A spectral description of Fredholm operators Let $L:H \to H$ be a bounded operator on a Hilbert space $H$, with finite dimensional kernel, and whose adjoint also has finite dimensional kernel.  Is it true that $L$ is Fredholm if and only if its spectrum is norm bounded below by a non-zero constant?
 A: I assume when you say 'the spectrum is bounded below' you mean that there exists $c>0$ so that no $\lambda$ with $0 < |\lambda| \leq c$ is in the spectrum. In fact, for any bounded operator $L$ on a Hilbert space, this condition is equivalent to the demand that $L$ has closed range. 
Each of these conditions for $L$ are equivalent to the same condition for $L^*L$, so we may as well assume $L$ is self-adjoint. Second, both conditions are preserved by the addition of a map $\epsilon p: H \to H$ which factors as an orthogonal projection $p: H \to \text{ker}(L) \hookrightarrow H$ (the first condition is preserved so long as $\epsilon$ is taken sufficiently small, though of course the constant $c$ will change). So we may as well assume that $L$ is injective.

From here on $L$ is an injective self-adjoint operator.
If $L$ has closed range, then the identification $\text{coker}(L) \cong \text{ker}(L^*) \cong \text{ker}(L) = 0$ and the closed graph theorem imply $L$ is an isomorphism; because $\|L^{-1}v\| \leq C\|v\|$, we see that $\|Lv\| \geq \frac 1C \|v\|,$ and in particular $|\text{Spec}(L)| \geq 1/C$.
Conversely, suppose $L$ has no eigenvalues near zero, and suppose $Lv_n \to w \in \overline{\text{Im}(L)}$. Then because $\frac 1C \|v_n\| \leq \|Lv_n\|$, we see that $$\limsup \frac 1C \|v_n\| \leq \|w\|,$$ and in particular $\|v_n\|$ is bounded, so $v_n$ converges in the weak* topology to some $v'$; correspondingly, we see that $Lv_n \to Lv'$ in the weak* topology, and hence $w = Lv'$. Thus the range of $L$ is closed, as desired. 

Adding on the conditions about the kernel of $L$ and its adjoint you get your result. I'm sure you can get the same answer for Banach spaces, but the argument would need to be different.
