[Math] Dense subset of Hilbert space has trivial orthogonal complement

functional-analysishilbert-spaces

If $D$ is a dense subset (not subspace) of a Hilbert space then is the orthogonal complement of $D$ equal to $\{ 0 \}$?

This is true if $D$ is a subspace but if you only know that $D$ is a subset you cannot apply the decomposition theorem. Any ideas?

Best Answer

Fix a vector $v$ in the orthogonal complement. The function $x\mapsto \langle x,v\rangle$ is continuous on the Hilbert space. Since it vanishes on a dense subset, it vanishes identically. In particular, it vanishes at $v$.

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