Compact operator on $l^2$

Question

Let $A$ be a bounded linear operator on $l^2$ defined by $A(a_n)= \left(\frac{1}{n} a_n\right)$. Would you help me to prove that $A$ is compact operator. I guess the answer using an approximation by a sequences of finite range operator.

Answer 1

Let $$T_k(a_n)_j=\begin{cases} \frac 1ja_j&\mbox{ if }j\leq k;\\ 0&\mbox{ otherwise.} \end{cases}$$ This gives a linear operator, and the range of $T_k$ is generated by $e_1,\dots,e_k$, a finite dimensional space. For $a\in\ell^2$, $$\lVert T(a)-T_k(a)\rVert=\sum_{j\geq k+1}\frac 1j|a_j|\leq\frac 1{k+1}\sum_{j\geq 1}|a_j|,$$ so $\lVert T-T_k\rVert\leq\frac 1{k+1}$ and $T$ is compact as the limit in operator norm of compact operators.

Note that more generally, we can define $T(a)(k):=d_ka_k$, where $d_k\to 0$, and $T$ from $\ell^p$ to $\ell^p$, where $1\leq p<\infty$, and this will give a compact operator by the same argument.

Answer 2

There was recently a duplicate question about this and I wrote a proof that does not use the fact that a limit of compact operators is compact. I'll just post it here:

Notations: I will denote by $a(n)$ a sequence in $l^2 = l^2(\mathbb{N})$, so now $a_m$ is a sequence of elements (sequences) in $l^2$.

The operator $T$ is compact if and only if the image of the unit ball is relatively compact. meaning that if $a_m$ are in the unit ball of $l^2$, then $T(a_m)$ admits a subsequence which converges in $l^2$.

By definition, $\|a_m\|_{l^2}\leq 1$ for all $m$, which (by taking a square) implies that $\sum_n |a_m(n)|^2 \leq 1$. In particular, $|a_m(n)|\leq 1$ for every $m,n$.

We will now construct a convergent subsequence for $a_m$ inductively:

Consider the sequence $a_m(1)$, since $|a_m(1)|\leq 1$, by Bolzano-Weirstrass there exists a convergent subsequence $a_{m^1_k}(1)\rightarrow a_\infty(1)$. Now look at $a_{m_k^1}(2)$, again it is bounded so there exists a subsequence, of the subsequence $m_k^1$ which converges $a_{m^2_k}(2)\rightarrow a_\infty(2)$, and since $m_k^2$ is a subsequence of $m_k^1$ we also have $a_{m^1_k}(1)\rightarrow a_\infty(1)$.

Continue by induction, you can construct for every $i$ a sub-sub-sub...-sub sequence $m_k^i$ and numbers $a_\infty(1),...,a_\infty(i)$ so that $a_{m_k^i}(1)\rightarrow a_\infty(1),...., a_{m_k^i}(i)\rightarrow a_\infty(i)$.

Now we claim that $T(a_{m_k^k})$ converges to the sequence $a(n):=\frac{a_\infty(n)}{n}$ in $l^2$. We need to prove two things:

1. $a$ is in $l^2$. To see this recall that $a_m(n)$ is bounded by $1$ and therefore so is $a_\infty(n)$, it then follows that $|a(n)|\leq \frac{1}{n}$ and therefore $$\|a\|_{l^2} = \sum_n |a(n)|^2 \leq \sum_n \frac{1}{n^2} \leq \infty$$

2. $\|T(a_{m_k^k}) - a\|_{l^2}\rightarrow 0$ as $k\rightarrow \infty$.

Let $\varepsilon>0$, we need to find $K$ so that for all $k>K$ we have $\|T(a_{m_k^k}) - a\|_{l^2}\leq \varepsilon$.

First, observe that $\|T(a_{m_k^k}) - a\|_{l^2}\|_{l^2} = \sum_n \left( \left| \frac{a_{m^k_k}(n) - a_\infty(n)}{n}\right|^2\right) \leq \sum_n \frac{2}{n^2}<\infty$

Therefore, there exists $N$ sufficiently large and independent on $k$, so that $$\sum_{n=N+1}^\infty \left( \left| \frac{a_{m^k_k}(n) - a_\infty(n)}{n}\right|^2\right) < \varepsilon/2$$

Now for all $1\leq n \leq N$, choose $k$ sufficiently large so that $m_k^k$ is a subsequence of $m_k^1,....,m_k^N$ and furthermore, $|a_{m_k^k}(n) - a_\infty(n)|<\sqrt{\varepsilon/2N}$. Then,

$$\|T(a_{m_k^k}) - a\|_{l^2} = \sum_n \left( \left| \frac{a_{m^k_k}(n) - a_\infty(n)}{n}\right|^2\right) = \sum_{n=1}^N \left( \left| \frac{a_{m^k_k}(n) - a_\infty(n)}{n}\right|^2\right) + \sum_{n=N+1}^\infty \left( \left| \frac{a_{m^k_k}(n) - a_\infty(n)}{n}\right|^2\right) < \sum_{n=1}^N \varepsilon/2N + \varepsilon/2 = \varepsilon.$$

This completes the proof.