Fredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$

Question

Let $X$ and $Y$ be Hilbert spaces with respective inner products $\langle , \rangle_{X,Y}$. Let $A:X \rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $(,)$ on $Y$. Take $y \in Y$, and define the map $l_y : x' \in X \mapsto (A x', y)$. Assume this last map is bounded, then by Riesz's theorem, there exists a unique $x \in X$ such that $l_y(x') = \langle x' ,x \rangle_X$. We define the formal adjoint of $A$, denoted as $A^*$ by $A^*(y) = x$. Note that this coincides with the adjoint of $A$ when $(,) = \langle , \rangle_{Y}$.

Is it true that if $AA^*$ is a Fredholm operator then $A$ is Fredholm as well?

A particular case I am interested take $d: L^2_k([0,1]) \rightarrow L^2_{k-1}([0,1])$ (where $L^2_k$ denotes the Sobolev space with $k$ weak derivatives in $L^2$) and take $(,)$ as the $L^2([0,1])$ inner product.

I have asked this question on mathexchange but didn't receive any answers. https://math.stackexchange.com/q/2864150/37986

Answer 1

Consider $H = L^2[0,\infty)$ with $(,) = \langle,\rangle$, $A: H \to H$ the shift operator $A f(t) = f(t+1)$, so that $$ A^* f(t) = \cases{f(t-1) & if $t \ge 1$\cr 0 & otherwise\cr} $$ Then $A$ is not Fredholm, but $A A^* = I$.