Let $H$ be a Hilbert space. $A: H \rightarrow H$ is linear and continuous, $x \in H$ and r is a real number s.t. $r>\text{dist}(x,A(H))$. Show that there exists a unique $y_0 \in H$ such that:
- $\|x-A(y_0)\| \leq r$
- $\|y_0\|=\inf\{\|y\|: y\in H, \|x-A(y)\| \leq r\}$.
Can $\inf$ be substituted with $\min$?
My attempt at solution:
Since $A$ is linear it is easy to show that $A(H)$ is convex. We use the theorem of distance to convex sets in Hilbert spaces to show that for $x$ there exists a unique $a \in A(H)$ such that $\|x-a\| = \text{dist}(x,A(H))$. If this is a good point to start, how do I show that $a$ has a unique inverse image using the properties provided in the problem?