Why do we need the extra condition of being 'Fredholm of index zero' when showing that an operator has a bounded inverse?

Question

In finite dimensional linear algebra if we have $A x = 0$, and $\ker(A) = 0$, then $A$ is invertible. But it seems having a trivial kernel is not a sufficient condition for invertibility when dealing with infinite dimensional operators. Consider the following theorem:

Theorem

Let $S_D: L^2(\partial D) \to W_1^2(\partial D)$ be the single layer potential for the Laplacian in $\mathbb{R}^3$ where $D$ is a bounded Lipschitz domain in $\mathbb{R}^3$. Let $\phi \in L^2(\partial D)$ satisfy $$S_D[\phi] = 0 \quad \text{on} \quad \partial D.$$ It can be shown then that $\phi = 0$. Now we want to prove that $S_D$ has a bounded inverse.

Proof

As $W_1^2(\partial D) \hookrightarrow{} L^2(\partial D)$ is compact, and $S_D$ maps $L^2(\partial D)$ into $W_1^2(\partial D)$ boundedly, we have that $S_D$ is Fredholm with index zero. And as we have that $\ker(S_D) = \{0\}$, this means that $S_D$ has a bounded inverse.

Question

Unlike in a finite dimensional matrix problem, in this case having a trivial kernel $\ker(S_D) = \{0\}$ was not enough to show that $S_D$ is invertible. We also needed to show that $S_D$ is Fredholm with index zero. Why is this? I am not very experienced with Fredholm theory. I know that being Fredholm of index zero means that an operator has

• finite dimensional kernel
• finite dimensional cokernel
• closed range
• $\dim(\ker) - \dim(\text{coker}) = 0$ (for the index zero property)

So why do we need these extra properties when dealing with the invertibility of an infinite dimensional operator as opposed to a finite dimensional matrix?

Answer 1

Consider the right-shift in $\ell^2$, i.e., the operator $R$ which maps $$(x_1, x_2, x_3, x_4, \ldots) \mapsto (0, x_1, x_2, x_3).$$ Then, it is easy to see that this operator is Fredholm, but has index $\pm 1$ (I am not sure about the sign in the definition of the index).

The left-shift has similar properties: it has finite-dimensional kernel and co-kernel, closed range and is surjective. In finite dimensions, this is enough to guarantee the invertibility.

Answer 2

In finite dimensional vector space we have the rank-nullity theorem that ensures that it is enough to require one of two:

• $A$ is ono-to-one ($\ker A=\{0\}$) or
• $A$ is onto ($\text{Im}(A)=Y$)

to ensure that $A\colon X\to Y$ is invertible.

In Banach spaces the situation is a bit more complicated, but we still have Banach theorem about bounded inverse that says that it is enough to require that $A$ is both ono-to-one and onto for invertibility. The theorem basically claims that if $A$ is a bijection, i.e. there exists algebraically inverse, then this algebraically inverse must be continuous (bounded).

In the theorem above we have Hilbert spaces and we know that $\ker(S_D)=\{0\}$. For invertibility we need only to check that the image of $S_D$ is the whole space $W_1^2(\partial D)$. To do it one may use Fredholmness. If we can prove that $S_D$ is Fredholm with index zero then we are done, because it means that $\dim\ker=\dim\text{coker}$, which together with $\dim\ker=0$ gives $\dim\text{coker}=0.$ The cokernel is the quotient space $W_1^2(\partial D)/\text{Im}(S_D)$. Its dimension is zero together with $\text{Im}(S_D)$ being closed susbace (Fredholm) gives the necessary $\text{Im}(S_D)=W_1^2(\partial D)$, i.e. onto.

P.S. I have no idea how to conclude in a simple way that $S_D$ is Fredholm with the index zero from $S_D$ being bounded and the compact embedding.