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

up vote 2 down vote accepted

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.

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.