Fix an integer $n\ge 2$. Let $[a,b]$ be an interval and $f: [a,b]\to \mathbb R$ be a continuous function and for $x\_1,...,x\_n$ being the Gaussian Quadrature nodes in $[a,b]$, and Gaussian Quadrature wights $w\_1,...,w\_n$ in $[a,b]$ , (everything being calculated with weight function $\omega(x)=1$ ) (see <https://en.wikipedia.org/wiki/Gaussian_quadrature> for details ), let $T\_{n,[a,b]} (f) := \sum\_{i=1}^n w\_if(x\_i)$. 


Now, for every $f\in C[0,1]$, let $T\_n(f)(0)=0$ and $T\_n(f)(x)=T\_{n,[0,x]}(f), \forall x \in (0,1]$. Let ${\mathbb R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $\mathbb R$. 


Then of-course $T\_n$ is a linear map from $C([0,1])$ to ${\mathbb R}^{[0,1]}$. My questions are the following : 


(1) For which $n$, is it true that $T\_n (C[0,1]) \subseteq C[0,1]$ ? 


(2) Let $\mathcal B([0,1])$ be the set of all bounded real-valued functions on $[0,1]$. For which $n$, is it true that $T\_n(C[0,1]) \subseteq \mathcal B([0,1])$ ? 


(3) For which $n$, is $T\_n$ an injective function ?


(4) If $n\ge 2$ is an integer for which $(2) $ (or (1)) holds, then is it true that $\sup\_{x\in [0,1]} |T\_n (f)(x)| \le M\_n \sup\_{x\in [0,1]} |f(x)|, \forall f \in C([0,1]) $, where the constant $M\_n$ only depends on $n$ ? 


(5) Is it true that $T\_n (f) \to T(f), \forall f \in C[0,1]$ ? Or at least does a subsequence of $\{T\_n\}$ converge to $T$ pointwise on $C[0,1]$ ? Where $T$ is the operator $T:C[0,1] \to C[0,1]$ , defined as $T(f)(x):=\int\_0^x f(t) dt$ 



On the continuity and injective-ness of Gauss quadrature scheme for numerical integration, with weight function identically $1$
The key here is the simple change-of-interval/rescaling formula, found e.g. at the link in the OP, according to which 
\begin{equation}
 T\_n(f)(x)=T\_{n,[0,x]}(f)=x\sum\_1^n w\_i f(xx\_i), \tag{\*}
\end{equation}
where the $w\_i$'s and $x\_i$'s are such that $T\_n(f)(1)=T\_{n,[0,1]}(f)=\sum\_1^n w\_i f(x\_i)$ for all $f$. 


From here, one immediately has the affirmative answer to Questions (1), (2), and (4). In particular, in (4), one may take $M\_n=\sum\_1^n|w\_i|=\sum\_1^n w\_i =1$. 


The answer to Question (5) is also positive. Indeed, recall that for any polynomial $p$ of degree $\le2n-1$ we have $T\_n(p)(1)=\int\_0^1 p$ and hence, by the rescaling, $T\_n(p)(x)=\int\_0^x p$ for all $x\in[0,1]$. Take now any $f\in C[0,1]$ and any real $\delta>0$ and let $p$ be a polynomial, of degree $\le2n-1$ for some natural $n$, such that $p-\delta\le f\le p+\delta$ on $[0,1]$. The operator $T\_n$ is positive, since $w\_i\ge0$ for all $i$. So, for $x\in[0,1]$
\begin{multline}
 T(p)(x)-\delta\cdot x=\int\_0^x (p-\delta)=T\_n(p-\delta)(x)\le T\_n(f)(x) \\ 
 \le T\_n(p+\delta)(x)=\int\_0^x (p+\delta)
 =T(p)(x)+\delta\cdot x,
\end{multline}
whence
\begin{equation}
 |T\_n(f)(x)-T(f)(x)|\le\delta\cdot x\le\delta, 
\end{equation}
so that $T\_n(f)\to T(f)$ uniformly on $[0,1]$, for any $f\in C[0,1]$. 


