Let H be a Hilbert space and $A$ be a function from $D(A)=H \to R(A) \subset H$ be a bounded linear operator. Let $f \in H$ and consider the following equations:
\begin{equation}Ax=f\end{equation} \begin{equation}Ax=0 \end{equation}
$Z_A$ is the set of all solutions of the equation $Ax=0$. Then $Z_A$ is a vector space.$\\$
a) Let $x_0$ be a solution for $Ax=f$, We need to show that the general solution for the equation $x=x_0+z$, where $z$ is an arbitrary element in $Z_A$.
My attempt:
I can just substitute $x=x_{0}+z$ into $Ax$ to get $A(x_0+z)=Ax_0=f$. But I would like to derive the expression x as a direct sum.
Let the solution set of A be defined as $$S=\{ x | Ax=f \}$$.
Let $y \in \overline{S}$, then there exists $(x_n) \in S$, such that $x_n \to x$.$\forall n$, $Ax_n=f$, since A is continuous, $Ax_n$ converges to $Ax$, and since $Ax_n=f$ for all n, $Ax=f$ as well. Thus, the solution set is closed. A closed set in a Hilbert space is complete. Hence, the solution set is complete. By the direct sum theorem for Hilbert spaces, I would like to decompose $S$ into direct sum.
The $N(A)$ is closed(complete). So I take the projection of every $x \in S$ on $N(A)$. I am stuck here, maybe I can write
$$S=N(A) \oplus N(A)^{\perp}$$ But am unable to proceed. Can someone please help?