Showing image of an integral operator is continuous Let $K(x,y)$ be continuous on the unit square $[0,1] \times [0,1]$ and let $\| K \|  = \max | K(x,y) |$. For $\phi (x) \in C([0,1])$, define
$$T\phi (x)= \int_{0}^{1}K(x,y)\phi(y)dy$$
(a) Show that $T\phi \in C([0,1])$ , i.e. $T: C([0,1]) \to C([0,1])$.
(b) Show that T is continuous by showing that there exists $C>0$ such that $ T\phi\leq C||\phi||$ for all $\phi \in C([0,1])$.
(c) If {$ \phi_{n}  $} , $ n \in (1,\infty)$ is a bounded sequence in $C([0,1])$, show that the sequence {$ T\phi_{n} $} , $ n \in (1,\infty)$  has a convergent subsequence. (An operator with this property is called a compact operator.)
 A: You can explain what's difficult if you can't fill in the details in the following! It's difficult to answer without knowing what you find difficult.

$$[T\phi](x+\delta) - [T\phi](x) = \int_0^1 \left[K(x+\delta,y) - K(x,y)\right] \phi(y) \,\mathrm{d}y$$
We can easily bound the magnitude of this by the integral of the largest value the integrand can take,
$$\sup_{y\in Y}| K(x+\delta,y) - K(x,y)| \times \sup_{y\in Y} |\phi(y)| \equiv f(x,\delta)$$
but this is a continuous function of $\delta$, as you should be able to prove, so since $f(x,0)=0$  we are done.

For the next part, use the same sort of bound for the integral as in the previous part, and work out what $C$ is in terms of $\lVert K \rVert$.

Finally, the two main ways to prove this result are (i) using the Arzela-Ascoli theorem; or (ii) by approximating $T$ by finite-rank operators, i.e. replacing $K$ by simple approximating functions. I prefer (i).
The only interesting part of this is equicontinuity. But once more the same inequality saves the day! Note that the $\phi_n$-dependent term factors off, and then the (uniform) boundedness of the $\phi_n$, and (uniform) continuity of $K$ is all we need.
A: To prove part a) note that
$$|T(\phi)(x + h) - T(\phi)(x)| \le \int_0^1|k(x + h, y) - k(x,y)||\phi(y)|\ dy.$$
Being $|k(x+h,y) - k(x,y)|\phi(y)| \le |2||K||_{\infty}||\phi||_{\infty} \in L^1([0,1])$, applying Lebesgue dominated convergence theorem it follows that $T(\phi)$ is continuous.
To prove part b) note that $T$ is linear and that
$$|T(\phi)| \le ||K||||\phi||.$$
To prove part three ypu should know that you can approximate $K$ with polinomial function, that integral operators with polinomial kernels are compact being of finite-rank, and that the uniform limit of compact operators is still compact. Reed ans Simon Functional Analysis is a good reference for this part.
