# On the continuity and injective-ness of Gauss quadrature scheme for numerical integration, with weight function identically $1$

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 $$T(f)(x):=\int_0^x f(t) dt$$

The key here is the simple change-of-interval/rescaling formula, found e.g. at the link in the OP, according to which $$$$T_n(f)(x)=T_{n,[0,x]}(f)=x\sum_1^n w_i f(xx_i), \tag{*}$$$$ where the $$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 $$$$|T_n(f)(x)-T(f)(x)|\le\delta\cdot x\le\delta,$$$$ so that $$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 $$$$f(y):=\frac{x_n}{w_n y}\,g\Big(\frac y{x_n}\Big)$$$$ for $$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]$$, and $$$$\|f\|_{[0,1]}\le M\|g\|_{[0,1/x_n]}=M\|h\|_J,$$$$ where $$M:=\frac1{w_ny_*}<\infty$$. Next, for $$x\in(0,1]$$, xw_i f(xx_i)=xw_i \frac{x_n}{w_n xx_i}\,g\Big(\frac{xx_i}{x_n}\Big)= \left\{ \begin{aligned} 0&\text{ if }i\le n-1,\\ g(x)&\text{ if }i=n, \end{aligned} \right. because $$\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.

• Thank you very much for your answer. Say ... do you think the $T_n$'s are compact operators ? – user521337 Nov 18 '18 at 0:02
• @user521337 : I have added a proof that $T_n$ is not compact. – Iosif Pinelis Nov 18 '18 at 15:45
• thanks ... btw do you have any idea about the injectivity of the $T_n$ s that I also asked in the original question ? – user521337 Nov 19 '18 at 2:00
• No, at this point I don't have a good idea about the injectivity. – Iosif Pinelis Nov 19 '18 at 12:32
• D you have any idea about whether any $T_n$ is bijective or not ? – user521337 Nov 25 '18 at 0:37