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Let $H$ be a Hilbert space, $A:H\rightarrow H$ a continuous, linear map. $x\in H$ and $ 0dist(x,A(H))$. Show that there exists exactly one vector $y_0\in H$, such that:

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

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?

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    I think orthogonal projections are the way to go. – 2017-02-18
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    I have one theroem on orthogonal projection: Let $H$-HIlbert, $M=\overline{M}$-subspace of $H$. Then: $1. \ \ h\in H, h_0\in M \implies \Vert h-h_0\Vert=dist(h,M)\iff h-h_0 \perp M \\$ $ 2. \ \ \forall \ h\in H \ \exists ! \ (h_0,h_1)\in H\times H : \ h_0\in M, h_1\perp M, h=h_0+h_1.$ but I don't see even how to start. – 2017-02-18
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    I think the following will help you. Without loss of generality, we can assume that $x= 0$. First, consider the set $ A_0 := A(H)$. This is clearly a non-empty closed convex set. Now assume $0 \notin A_0$, we want to show there exists $y_0 \in A_0$ s.t. $||y_0|| = \inf_{y \in A} ||y||$. To do this, there is a sequence $y_n \in A_0$ s.t. $\lim ||y_n||$ equals the inf. Now use parallelogram theorem to show that $y_n$ is a Cauchy sequence. (Warning, this may be incorrect as I don't have a pen and paper, but I think this is the way to go) – 2017-02-19

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