Show that $F(\{ f \in L^{2}([0,1]): \vert \vert f \vert \vert_{2}\leq 1\})$ is relatively compact

Question

Let $F: L^{2}([0,1])\to C([0,1]), f \mapsto F(f)$ where $(Ff)(t)=\int_{0}^{1}(t^{2}+s^{2})(f(s))^{2}ds$

Show that $F(\{ f \in L^{2}([0,1]): \vert \vert f \vert \vert_{2}\leq 1\})$ is relatively compact.

My steps:

let $(f_{n})_{n}$ be an arbitrary sequence in $\{ f \in L^{2}([0,1]): \vert \vert f \vert \vert_{2}\leq 1\}$ now via the diagonalization method (I have seen this for normal sequences, but not for function sequences) I want to find my subsequence $(f_{n_{k}})_{k}$ and the proposed limit $f$.

note that for any $f_{n}\in \{ f \in L^{2}([0,1]): \vert \vert f \vert \vert_{2}\leq 1\}\Rightarrow \forall t \in [0,1]:\vert Ff_{n}(t)\vert=\vert\int_{0}^{1}(t^{2}+s^{2})(f(s))^{2}ds\vert\leq \int_{0}^{1}\vert(t^{2}+s^{2})(f_{n}(s))^{2}\vert ds\leq \vert \vert t^{2}+s^{2}\vert\vert_{\infty}\vert\vert f_{n}\vert\vert_{2}<\infty$

What can I say from this? In the case of simple sequence spaces I could use Bolzano Weierstraß to show that the exists a subsequence etc. But now I am dealing with functions, and I do not know what to do

Answer 1

$$(Ff)(t) = {t^2}\int_0^1 {f{{(s)}^2}ds} + \int_0^1 {{s^2}f{{(s)}^2}ds}$$ The image is spanned by $\{1,t^2\}$. When $\|f\|\leq 1$, the two integrals are bounded. Hence image under $F$ is bounded and finite rank, so relatively compact.