Moreover, if two functions, say $f$ and $g$ in $C[0,1]$, are uniformly close to each other, than any polynomial $p$ uniformly close to $f$ will also be uniformly close to $g$. Hence, it follows from the above reasoning that $T\_n\to T$ in the $(\|\cdot\|\_\infty,\|\cdot\|\_\infty)$ operator norm. 


The answer to Question (3) is this: the linear operator $T\_n$ is not injective for any natural $n$. Indeed, letting, as usual, the nodes $x\_i$ be increasing in $i$, we have $x\_n<1$. On the other hand, by (\*), $T\_n(f)=0$ for any $f\in C[0,1]$ such that $f=0$ on $[0,x\_n]$. So, letting e.g. $f(x):=\max(0,x-x\_n)$ for $x\in[0,1]$, we have $f\in C[0,1]$ and $f\ne0$, whereas $T\_n(f)=T\_n(0)$. 




---


**Added in response to a comment by the OP:** No, the operator $T\_n$ is not compact for any natural $n$. Indeed, take any compact intervals $J$ and $K$ such that $\emptyset\ne J^\circ\subset J\subset K^\circ\subset K\subset (x\_{n-1}/x\_n,1)$, where ${}^\circ$ denotes the interior and $x\_{n-1}:=0$ for $n=1$. Take any $h\in C(J)$. Then $h$ equals $g|\_J$, the restriction to $J$ of some function $g\in C[0,1/x\_n]$ such that the support $\text{supp}\,g$ of $g$, defined as the closure of the set $\{x\colon g(x)\ne0\}$, is contained in $K$ and $\|g\|\_{[0,1/x\_n]}=\|h\|\_J$, where $\|u\|\_I:=\max\_{x\in I}|u(x)|$ for any interval $I$ and any $u\in C(I)$. 
Let now 
\begin{equation}
 f(y):=\frac{x\_n}{w\_n y}\,g\Big(\frac y{x\_n}\Big)
\end{equation}
for $y\in(0,1]$, with $f(0):=0$. Then $\text{supp}f\subseteq x\_n\,K\subset(x\_{n-1},x\_n)\subset(0,1)$, whence $y\_\*:=\min\text{supp}f\ge\min(x\_n K)=x\_n\min K>0$, $f\in C[0,1]$, and 
\begin{equation}
 \|f\|\_{[0,1]}\le M\|g\|\_{[0,1/x\_n]}=M\|h\|\_J, 
\end{equation}
where $M:=\frac1{w\_ny\_\*}<\infty$. 
Next, for $x\in(0,1]$, 
\begin{equation}
 xw\_i f(xx\_i)=xw\_i \frac{x\_n}{w\_n xx\_i}\,g\Big(\frac{xx\_i}{x\_n}\Big)=
 \left\{
 \begin{aligned}
 0&\text{ if }i\le n-1,\\
 g(x)&\text{ if }i=n,
 \end{aligned}
 \right.
\end{equation}
because $\text{supp}\,g\subseteq K\subset (x\_{n-1}/x\_n,1)$ and hence $g=0$ on $[0,x\_{n-1}/x\_n]$. 
So, $T\_n(f)=g$ on $[0,1]$ and hence $T\_n(f)=h$ on $J$, that is, $(R\_JT\_n)(f)=h$, where $R\_J\psi:=\psi|\_J$, the restriction of $\psi$ to $J$. So, $(R\_JT\_n)(MB\_{[0,1]})\supseteq B\_J$, where $B\_I$ denotes the unit ball in $C(I)$. Since $B\_J$ is not compact, it follows that the operator $R\_JT\_n$ is not compact. Therefore and because the operator $R\_J$ is continuous, it follows that the operator $T\_n$ is indeed not compact. 


