Ker(I-L) Finite-Dimensional: Proof

[Fermat1]

Let L be an compact operator on a compact space K , and Let I be the identity on K.
Show that Ker(I-L) is finite-dimensional.

My attempt: Let $x_{n}$ be a sequence in the unit ball. K is compact so $(I(x_{n}))=(x_{n})$ has a convergent subsequence and L is compact operator so $L(x_{n})$ has a convergent subsequence. Thus, $(I-L)(x_{n})$ has a convergent subsequence so it is compact. $I-L$ restricted to the kernel is the identity, so since the identity is compact iff the space is finite dimensional, the kernel is finite dimensional. Is this correct?

 

[jvicens]

Yes, your approach is correct. Here is a more detailed explanation:

Since $K$ is compact, it is also bounded. This means that for any sequence $(x_n)$ in $K$, there exists a convergent subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} x_{n_k}=x\in K$.

Now, since $L$ is a compact operator, it maps any bounded sequence in $K$ to a sequence with a convergent subsequence. So, $(L(x_{n_k}))$ has a convergent subsequence, say $(L(x_{n_{k_j}}))$. Let $\lim_{j\to\infty} L(x_{n_{k_j}})=y\in K$.

Since $I$ is the identity operator, we have $I(x_{n_{k_j}})=x_{n_{k_j}}$ for all $j$. So, $(I(x_{n_{k_j}}))$ is a subsequence of $(x_{n_k})$ and hence, it also converges to $x$.

Now, let $x_{n_{k_j}}\in\ker(I-L)$. Then, $(I-L)(x_{n_{k_j}})=I(x_{n_{k_j}})-L(x_{n_{k_j}})=x_{n_{k_j}}-L(x_{n_{k_j}})$. Since $(x_{n_{k_j}})$ and $(L(x_{n_{k_j}}))$ both converge to $x$ and $y$, respectively, we have $\lim_{j\to\infty} (x_{n_{k_j}}-L(x_{n_{k_j}}))=x-y$.

Since $K$ is compact, it is also closed. So, $x-y\in K$. But $x-y$ is a limit point of the sequence $(x_{n_{k_j}}-L(x_{n_{k_j}}))$ which is contained in $\ker(I-L)$. So, $x-y\in\ker(I-L)$. This shows that every limit point of $\ker(I-L)$ is also contained in $\ker(I-L)$. Hence, $\ker(I-L)$ is closed.

Now, since $\ker(I-L)$ is a closed subset of the compact set $K$, it