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>



Fredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$
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 $A A^\* = I$.


