Proving Toeplitz matrix defines bounded operator on $ l^2 $

Question

I should first mention this: I have asked this question previously but I only got a partial answer that does not suit the actual assumptions but only the related ones, it reads as follows:

Define the infinite matrix $ A = [a_{ij}]_{i,j=1}^{\infty} $ using a sequence $ \{ \alpha_n \}_{n=-\infty}^{\infty} $ such that $ a_{i,j} = \alpha_{i-j} $ and also we know $ 0 < \sum_{n=-\infty}^{\infty}{|\alpha_n|} < \infty $.

a. We need to show the matrix A defines a bounded operator on the space of sequences $ \mathcal{l}^2 $ and to find it's operator norm

b. Then we need to check if this operator is compact or not

Hint : Try to think of A as the matrix representing an operator acting on a subspace of $ L^2[-\pi,\pi] $

Here is a link to the question and the solutions I got

As you can see the problem here is that my matrix has only positive indices so it is not negative infinite as well and as you can see this doesn't change much but the essence is of course that the matrix looks like the representation matrix of the product operator with a well defined function but because of the basis enumeration $ \{ e^{inx} \} $ including negatives this is not completely the case. The question said a subspace of $ L^2[-\pi,\pi] $ so maybe change some assumptions to make the method work. Help needed and appreciated

Answer 1

I had not noticed before that your indices on the matrix were restricted to positive entries. This matrix has the form $$ \begin{pmatrix} \alpha_0 & \alpha_{-1} & \alpha_{-2} & \alpha_{-3} & \cdots \\ \alpha_1 & \alpha_{0} & \alpha_{-1} & \alpha_{-2} & \cdots \\ \alpha_2 & \alpha_{1} & \alpha_{0} & \alpha_{-1} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}. $$ The subspace $H^{2}$ of $L^{2}$ of interest consists of all $f \in L^{2}[-\pi,\pi]$ for which $$ (f,e^{-inx})=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx}dx = 0,\;\;\; n=1,2,3,\cdots. $$ Every $f \in H^{2}$ has the form $$ \sum_{n=0}^{\infty}f_n e^{inx},\;\;\; \|f\|^{2}=\sum_{n}|f_n|^{2} < \infty. $$ Let $P$ denote the orthogonal projection of $L^{2}$ onto $H^{2}$ given by $$ Pf = \sum_{n=0}^{\infty}f_n e^{inx} $$ In other words, $P\sum_{n=-\infty}^{\infty}f_ne^{inx} = \sum_{n=0}^{\infty}f_ne^{inx}$.

Define a function $a(z)=\sum_{n=-\infty}^{\infty}\alpha_n e^{inx}$. If you multiply $a$ by $f \in H^{2}$, and then project, you get \begin{align} P(a(z)f(z)) & = (\alpha_0 f_0 + \alpha_{-1} f_1 + \alpha_{-2} f_2 +\cdots) \\ & + (\alpha_1 f_0 + \alpha_0 f_1 + \alpha_{-1}f_2+\cdots)e^{ix} \\ & + (\alpha_2 f_0 + \alpha_1 f_1 + \alpha_0 f_2 +\cdots) e^{2ix} \\ & + \cdots \end{align} The first term is where the sum of the indices is always 0. The second term is where the sum of the indices is always 1. Etc.. The negative powers are truncated by the projection $P$. This operator has the desired matrix representation. The only question is whether or not this operator maps back into $H^{2}$, which is to say that the sum of the squares of the coefficients of $P(af)$ is finite. Check this $$ \|P(af)\|^{2}=\sum_{k=0}^{\infty}|\sum_{n=k}^{\infty}\alpha_{k-n}f_{n}|^{2} $$ You can apply Cauchy-Schwarz to the inner sum after writing $$ |\alpha_{k-n}f_{n}|=|\alpha_{k-n}|^{1/2}(|\alpha_{k-n}|^{1/2}||f_{n}|). $$ Let $\|a\|_{1} = \sum_{k=-\infty}|\alpha_k|$. The result is \begin{align} \|P(af)\|^{2} & \le \sum_{k=0}^{\infty}\left(\sum_{n=0}^{\infty}|\alpha_{k-n}|\sum_{n=0}^{\infty}|\alpha_{k-n}||f_{n}|^{2}\right) \\ & \le \|a\|_1\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}|\alpha_{k-n}||f_{n}|^{2} \\ & = \|a\|_1\sum_{n=0}^{\infty}|f_{n}|^{2}\sum_{k=0}^{\infty}|\alpha_{k-n}| \\ & \le \|a\|_1^{2}\|f\|^{2}. \end{align} So $P(af)$ is a bounded operator, with $\|P(af)\|\le \|a\|_1\|f\|$. This isn't the tightest bound. For example, if $\|a\|_{\infty}=\sup_{x}|a(x)|$, then $\|P(af)\| \le \|af\| \le \|a\|_{\infty}\|f\|$. Clearly $\|a\|_{\infty}\le \|a\|_1$.