Hankel Operator is compact

Question

I am currently working on compact operators. I am trying to solve the following exercise

Problem Let $(a_j)_{j \in \mathbb{N}}$ be a sequence of complex numbers in $\ell_1$, i.e. $\sum_j |a_j| < \infty$. Define the following operator $A$ on $\ell_2$ with the following matrix representation with respect to the standard basis of $\ell_2$

$$\left[\begin{array}{l} a_1&a_2&a_3& \dots \\ a_2&a_3&a_4&\dots \\ a_3&a_4&a_5 & \dots \\ \vdots& \vdots& \vdots&\ddots \end{array}\right]$$

Prove that $A$ is a compact operator.

I have so far tried to construct a sequence of finite rank operators by truncating the matrix in a right way. However I run into some nasty sums which fail to reveal any convergence. Can someone please help me?

Answer 1

Solution 1 By the Schur test, with $p_n=1,$ we have $\|A\|_{\ell^2\to \ell^2}\le \sum_{n=1}^\infty|a_n|.$ Let $A_N$ denote the truncated matrix, i.e. $$A_N(i,j)=\begin{cases} A(i,j) & i+j\le N\\ 0 &{\rm otherwise} \end{cases} $$ Then by the same test $\|A-A_N\|_{\ell^2\to \ell^2}\le \sum_{n=N+1}^\infty |a_n|.$ As $A_N$ is finite dimensional the operator $A$ is compact.

Remark The Schur test is formulated for matrices with nonnegative entries. However the operator norm of the matrix with entries $|a_{ij}|$ is greater or equal than the norm of the matrix with entries $a_{ij}.$

Solution 2 We will use the fact that we deal with the Hankel matrix. It will be more convenient to denote $b_n=a_{n+1}.$ We can replace the assumption by a weaker one, namely: the series $h(x):=\sum b_ne^{inx}$ is uniformly convergent. For $u,v\in \ell^2(\mathbb{N}_0)$ let $$f(x)=\sum_{n=0}^\infty u_ne^{-int},\ g(x)=\sum_{n=0}^\infty v_ne^{int}$$ Then $$\langle Bu,v\rangle ={1\over 2\pi}\int\limits_0^{2\pi}f(x)\overline{g(x)}h(x)\,dx $$ Therefore by the Cauchy-Schwarz inequality and the Parseval identity we get $$|\langle Bu,v\rangle |\le \|h\|_\infty\|fg\|_1\le\|h\|_\infty\|f\|_2\|g\|_2\\ = \|h\|_\infty\|u\|_2\|v\|_2$$ Hence $$\|B\|_{\ell^2\to\ell^2}\le \|h\|_\infty$$ Let $h_N(x)=\sum_{n=0}^Nb_ne^{inx}$ and $B_N$ the corresponding truncated matrix. Then $$|B-B_N\|_{\ell^2\to\ell^2}\le \|h-h_N \|_\infty \to 0$$ Hence $B$ is a compact operator as the limit of finite dimensional operators $B_N.$

Answer 2

We define the operator $A_n$, such that $$ (A_nx)_k=\left\{ \begin{array}{ccc} a_nx_k & \text{if} & k\le n,\\ 0 & \text{if} & k> n,\\ \end{array} \right. $$ where $x=(x_n)$. Clearly, $A_n$ is a bounded operator of finite rank, and hence a compact one. It suffices to show that $\|A_1+\cdots+A_n-A\|\to 0$. Here $\|\cdot\|$ is the operator norm.

Note that, the set compact operators $K:H\to H$, where $H$ is a Banach space, is a closed subset of the set $B(H)$ of bounded operators on $H$. It therefore suffices to show that $A_1+\cdots+A_n\to A$, in the operator norm.

Clearly, $\|A_n\|=|a_n|$, and hence the series $\sum_{n=1}^\infty A_n$ converges, in the operator norm sense, and also, for every $x=(x_n)\in\ell^2(\mathbb N)$, $$ \lim_{n\to\infty}\left(\sum_{j=1}^n A_jx\right)_{\!\!k}= \lim_{n\to\infty}\left\{ \begin{array}{ccc} \sum_{j=1}^{n-k}a_{j+k} x_j & \text{if} & k\le n \\ 0 & \text{if} & k>n \end{array} \right. = Ax, $$ and thus $\sum_{n=1}^\infty A_n=A$.