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We have a Hilbert space $H$ and $x_0\in H$. We need to show that if we let $V$ be a closed subspace of $H$, then $$ \min\{\|x-x_0\|\,:\,x\in V\}=\max\{|\langle y,x_0\rangle|\,:\,y\in V^\perp,\,\|y\|=1\}. $$

I have a good intuitive idea of why this is true, but I am having a hard time deciding upon a proof strategy. Things I have been mulling over:

1.) Since $V$ is a closed subspace, then it is convex and it is well-known for a closed convex subset, we can always find $x^*\in V$ such that $\|x^*-x_0\|=\min\{\|x-x_0\|\,:\,x\in V\}$ and that $x^*-x_0\in V^\perp$.

2.) It is also known that the inner product $\langle y,x_0 \rangle$ is maximized when the unit vector $y$ and the arbitrary vector $x_0$ "pointing" in the same direction.

3.) I am pretty sure this all ties together considering the projection of $x_0$ into $V^\perp$ as $x^*-x_0$.

I keep chasing my tail and cannot tease a solution from this. Can someone help point me in the right direction?

Laars Helenius
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It is like you are saying. If you write $P_V$ for the orthogonal projection onto $V$, then $$ \min\{\|x-x_0\|:\ x\in V\}=\|P_Vx_0-x_0\|=\|(I-P_V)x_0\|=\|P_{V^\perp}x_0\|=\max\{|\langle P_{V^\perp}x_0,y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\} =\max\{|\langle x_0,y\rangle|:\ y\in V^\perp,\ \|y\|=1\} $$

Martin Argerami
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