Q: existence of a unique vector in a Hilbert space Let $H$ be a Hilbert space, $A:H\rightarrow H$ a continuous, linear map. $x\in H$ and  $ 0<r\in\mathbb{R}$,  such that $r>dist(x,A(H))$. Show that there exists exactly one vector $y_0\in H$, such that:


*

*$\Vert x-A(y_0)\Vert\le r$

*$\Vert y_0\Vert=\inf\left\{\Vert y\Vert: \ y\in H, \ \Vert x-A(y)\Vert\le r\right\}$


Can we substitute $\inf$ by the $\min$?
I tried to use some strong tools like Hahn-Banach or Banach-Steinhaus but I really can't see how to apply them? Or maybe this problem goes in the other way?
Any hint?
