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?
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$\begingroup$ You should be more precise when you state the spectrum condition: after all $L$ is not necessarily invertible, so $0$ may be in the spectrum. $\endgroup$– fedjaCommented May 21, 2019 at 2:48
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$\begingroup$ I guess if the kernel of $L$ is not empty, then $0$ is in the spectrum. Also, the complement of the range of $L$ is inside the kernel of the adjoint $L^*$. Therefore, if the kernel of $L^*$ is finite, then so is the complement of the range of $L$. So the only think left is to show that the range of $L$ is closed. $\endgroup$– KerrCommented May 21, 2019 at 3:54
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$\begingroup$ (I deleted my previous comment since it was, unfortunately, non-sense.) $\endgroup$– Jochen GlueckCommented May 21, 2019 at 16:24
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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.
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1$\begingroup$ To be clear: we always have a canonical isomorphism $(H/\overline{\text{Im}}(L))^* \cong \ker(L^*)$, and the former is $\text{coker}(L)$ if and only if the range of $L$ is closed. In particular, it is not automatic from the assumption that $\ker(L^*)$ is finite-dimensional that the same is true of $\text{coker}(L)$. $\endgroup$– mmeCommented May 21, 2019 at 12:40
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1$\begingroup$ Thanks for adding this comment, which shows that my comment below the question was actually non-sense (so I deleted it). $\endgroup$ Commented May 21, 2019 at 16:27
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1$\begingroup$ @JochenGlueck no problem, I made the same mistake when writing this down... the test case that made things clear to me was $-i \cdot \int: L^2(S^1) \to L^2(S^1)$, equivalent to multiplication by $(\cdots, 1/n, \cdots)_{n \in \Bbb Z})$ on $\ell^2(\Bbb Z)$ by Fourier transform; this is self-adjoint and hence $\text{ker}(L^*) = \text{ker}(L) = 0$ - but of course the image is dense but not closed (it lies in the Sobolev space $H^1(S^1)$, whose norm transferred to a subspace of $\ell^2(\Bbb Z)$ is $\|(a_n)_{n \in \Bbb Z}\|_{H^1} = \sum_{n \in \Bbb Z} |na_n|^2$. $\endgroup$– mmeCommented May 21, 2019 at 16:57
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$\begingroup$ I think one needs to assume normality for this answer to be true. Consider for example the operator $(a_1, a_2, a_3, a_4, a_5, a_6, \dots ) \mapsto (0, a_1, 0, \frac 1 3 a_3, 0, \frac 1 5 a_5, \dots)$ on $\ell^2(\mathbb N)$, which is bounded, has spectrum $\{0\}$, but does not have closed image. The 'bounded below' condition holds for this operator $L$, but not for $L^* L$. $\endgroup$– PhilippCommented May 10, 2022 at 16:25