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I have this question with a given clue I cannot decipher in Functional Analysis stating:

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] $

Now I think I know how A acts on a sequence $ \{x_j\}_{j=1}^{\infty} \in \mathcal{l}^2 $ we have a sequence $ \{y_i\}_{i=1}^{\infty} $ defined by $ y_i = \sum_{j=1}^{\infty} a_{ij}x_j $

In order to show the operator is bounded I need to look at a general sequence of $ \mathcal{l}^2 $ norm of value 1 and then check indeed that its image under A is an $ \mathcal{l}^2 $ sequence of bounded finite constant norm, and find this norm of operator A, and in the second part I need to check if closure of image of unit ball in $ \mathcal{l}^2 $ is compact in $ \mathcal{l}^2 $. All this I know in theory but I cannot do it practically as the clue I just cannot understand although I think it has something to do with Fourier coefficients, although I am not sure. Could someone guide me to solve both parts please? Thanks

SIGNIFICANT PROGRESS: I figured using idea of representation matrix that we could get $ \sum_j <Te_j,e_i><x,e_j> = <Tx,e_i> $ so under standard complex exponential basis $ e^{ik} $ ; $ k \in \mathcal{Z} $ and the product operator by g(x) we have that the entries of the matrix are $ a_ij = <Te_j,e_i>= <g(x)e_j,e_i> = 1/{2\pi}\int_{-\pi}^{\pi} g(x)e^{-(i-j)x}$ which is i-j complex Fourier coefficient of g(x) so if we know there is a g(x) whose Fourier coefficients are exactly $ \alpha $ then the representation matrix of product with g is exactly A so now using the representation matrix meaning each member of the result sequence y is the sum of products of a row of the A matrix with each member of x sequence and the result is the sequence of Fourier coefficients of product with g(x). I made this progress thanks to random123's comment but I still cannot get further help please I really need the help to continue, thanks.

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Croc2Alpha
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    Consider multiplication by a function which has as its fourier coefficients a_i,j. Under the fourier transform multiplication turns into convolution. If you need i can provide the details. – random123 Sep 21 '15 at 17:12
  • @random123 : Could you please provide your details as you have mentioned please? – Croc2Alpha Sep 21 '15 at 19:52

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Write $$ f = \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx} $$ What happens if you mutliply $f(x)$ by $a(x)=\sum_{n=-\infty}^{\infty}\alpha_ne^{inx}$? The function $a$ is continuous, periodic and bounded. So this multiplication operator $M_{a}$ is a bounded linear operator on $L^{2}$. $$ M_{a}f = \sum_{n=-\infty}^{\infty}\alpha_ne^{inx}\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}. $$ The coefficient of $e^{ikx}$ in the final product is the sum of all terms where the indices sum to $k$: $$ \sum_{n=-\infty}^{\infty}\alpha_{k-n}\hat{f}(n) = \sum_{n=-\infty}^{\infty}a_{k,n}\hat{f}(n). $$

Disintegrating By Parts
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    Every $f \in L^{2}[-\pi,\pi]$ is $f = \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}$ with $|f|{L^{2}}^{2}=\sum{n}|\hat{f}(n)|^{2}$ if you use the normalized $L^{2}$ inner product $(f,g)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt$. $f \mapsto \hat{f}$ is a unitary map of $L^{2}$ onto $\ell^{2}(\mathbb{Z})$. – Disintegrating By Parts Sep 21 '15 at 21:30
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    @zbigniew2015 : The norm of $M_{a}f=a(x)f(x)$ on $L^{2}[-\pi,\pi]$ is $\sup_{x}|a(x)|$. So the norm of your matrix operator is $\sup_{x}|\sum_{n=-\infty}^{\infty}\alpha_n e^{inx}|$. – Disintegrating By Parts Sep 21 '15 at 21:36
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    @zbigniew2015 : Multiplication by the constant function $1$ is the identity operator, which is not compact. $f \mapsto \hat{f}$ is isometric by Parseval's identity. The inverse is also isometric. Norm properties transfer exactly. – Disintegrating By Parts Sep 21 '15 at 21:40
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    @zbigniew2015 It's counterpart is the identity operator. The only operator unitarily equivalent to the identity operator is the identity operator. – Disintegrating By Parts Sep 21 '15 at 22:04
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    @zbigniew2015 : The example you have given from another post is not the same. That operator is defined on $\ell^2(\mathbb{N})$. – Disintegrating By Parts Sep 21 '15 at 22:07