The following question has two parts. I was able to do the first part and proceeded with some ideas for the second part.
(a) Let $X$ be a Banach space, $X^{*} $ be its dual, $x_{0} \in X$ and $\phi_{0} \in X^{*}$. Define $T : X^{*}\rightarrow X^{*}$ by $T(\psi) = \psi (x_{0})\phi_{0}$ for $\psi\in X^{*}$. Prove that T is compact.
(b) Using part (a) or otherwise, prove that given a two-variable polynomial function $a$, the operator $A : L^{\infty} ([0, 1], m) \rightarrow L^{\infty} ([0, 1], m)$ (where $m$ denotes the Lebesgue measure) defined by $$Af(x) = \int_{0}^{1}a(x, y)f(y)dy$$ is compact.
For part (a), I took a bounded sequence and showed that its image has a convergent subsequence using completeness of $\mathbb{C}$. For part (b), I proceeded as follows:
We have $$A(f)=\int_{0}^{1}a(.,y)f(y)dy$$ Now $L^{\infty}$ is isomporphic to $(L^{1})^{*}$ via the map $f\rightarrow I_{f}$, where $I_{f}(g)=\int fgdm$. So $Af(x)=I_{a(x,.)}(f). $
I can't figure out what to do next and if I am proceeding in the right direction.