compact integral operator and finite dimension of eigenspace

Question

Let $K:f\mapsto Kf$ where $Kf(x) := \int_{0}^1 k(x,t)f(t)dt$ where $f$ is continuous on $[0,1]$ and $k$ is continuous on $[0,1]\times [0,1]$.

We know that $K$ is a compact operator (by Ascoli).

Let $\lambda\neq 0$, how to show that Ker$(K-\lambda Id)$ has finite dimension ?

Answer 1

Suppose that $\ker(K - \lambda I)$ is infinite dimensional. Then (by Riesz's lemma, for instance) we can select a bounded sequence $(f_i) \subset \ker(K - \lambda I)$ with no convergent subsequence. Conclude that $(Kf_i)$ has no convergent subsequence, which means that $K$ is not compact.

Answer 2

Here is a direct proof (of course it is more general to do this with compact operators but it is advantageous to know alternatives, too. Also this is an exercise in Lang's Real and Functional Analysis book in a chapter before compact operators were introduced):
Consider the space of continuous functions on $[0,1]$ together with the $L^2$ inner product. Let $f_1,\dots,f_n$ be an orthonormal family of eigenfunctions of $K$ for the eigenvalue $\lambda$. By Bessel's inequality we have $$\sum_{i=1}^n|\underbrace{\langle f_i,\overline{k(x,-)}\rangle}_{=(Kf_i)(x)=\lambda f_i(x)}|^2\leq \| k(x,-)\|^2$$ for all $x\in[0,1]$ and hence: $$|\lambda|^2\sum_{i=1}^n |f_i(x)|^2\leq \|k(x,-)\|^2$$ Integrating over $[0,1]$ then gives us:

$$n=\sum_{i=1}^n\|f_i\|^2\leq|\lambda|^{-2}\int_0^1\|k(x,-)\|^2dx$$ Hence $n$ is bounded and therefore $\dim\ker(K-\lambda\operatorname{id})<\infty$.