Prove that the following operator is compact. Let $A$ be defined on $l_2$ by $y=Ax$ where $y_i=\sum_j^\infty a_{ij} x_j$ and where $\sum_{ij}^\infty a_{ij}^2$ is finite.
Prove that $A:l_2\rightarrow l_2$ and that $A$ is compact, without using notions of precompactness.
I have seen the same question answered here but I am wondering if it can be done without the precompact arguments.
Thanks! 
 A: You can use the following result to prove it:

Theorem: If $L:H\rightarrow H$ is a bounded linear operator on a separable Hilbert space $H$, $(e^{n})_{n}$ is an orthonormal basis, and $\sum_{n=1}^{\infty}\Vert Le^{n}\Vert^2<\infty$, then $L$ is compact.

Proof: Define the operator $L_{N}$ by $L_{N}x=\sum_{n=1}^{N}\langle x,e^{n}\rangle e^{n}$. We want to show that $L_{N} \rightarrow L$ with respect to the operator norm topology in order to conclude that $L$ is compact (a convergent sequence of compact operators converges to a compact operator). Note that
\begin{align*}
\left\Vert L_{N}x-Lx\right\Vert ^{2} & =\left\Vert \sum_{n\geq N+1}\left\langle x,e^{n}\right\rangle Le^{n}\right\Vert ^{2}\\
 & \leq\left(\sum_{n\geq N+1}\left|\left\langle x,e^{n}\right\rangle \right|\left\Vert Le^{n}\right\Vert \right)^{2}\\
 & \leq\left(\sum_{n\geq N+1}\left\langle x,e^{n}\right\rangle ^{2}\right)\left(\sum_{n\geq N+1}\left\Vert Le^{n}\right\Vert ^{2}\right)\\
 & \leq\left\Vert x\right\Vert \sum_{n\geq N+1}\left\Vert Le^{n}\right\Vert ^{2}.
\end{align*}
Now, for any $\epsilon>0$, we can pick $N$ large enough so that $\sum_{n\geq N+1}\Vert Le^{n}\Vert^{2}<\epsilon$, from which the desired result follows.
A: The crux of the problem is that $\{a_i\}_{i \in \mathbb{N}} \in c_0$.
Suppose $a_n \to 0$, so that $a \in c_0$ (we call the sequence $"a"$). Then for all $\epsilon>0$ there is sufficiently large $k$ such that $|a_k| < \epsilon$. So approximate $A$ by its finite rank truncations $A_N$ with $A_Nx = \{a_nx_n\}_{x=1}^{N}$. It is well known that finite rank operators are compact (I urge you to think about this, if you haven't seen this fact). So then we want to show that $\|A_n - A_m\|$ is Cauchy, so then by the Cauchy Criterion $\{A_n\}_{n \in \mathbb{N}}$ will be uniformly convergent. 
So, we have for $n>m >k$,
$\|A_n - A_m\|^2 = \sup_{\|x\|_2 =1} \sum_{i=m+1}^n |a_ix_i|^2 <\sup_{\|x\|_2 =1}  \epsilon^2 \sum_{i=m+1}^n |x_i|^2 \le \epsilon^2$
since $\|x\|_2=1$, and dropping squares we see this proves uniform convergence by the Cauchy Criterion. To see that $A_n \to A$ in norm, we have that $\|A-A_n\|^2 \le \displaystyle \sup_{\|x\|_2 = 1} \sum_{i=n+1}^\infty |a_ix_i|^2 < \sup_{\|x\|_2=1} \epsilon^2 \|x\|_2^2 = \epsilon^2$. Then $A$ is the uniform limit of compact finite rank operators and thus $A$ is as well (the uniform limit of compact operators is compact).
