Let $T: X \to Y$ be a bounded linear map between Hilbert spaces $(X, \langle \cdot , \cdot \rangle_X)$ and $(Y, \langle \cdot , \cdot \rangle_Y)$ (the Hilbert spaces may be complex or just real spaces). Suppose that $y$ is in the range $\mathcal{R}(T)$ of our operator, so that $T^{-1}(y)$ is a nonempty, closed subset of $X$.
Let $ \alpha = \inf \{\|x\|_X : x\in T^{-1}(y) \} \ge 0.$
I would like to show that there exists a unique $x_0 \in T^{-1}(y) $ such that $\|x_0\| = \alpha$.
My attempt so far:
We may choose a sequence $\{x_n\} \subseteq T^{-1}(y)$ so that $\|x_n \| \to \alpha $. This sequence is bounded in norm, hence it has subsequence $\{x_{n_k}\}$ that is weakly convergent to some $x' \in X$. My hope is that I can show $x'$ is the desired element $x_0$ that I seek. But I have gotten stuck.
One other observation I have made is that, while $T^{-1}(y)$ is not necessarily a subspace of $X$, it is a convex subset of $X$. That is, for $x_1, x_2 \in T^{-1}(y)$ and $\lambda \in (0,1)$, we have $T(\lambda x_1 + (1 -\lambda) x_2) = \lambda T( x_1) + (1 -\lambda) T( x_2) = y.$
Hints or solutions are greatly appreciated.
You can see it directly using the orthogonal decomposition $$ X = \mathcal{N}(T)\oplus\mathcal{N}(T)^{\perp}. $$ If $Tx=y$, then $x=x'+x''$ for unique $x'\in\mathcal{N}(T)$, $x''\in\mathcal{N}(T)^{\perp}$. And $Tx=Tx''$. If $Tw=y$ also holds then $w''=x''$ because $w''-x''\in\mathcal{N}(T)^{\perp}\bigcap\mathcal{N}(T)$. So, with a small abuse of notation, $$ T^{-1}(y) = x''\oplus\mathcal{N}(T), $$ from which it follows that $x''$ is the unique element of $T^{-1}(y)$ of smallest norm.