Fredholm Alternative as seen in PDEs, part 2 ...continued from part 1.
I have more questions concerning more of Evans' proof in the Fredholm Alternative. As stated in my previous question, I do not have functional analysis background, but I want to still understand the proof. So my five questions will be fundamental and basic. Any answers and clarifications are welcomed.
NB: For anyone who is about to suggest me to read suggested books or point out that I have a lack of background in functional analysis (and even theoretical linear algebra), please don't. Because I already know that. As suggested by a user on my previous question, I plan to read the Haim Brezis textbook over the summer when I don't have classes in the regular school semesters.
With that said, I am printing the theorem below, along with the second excerpt of the textbook's proof and my basic questions to follow.
(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 second excerpt of the textbook proof:



*Assertion (iii) is now a consequence of (ii) and the general fact that $$\overline{R(A)}=N(A^*)^\perp \text{ for each bounded linear operator }A: H \to H.$$

a. Can the overline in $\overline{R(A)}$ be explained? Why do we not need the overline for statement (iii), $R(I-K)=N(I-K^*)^\perp$?



*To verify (iv), let us suppose to start with that $N(I-K)=\{0\}$, but $H_1=(I-K)(H) \subsetneq H$. According to (ii), $H_1$ is a closed subspace of $H$. Furthermore $H_2 \equiv (I-K)(H_1) \subsetneq H_1$, since $I-K$ is one-to-one. Similarly if we write $H_k \equiv (I-K)^k(H)$ ($k=1,\ldots$), we see that $H_k$ is a closed subspace of $H$, $H_{k+1} \subsetneq H_k$ ($k=1,\ldots$). Conversely, choose $u_k \in H_k$ with $\|u_k\|=1$, $u_k \in H_{k+1}^\perp$. Then $Ku_k-Ku_l=-(u_k-Ku_k)+(u_l-Ku_l)+(u_k-u_l)$. Now if $k > l$, $H_{k+1} \subsetneq H_k \subseteq H_{l+1} \subsetneq H_l$. Thus $u_k-Ku_k$, $u_l-Ku_l$, $u_k \in H_{l+1}$. Since $u_l \in H_{l+1}^\perp$, $\|u_l\|=1$, we deduce $\|Ku_k-Ku_l\| \ge 1$ ($k,l=1,\ldots$). But this is impossible since $K$ is compact.


b. I know that $I$ and $K$ considered linear transformations because $I$ is the identity operator (is this true?) and $K$ is given to be a compact linear operator. Hence, $I-K$ is a linear transformation, right? Due to a theorem from linear algebra, is it because $I-K$ established to be one-to-one because the nullspace $N(I-K)=\{0\}$ as initially given?
c. Why does the one-to-one property $I-K$ implies that we can write $H_2 \equiv (I-K)(h_1) \subsetneq H_1$, which would allow us to write after $k$ iterations $H_k \equiv (I-K)^k(H)$?
d. How does $u_l \in H_{l+1}^\perp$ and $\|u_l\|=1$ suggest $\|Ku_k - Ku_l\| \ge 1$, which would make $\{Ku_k\}$ not Cauchy and contradict the compactness of $K$? Ultimately, I cannot right now see the connection that allows us to conclude that the rank $R(I-K)=H$, thereby proving the forward direction of (iv)...



*Now conversely assume $R(I-K)=H$. Then owing to (iii), we see that $N(I-K^*)=\{0\}$. Since $K^*$ is compact, we may utilize step 5 to conclude $R(I-K^*)=H$. But then $N(I-K)=R(I-K^*)^\perp=\{0\}$. This conclusion and step 5 complete the proof of assertion (iv).


e. Given $R(I-K)=H$, because of (iii) we have $N(I-K^*)^\perp=H$. But how does the transpose of $N$ equivalency with $H$ imply $N(I-K)=\{0\}$?
Follow-up part 3...
 A: I remember I got stuck on the same questions and hoped that if I read it carefully, it would become cristal clear...
daw already answered a) and b) but I want add one more comment about b): (iv) means that being injective is equivalent to being surjective (for a compact operator), and he says that this is true for finite dimensional maps.


*

*finite dimensional linear maps are actually completely determined by the images of each of the basis vectors (in finite numbers). so we see that this statement is almost like the equivalence between injective and surjective for a maps on finite sets.

*one definition of "infinite" (by either cantor or dedekind...) says that such a set is characterized by the existence of a bijection with a proper subset of itself. Hence we can imagine that being injective will not imply bijective in general.


Now the other questions:
c. Let $x\in (H_1)^{\perp}$ which is not trivial by assumption. Its image $(I-K)(x)$ is in $H_1$ but if $(I-K)H_1 = H_1$ then any element $b\in H_1$ can be written $(I-K)(a)$ for  $a\in H_1$. Now we have two pre-images of $b:= (I-K)(x)$,  namely $x$ and $a$ which belong to orthogonal subspaces (i.e. we can be sure they are different). This contradicts injectivity of $(I-K)$
d. the equality is of the form $K u_k +K u_l = (something) - u_l$. The something belongs to $H_{l+1}$ while $u_l \in H_{l+1}^{\perp}$ and has norm 1. Use Pythagoras' theorem.
The general structure of the proof of (iv, $\Rightarrow$) is absurd. Assume the negation of the conclusion, then there is a contradiction to $K$ compact.
e. this is definiteness of the norm: let $x\in N(I-K)$ then for all $y\in H$
$$ \langle x,y\rangle= 0$$
Take $y:= x$...
A: a) The range of a linear operator is not closed in general, but the annihilator of a subspace is. In the equation $\overline{ R(A)} =N(I-K^*)^\perp$ the space on the right-hand side is always closed. (ii) proves that $R(I-K)$ is closed, hence we do not need to take the closure in the statement of (iii)
b) $I,K,I-K$ are linear operators. However, the statement: $N(I-K)=\{0\}$  implies $R(I-K)=H$ is not true in infinite-dimensional spaces. Hence the lengthy argument.
