Proving $R_xy=(x_1y_1,x_2y_2,\dots)$ with $R_x$ over $\ell^{2}$, is compact if, and only if $\lim_{x \rightarrow \infty} x_n = 0.$

Question

I am trying to prove that if i have a real sequence $x_n$, and i define the operator $R_x$ over $\ell^{2}$ as $R_xy=(x_1y_1,x_2y_2,\dots)$, then $R_x$ is compact if, and only if $\lim_{x \rightarrow \infty} x_n = 0$. It seems like this is an easy excercise but i've been trying to do it and with every idea i had i couldn't continue at some point. Can you help me?

Answer 1

I assume that by $\ell^2$ you are refering to square summbale sequences of real numbers.

Suppose that $\lim_{n\to\infty}x_n=0$. For every $n\in\mathbb{N}$ We consider the operator $R_n:\ell^2\to\ell^2$ defined by $R_n(y)=(x_1y_1,\dots,x_ny_n,0,0,0,\dots)$. It is immediate that each $R_n$ is a finite rank operator (its image is a finite dimensional space). Now if we show that $R_n\to R_x$ in operator norm, then $R_x$ will be compact as a limit of finite-rank operators. But indeed, if $\varepsilon>0$ we pick $n_0$ large enough such that for all $n\geq n_0$ $|x_n|<\varepsilon$. For $n>n_0$ it is $$\|R_n-R_x\|^2=\sup_{\|y\|\leq1}\sum_{k=n+1}^\infty|x_ky_k|^2\leq\sup_{\|y\|\leq1}\varepsilon^2\sum_{k=n+1}^\infty|y_k|^2\leq\varepsilon^2$$ so we have $R_n\to R_x$ in operator norm and this shows that $R_x$ is compact.

Conversely, suppose that $R_x$ is compact. Then, if $(z_n)\subset\ell^2$ is a bounded sequence of elements of $\ell^2$ (note: by this I mean that each $z_n$ is an element of $\ell^2$, so each $z_n$ is a square summable sequence!), then $(R_x(z_n))_{n=1}^\infty$ has a convergent subsequence in $\ell^2$ (this is the definition of compactness).

Suppose that $x_n\not\to0$. Then there exists $\varepsilon>0$ and a subsequence $(x_{n_k})\subset (x_n)$ such that $|x_{n_k}|\geq\varepsilon$ for all $k\in\mathbb{N}$. Consider $e_{n_k}=(0,\dots,0,1,0,\dots,0,\dots)$ the sequence that is $0$ everywhere except the $n_k$ slot, where it is $1$. Then $(e_{n_k})_{k=1}^\infty\subset\ell^2$ and since $\|e_{n_k}\|=1$ for all $k$, $(e_{n_k})$ is a bounded sequence. Since $R_x$ is compact, as we explained, $(R_x(e_{n_k}))$ must have a convergent subsequence, say $R_x(e_{n_{k_l}})\to z\in\ell^2$. But it is $\|R_x(e_{n_k})\|=|x_{n_k}|\geq\varepsilon$, so it is also $\|R_x(e_{n_{k_l}})\|\geq\varepsilon$ for all $l$ and taking limits as $l\to\infty$ gives $\|z\|\geq\varepsilon$.

Observe now that for any $y=(y_1,y_2,\dots)\in\ell^2$ it is $$\lim_{n\to\infty}\langle e_n,y\rangle=\lim_{n\to\infty}y_n=0.$$

But we have that $$\langle z,y\rangle=\lim_{l\to\infty}\langle R_x(e_{n_{k_l}}),y\rangle=\lim_{l\to\infty}x_{n_{k_l}}y_{n_{k_l}}=\lim_{l\to\infty}\langle e_{n_{k_l}}, R_x(y)\rangle=0$$

Thus $\langle z,y\rangle=0$ for any $y\in\ell^2$. For $y=z$ we have $\|z\|^2=0$, a contradiction.