# Is this converse of Parseval's inequality true?

Let $H$ be a real Hilbert space, $e_n\in H$ an orthonormal sequence of vectors, let $E$ be the closure of the linear span of the $e_n$ and let $x\in H$ be some vector. Now suppose that

$$\sum_{n=0}^\infty \langle x,e_n\rangle^2=\lVert x\rVert^2$$

Does it follow that $x\in E$? I think it does, because if $y$ is the projection of $x$ onto $E$, then $\langle y,e_n\rangle=\langle x,e_n\rangle$ for all $n$ (right?), so, applying Parseval's theorem to the point $y$ in the Hilbert space $E$, we get $\lVert x\rVert=\lVert y\rVert$, but we know from the triangle inequality (rearranged slightly) that $\lVert x-y\rVert-\lVert y\rVert\leq\lVert x\rVert$, so $\lVert x-y\rVert=0$.

• Why $\lVert x-y\rVert-\lVert y\rVert\leq\lVert x\rVert$, implies $\|x-y\|=0$ ? – Tsemo Aristide Oct 22 '16 at 19:54
• @TsemoAristide You're right, there's no reason. I was trying to justify my intuition that if $x$ has the same norm as its projection onto a subspace, then it should be equal to its projection. But maybe that's still true, even if it takes something more complicated to prove it. – Jack M Oct 22 '16 at 20:00
• It is quite easy, it just follows from Pythagoras: $y$ is perpendicular to $x - y$, hence, $\|x\|^2 = \|y\|^2 + \|x - y\|^2$. – gerw Oct 22 '16 at 20:03
• @gerw Perfect. So then, with that amendment, is my proof correct? – Jack M Oct 22 '16 at 20:04

Yes, $x \in E$ follows. However, (as outlined in the comments) your final argument is flawed.
It can be repaired by using Pythagoras: $y$ is perpendicular to $x−y$, hence, $$‖x‖^2=‖y‖^2+‖x−y‖^2.$$ This implies $\|x-y\| = 0$.