I want to prove the following proposition:
Suppose $k \in L^2 ([a,b] \times [a,b])$, then the linear operator $K: L^2 ([a,b]) \to L^2 ([a,b])$ defined as: $(Kf)(t) = \int_a^b k(t,s)f(s)ds$ is compact.
I came up with this proof, but I am not entirely sure it is correct, I would be very grateful if someone could proofread it. I am assuming the result is true for $k$ continuous which I know follows from Arzelà-Ascoli.
Let $(f_n)_{n=1}^{\infty} \in L^2 ([a,b])$ be bounded, say $||f_n||_2 \leq M$. If I remember, I believe continuous functions are dense in $L^2$, so I let $(k_i)_{i=1}^{\infty}$ be a sequence of continuous functions converging to $k$ in $||.||_2$. For all $i$, let $K_i$ be the operator associated to $k_i$ (I know $K_i$ is compact).
Then I use a diagonal argument: using that $K_i$ is compact, I find a convergent subsequence $f_{1,n}$ such that $K_1 f_{1,n}$ converges, then I find a convergent subsequence $f_{2,n}$ of $f_{2,n}$ such that $K_2 f_{2,n}$ converges... Let $g_n = f_{n,n}$, then $K_i g_n$ converges in $L^2$ for all $i$.
Let $h_i = lim_{n \to \infty} K_i g_n$. I want $h_i$ to converge. $$ ||h_i-h_j||_2 \leq |h_i-K_i g_n||_2 + ||K_i g_n - K_j g_n||_2 + ||K_j g_n - h_j||_2 (1)$$ and $$ ||K_i g_n - K_j g_n||_2 = \int_a^b \left| \int_a^b (k_i(t,s)-k_j(t,s))g_n(s) ds \right|^2dt \leq \int_a^b \left( \int_a^b \left| (k_i(t,s)-k_j(t,s))g_n(s) \right| ds \right)^2dt \leq \int_a^b \int_a^b |k_i(t,s) - k_j(t,s)|^2 ds \int_a^b |g_n(s)|^2ds dt \leq \int_a^b \int_a^b |k_i(t,s) - k_j(t,s)|^2 dsdt \cdot M^2 $$ Using Cauchy-Schwartz inequality. Since $(k_i)$ converges in $L_2$ I can choose $i,j$ such that $||K_i g_n - K_j g_n||_2 < \varepsilon / 3$ (note this does not emend on $n$), and then I can choose $n$ large enough to make the two other terms in (1) less than $\varepsilon / 3$, proving that $(h_i)$ is Cauchy, and hence it converges to some $h \in L^2$.
Finally, I claim that $Kg_n$ goes to $h$: $$ ||Kg_n - h||_2 \leq ||K g_n - K_i g_n||_2 + ||K_i g_n - h_i||_2 + ||h_i-h||_2$$ First I can choose $i$ to make first and third term small, then once $i$ is fixed I can choose $n$ to make the middle term small, hence proving the claim.
If anyone knows a shorter proof or a proof that doesn't assume the result for continuous functions, I am all ears.
The result you are using says that when $k$ is continuous the corresponding operator is compact when considered as an operator from $C[a,b] \to C[a,b]$. This result is not useful here.
You can prove this result by showing that $k$ can be approximated in the norm of $L^{2}([a,b] \times [a,b])$ by finite linear combinations of functions of the form $k_1(t)k_2(s)$. [Use the fact that limits in operator norm of finite rank operators is compact].