Two versions of the Fredholm alternatives Here are two version of the Fredholm alternatives among which I would like to know the relation:
One version is from the appendix of Evans's Partial Differential Equations:


  Here $H$ is a Hilbert space. 

Another version is from an old post in Terry Tao's blog:



Having gone through the details of each of the theorems above, I have the following questions:   


*

*Can one get Theorem 5 from Theorem 1 by making $\lambda=1$?

*Are these two versions of the Fredholm alternatives equivalent when
$X$ is assumed to be a Hilbert space in Theorem 1?

 A: Question 1:
Yes, you can.  Suppose that $T$ is a compact linear operator, and assume that the consequences in 5 all hold.  Then, suppose that $\lambda \neq 0$ fails to be an eigenvalue (we want to show that $T - \lambda$ has a bounded inverse).  
It would follow that the eigenspace $N(I - \frac 1{\lambda}T) = N(\lambda - T) = \{0\}$.  By (iv), we have 
$$
R(I - \frac 1{\lambda}T) = R(\lambda - T) = R(T - \lambda) = H
$$
Thus, the map $T - \lambda$ is bounded (since it is compact), is injective (since its kernel, the eigenspace, is $\{0\}$), and is surjective.  By the bounded inverse theorem, $(T - \lambda)^{-1}$ exists and is bounded.

Question 2:
It seems that Evans' version contains a bit more information about compact operators.  For example, nothing about Tao's version says anything about finite dimensional kernels.  That being said, let's assume $X = H$ is a Hilbert space, and try to go the other direction, filling in the blanks as we go.
Suppose that $K$ is a compact operator which has $\lambda = 1$ as an eigenvalue.  If $N(I - K)$ where infinite dimensional, then $K \mid_{N(I - K)}$ would be the identity operator on an infinite dimensional space, which fails to be compact.  So, $N(I - K)$ must be finite dimensional.
From here, we could presumably prove that $R(I - K) = N(I - K)^{\perp}$, from which we would get (ii) for free.  I'll skip that, but I expect that it's in Evans.  We would also need to prove (v) without help from Tao.
Because $R(I - K)^\perp = N(I- K)^{\perp \perp} = N(I - K) \neq \{0\}$, we can conclude that $R(I - K) \neq H$.
On the other hand, suppose that $\lambda = 1$ is not an eigenvalue.  Because $(T - I)^{-1}$ exists, we can conclude that $T-I$ is onto.  Thus, the required conditions for (i) (ii) (iii) (iv) and (v) hold trivially.  So, Tao's version certainly helps with one direction.
