epsilon, one can
identify delta, such that, difference between
the images remains less than epsilon, provided
the differences between the points is less
than delta. Now, if X naught is an arbitrary
point. If this is true for all trial, this
points, all the points of D T, then, we say
T is continuous over the J. So, we say T is
continuous, if T is continuous at every X,
belonging to domain D T . So, clear? So, where.
Now, if an operator is a continuous operator
and operator is also linear operator, operator
is bounded, these three concept we have introduced.
What is the relation between these three concepts?
The very interesting result is, we can now
say that, in case of the linear operator,
the continuity and boundedness comes out to
be the same, ok. Because, in general, in a
function, a function is continuous, it will
be bounded over a; if a function is continuous
over a closed interval, it will be a bounded
function. But if the function is bounded,
then, it need not be a continuous function,
ok. But here, the continuity and boundedness
will be identical, when T is a linear. And,
another point, which I alsom I want to make
it clear, the concept of the boundedness of
a linear operator and the concept of the boundedness
of a function is different. In case of the
function or mapping,we say a function is bounded,
when the corresponding range set is bounded.
But here, we do not talk about this thing;
what we say, a operator is bounded, when this
satisfy the certain condition. The condition
is the norm of T X is less than equal to C
times norm of X; that is the image set, the
image is norm of T X . So, length of this
vector divided by the norm of this original
vector, if we take the supremum of this, this
must be some number C, greater than 0; supremum
must exist for it. Then, we say, it is a bounded
operator ok.
So, we have now, a relation between this;
and the relation is in the form of result,
we say theorem. Let T be a operator from D
T to Y, be a linear operator, where D T lies
in X, X and Y both are normed spaces, normed
space. Then, one, T is continuous, if and
only if, T is bounded. Second one is, if T
is continuous 
at a single point, then, it is continuous,
it is continuous throughout the domain T;
it is continuous on D T, throughout the domain.
So, in case of the linear operator, this is
also, second result in testing, that to test
the continuity of the entire domain, just
simplifying the continuity at a single point;
because the T is linear, it will automatically
spread the continuity over the entire domain
D T. The proof.
Suppose, T is 0; case 1. Then, nothing is
proved; everything is very obvious; because
in case of the 0 operator, which is a linear
operator, continuity and Boundedness will
be the same. So, it will nothing do. So, it
is obvious. T is a statement is obvious or
trivial also. So, let T is not equal to 0,
ok. Now, assume… So, we wanted to proof
first thing, a. So, assume, T is bounded.
We wanted to show T is continuous; T is bounded;
given T is linear, this is known already,
linear. To show T is continuous, T is continuous.
So, suppose X naught be a point in the domain
D T. And, let epsilon greater than 0 be given.
So, for the continuity means, we have to find
a delta, such that, norm of T X minus T X
naught less than epsilon, only when X minus
X naught less than delta, ok. So, since T
is linear, linear, so, for every X, every
X, for every X, belonging to the domain D
T, such that, norm of X minus X naught less
than delta, where delta, I am choosing to
be epsilon over norm T.
We obtained, we get norm of T X minus X naught,
this is equal to norm of T X minus T X naught,
which is less than equal to norm of T into
X minus X naught, and this is less than equal
to norm of T and X minus X naught is less
than epsilon over norm T, so, it is less than
epsilon; it is basically equal, ok. It means
that, if I choose an arbitrary point X naught
and epsilon greater than 0, then, because
T is linear, this condition is, can be written.
T of X minus X naught can be written as T
of X minus X naught and because T is linear,
by this and bounded, is also, because of the
boundedness, we can write from here to here.
Because norm of T X is less than equal to
norm T into norm of X . So, from here, we
can write this thing. And again, epsilon is
chosen. So, we can find a delta in terms of
epsilon, which is equal to epsilon by norm
T. Substitute it, we get this thing. Therefore,
what we conclude is, the T is continuous at
X naught.
But X naught is an arbitrary point. So, we
can choose any point. Therefore, T is continuous
over the throughout, continuous on D T, throughout
the domain D T. Let us see the converse part.
Conversely, what is given now that, we wanted
the T to be bounded. So, T is given to be
continuous. Given T linear and and continuous,
continuous on, at an arbitrary point 
X naught belonging to D T, ok. So, by definition,
then, for given epsilon greater than 0, there
will exist a delta, which is depends on epsilon
greater than, such that, norm of X minus X
naught is less than delta; norm of X minus,
such that, norm of T X minus T X naught is
less than epsilon, for all X, for all X belonging
to D T, satisfying this condition. Norm of
X minus X naught less than delta; this is
by definition, ok.
So, now, choose the point y, choose any y,
different from 0 in D T. And, let x is suppose,
x naught plus delta over norm y, into y, since
y is not equal to 0, this is 0; y is not equal
to 0, so, we can choose like this. Now, from
here, obviously, x minus x naught norm y is
delta. This. So, by definition of continuity,
therefore, T x minus T x naught, this is equal
to norm of T x minus x naught, because the
T is linear, ok. Then, this will be equal
to norm of T x minus x naught. You can write
it, delta over norm y into y; and this will
be equal to… Now, since T is, T is given
to be a, this continuity is satisfying; T
is continuous. So, this has to be less than
epsilon, for whenever the norm of this thing
is there. So, we can say, this part is less
than epsilon.
So, we are taking this delta over norm y out,
into T of y. Now, this is to be less than
epsilon as T is continuous. So, from here,
we get norm of T y is less than equal to norm
of y, norm of y over delta into epsilon and
that is shows that, this will be equal to,
that is norm of T y is less than equal to
constant times norm y; because this is nothing,
but simply a constant. Therefore, T is bounded,
ok. So, this shows that T is bounded. So,
this proves the result. The second part is
very obvious. Second part, what it says is
that, if it is T is continuous at a single
point, it is continuous on this. So, if suppose,
T is continuous at the single point, then,
according to this, it must be bounded; bounded
means throughout the bounded. So, it is continuous
again at any arbitrary point, and this proves.
So, second part follows by the… Thank you.
Thanks.
