A problem connecting Hilbert spaces and normed linear spaces The problem is the following:
Let $X$ be a normed linear space and $Y$ be a Hilbert space. 
Let $ A : X \rightarrow Y$ be a linear operator.
For $y \in Y \, $, let $ \, S_y = \{ x \in X : \| Ax − y \| \le \|Au − y \|, \, \forall \, u \in X \} $ 
Show that $S_y$ is nonempty if and only if $ \, y \in R(A) + R(A)^{\bot}$ .
In addition, if $X$ is Hilbert and $N(A)$ is a closed subspace, show that for every $y \in R(A) + R(A)^{\bot}$,
there is a unique $x_y \in S_y$ such that
$ \| x_y \| = inf \{ \| x \|: x \in S_y \}$.

*

*My attempt:

I guess that if I prove that $S_y$ is a convex subset of $Y$, and as $Y$ is Hilbert, I can say that $ \exists ! \, x_y \in S_y $, as required in the second part.
For the first part. I'm not quite sure how to go about it.
 A: Assume that $y \in R(A)\dotplus R(A)^\perp$, so it can be written as $y = Ax+z$ for some $x \in X$ and $z \in R(A)^\perp$.
Now for any $u \in X$ we have
$$\|Ax-y\|^2 = \|-z\|^2 \le \|Au\|^2+\|-z\|^2 = \|Au-z\|^2$$
since $Au \perp -z$. Therefore $x \in S_y$ so $S_y$ is nonempty.
Conversely, assume $y \notin R(A)\dotplus R(A)^\perp$. Then since $\overline{R(A)}\dotplus R(A)^\perp = Y$, it has to be $y\in (\overline{R(A)}\setminus R(A))\dotplus R(A)^\perp$. We can therefore write $y = a+b$ with $a \in \overline{R(A)}\setminus R(A)$ and $b \in R(A)^\perp$. Pick a sequence $(x_n)_n$ in $X$ such that $Ax_n \to a$.
Assume that $x \in S_y$. Since $Ax \ne a$, we have $\|Ax-a\|>0$ so there exists $n\in\Bbb{N}$ such that $\|Ax_n-a\|<\|Ax-a\|$. Therefore
$$\|Ax_n - y\|^2 = \|Ax_n - a - b\|^2 = \|Ax_n-a\|^2+\|-b\|^2 \\
\le \|Ax-a\|^2+\|-b\|^2 = \|Ax - a - b\|^2 = \|Ax - y\|^2.$$
since $Ax_n-a, Ax-a \perp b$. This is a contradiction. Therefore $S_y = \emptyset$.
For the second part, we need to show that $S_y$ is a closed, convex subset of $Y$. Assume $(x_n)_n$ is a sequence in $S_y$ such that $x_n \to x_0 \in X$.
If we write $y = Ax+z$ with $x \in X$ and $z \in R(A)^\perp$, then (as @Berci suggested) for every $v \in S_y$ it holds $Av = Ax$. Indeed,
$$\|Av-Ax\|^2+\|-z\|^2 = \|Av-y\|^2 \le \|Ax-y\|^2 = \|-z\|^2$$
since $Av-Ax \perp -z$ and therefore $\|Av-Ax\| =0$.
Now we also have $Ax_n = Ax$ and hence $(x_n-x)_n$ is a sequence in $N(A)$. Since $N(A)$ is closed, we get that the limit $x_0-x$ is also in $N(A)$. Therefore $Ax_0 = Ax$. The first part of the proof showed $x \in S_y$ so for every $u \in X$ we have
$$\|Ax_0-y\| =\|Ax-y\| \le \|Au-y\|$$
and hence $x_0 \in S_y$.
For convexity let $x,x' \in S_y$ and let $\alpha, \beta \ge 0$ such that $\alpha+\beta = 1$. For any $u \in X$ we have
\begin{align}
\|A(\alpha x+\beta x') -y\| &= \|\alpha(Ax-y)+\beta(Ax'-y)\| \\
&\le \alpha \|Ax-y\| + \beta\|Ax'-y\| \\
&\le \alpha\|Au-y\|+\beta\|Au-y\| \\
&= \|Au-y\|
\end{align}
so $\alpha x + \beta x' \in S_y$. Hence $S_y$ is convex. Now use the fact that every closed, convex, nonempty subset of a Hilbert space has a unique element of minimal norm.
