Fredholm Alternative as seen in PDEs, part 1

Given my limited background in theoretical linear algebra and my lack of background of functional analysis, it is not easy for me to understand Fredholm's alternative. Yet, I would like to understand the theorem and its proof, in connection to the concept of weak solutions to certain PDEs. Therefore, I am reprinting part of the proof here, and I have (rather basic) questions to follow that I hope anyone can clarify for me.

(PDE Evans, Appendix D, Theorem 5)

THEOREM 5 (Fredholm Alternative). Let $K : H \to H$ be a compact linear operator. Then

(i) $N(I-K)$ is finite dimensional,

(ii) $R(I-K)$ is closed,

(iii) $R(I-K)=N(I-K^*)^\perp$,

(iv) $N(I-K)=\{0\}$ if and only if $R(I-k)=H$,

and

(v) $\dim N(I-K)=\dim N(I-K^*)$.

The first excerpt of the proof is as follows:

Proof. 1. If $\dim N(I-K)=+\infty$, we can find an infinite orthonormal set $\{u_k\}_{k=1}^\infty \subset N(I-K)$. Then $$Ku_k=u_k \quad (k=1,\ldots).$$ Now $\|u_k-u_l\|^2=\|u_k\|^2-2(u_k,u_l)+\|u_l\|^2=2$ if $k \not=l$, and so $\|Ku_k-Ku_l\|=\sqrt{2}$ for $k\not=l$. This however contradicts the compactness of $K$, as $\{Ku_k\}_{k=1}^\infty$ would then contain no convergent subsequence. Assertion (i) is proved.

a. Is it true that $\{u_k\}_{k=1}^\infty$ is orthonormal, it follows that, for $k\not=l$, $(u_k,u_l)=0$ and $\|u_k\|=\|u_l\|=1$?

b. Is $K$ compact because $K$ is considered to be a bounded linear operator? (This is not said explicitly, but I think this is so, based on that $Ku_k=u_k$.)

c. Does $\|Ku_k-Ku_l\|=\sqrt{2}$ imply that $K$ is not compact? In general, is $K$ only compact if $\|Ku_k-Ku_l\|=0$ and not any other number?

2. We next claim there exists a constant $\gamma > 0$ such that $$\|u-Ku\|\ge \gamma \|u\| \quad \text{for all }u\in N(I-K)^\perp. \tag{4}$$ Indeed, if not, there would exist for $k=1,\ldots$ elements $u_k \in N(I-K)^\perp$ with $\|u_k\|=1$ and $\|u_k-Ku_k\|<\frac 1k$. Consequently, $$u_k-Ku_k \to 0. \tag{5}$$ But since $\{u_k\}_{k=1}^\infty$ is bounded, there exists a weakly convergent subsequence $u_{k_j} \rightharpoonup u$. By compactness $Ku_{k_j} \to Ku$, and then (5) implies $u_{k_j} \to u$. We therefore have $u \in N(I-K)$ and so $$(u_{k_j},u)=0 \quad (j=1,\ldots).$$ Let $k_j \to \infty$ to derive a contradiction to (4).

d. Do we know that $\{u_k\|_{k=1}^\infty$ is bounded because we established earlier that $N(I-K)$ is finite dimensional? I'm also assuming there is a proof in linear algebra for this statement.

e. Where did the $\frac 1k$ come from in $\|u_k-Ku_k\|<\frac 1k$? Did they just pick something that converges to $0$? Can we say, for example, $\|u_k-Ku_k\|<\frac 3{k^2}$ and the proof would still work?

f. How come $\{u_k\}_{k=1}^\infty$ bounded implies there is only a weakly convergent subsequence, rather than a convergent subsequence?

g. Finally, when we let $k_j \to \infty$, then $u_{k_j} \to u$. So we obtain $(u,u)=0$, which I think means $\|u\|=0$. But currently I can't see how this contradicts (4), as in how it contradicts $\|u-Ku\|\ge \gamma \|u\|$.

3. Next let $\{v_k\}_{k=1}^\infty \subset R(I-K)$, $v_k \to v$. We can find $u_k \in N(I-K)^\perp$ solving $u_k - Ku_k=v_k$. Using (4) we deduce $$\|v_k-v_l\| \ge \gamma \|u_k-u_l\|.$$ Thus $u_k\to u$ and $u-Ku=v$. This proves (ii).

Follow-up part 2...

Solution 1:

a) orthonormal means $(u_l,u_k)=0$ if $l\ne k$ and $\|u_k\|=1$.

b) In this context, usually a compact operator is defined to be a linear and bounded operator.

c) An orthonormal sequence weakly converges to zero. Thus, $(Ku_k)$ contains a strongly converging subsequence, as $K$ is compact. Since $\|Ku_k - ku_l\|=\sqrt2$ for $k\ne l$, there is no such subsequence. A contradiction.

d) Here, the sequence $(u_k)$ is a completely (different from that in 1.) new sequence in $N(I-K)^\perp$ constructed such that $\|u_k\|=1$, hence it is bounded.

e) The expression $1/k$ is arbitrary. You can take any sequence $a_k$ with $a_k\to0$.

f) This is infinite-dimensional stuff: Bounded and closed sets are not compact in general, but only weakly compact.

g) We know already $u_{k_j}\to u$, $\|u\|=0$. Since by construction $\|u_{k_j}\|=1$ this is a contradiction.