Equivalent definition of distance between a point and set.

Question

If $A$ is a closed subspace of a Hilbert space $H$, prove that $$\text{dist}(x,A) = \sup\{|\langle x,y\rangle|: y \in A^\perp, ||y||=1\}.$$ I think the idea here is to show that the "standard" definition where we define the distance as the infimum of norms is both less than and greater than this equation above. Only I am not sure how to proceed. For example, if we consider $y \in A^\perp$ such that $||y||=1$, then we have $|\langle x,y \rangle| \leq ||x||$, taking supremum on both sides tells us that $\text{dist}(x,A) \leq ||x||.$ Of course this would not work, but is there something I am missing like a property about an inner product that could crack this?

Answer 1

You are missing the property that each element $x$ can be uniquely written as a sum $y + z$, $y \in A^\perp$, $z \in A$.