Theorem 1.1 of The mathematics of computerized tomography In the book "The mathematics of computerized tomography" by Natterer comes the theorem:
Theorem 1.1 $f=A^+g$ is the unique solution of $A^*Af=A^*g$ in $range(A^*)$.
where $A:H\rightarrow K$ is a linear bounded operator, $H$ and $K$ are Hilbert spaces, $g\in range(A)\oplus range(A)^\perp$ and $f=A^+g$ is the minimizer of $\| Af-g\|_K$. Here is supposed that the adjoint exists.
Now I copy exactly the proof in the book:
$f$ minimizes $\|Af-g\|$ if and only if $(Af-g,Au)=0$ for all $u\in H$, i.e., if and only if $A^*Af=A^*g$. Among all solutions of this equation the unique element with least norm is the one in $(ker(A))^\perp=\overline{range(A^*)}$.
My concern: I haven't been able to prove that this solution $f\in \ker(A)^\perp$ and is unique. I know $g=Af+v$ where $v\in range(A)^\perp=\ker(A^*)$ but I cant find a way to relate it with $(f,u)=0$ for all $u\in\ker(A)$ because I dont know how to introduce $A$ in this equation. Any suggestion is valuable.
 A: If $\tilde{f}$ is a fixed minimizer of $\|Af-g\|_K$ then it also minimizes $$Q(f)=\|Af-g\|_K^2.$$
But, if $h\in H$, then $$Q(f+h)=Q(f)+\langle Af-g,Ah\rangle_K+\langle Ah,Af-g\rangle_K+\|Ah\|^2_K\,\quad \tag{1}.$$ In particular $$Q(f+h)=Q(f),\qquad f\in H, \quad h\in \ker(A)\tag{2}.$$
This and $(1)$ means that $$Q(\tilde{f})\leq Q(\tilde{f}+h)=Q(\tilde{f})+2\text{Re}\langle A^*(A\tilde{f}-g),h\rangle_H+\|Ah\|_K^2,\qquad h\in H,$$ and follows that $$A^*(A\tilde{f}-g)=0\tag{3}.$$
Expression $(3)$ tell us that $(A\tilde{f}-g)\in \ker(A^*)=range(A)^\perp$.
Now, we can write $g=g_1+g_2$, with $g_1\in \ker(A^*)^{\perp}=\overline{range(A)}$ and $g_2\in \ker(A^*)$ in a uninique form. Therefore $g_1=A\tilde{f}\in range(A)$, and $g_2=g-A\tilde{f}$.
This means that, if $k\in H$ is another minimizer of $Q(f)$, then $Ak=A\tilde{f}=g_1$ and $g-Ak=g-A\tilde{f}=g_2$. It follows that $(\tilde{f}-k)\in ker(A)$, or $k=\tilde{f}+h$, $h\in ker(A)$.
In particular, $\tilde{f}=f_1+f_2$, with $f_1\in \ker(A)^{\perp}=\overline{range(A^*)}$  and $f_2\in \ker(A)$ in a uninique form too.
It follows from $(2)$ that each menimizer $k$ of $Q(f)$ has the form $k=f_1+(f_2+h)$, $f_1\in \ker(A)^{\perp}=\overline{range(A^*)}$  and $(f_2+h)\in \ker(A)$.
We can use the notation $f_1=A^+g=A^+g_1$, to denote the unique element of $\overline{range(A^*)}$ which is a minimizer of $Q(f)$.
Note:

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*You can find the definition to $A^+$, as $$A^+(u)=0,\quad u\in range(A.)^{\perp}$$ and $$A^+(u)=\left[A_{\vert \ker(A)^{\perp}}\right]^{-1}u,\quad u\in range(A),$$ with $$A_{\vert \ker(A)^{\perp}}:\ker(A)^{\perp}\to range(A),$$ in Section 3.1 and Definition 3.10 of the text Moore-Penrose inverse of some linear maps on infinite-dimensional vector spaces, for instance.


*You can find related results searching for "\(f=A^+g\)" on SearchOnMath.
