Is $\langle f - m, m \rangle = 0$ sufficient for $m$ to be an orthogonal projection?

Question

I was looking at https://en.wikipedia.org/wiki/Hilbert_projection_theorem and I wondered about the following statement, but I wasn't able to prove or disprove it.

Suppose that $M$ is a subspace of a Hilbert space $\mathcal{H}$ and that $f \in \mathcal{H}$. If $m \in M\setminus 0$ has $\langle f - m, m \rangle = 0$, then is it true that $\langle f - m, m' \rangle = 0$ for all $m' \in M$?

Answer 1

Counterexample:

Consider $\mathcal{H} = \mathbb{R}^{3}$ with the standard inner product $\langle (a,b,c), (d,e,f) \rangle = ad + be + cf$, and $M = \operatorname{span}\{(1,0,0), (0,1,0)\}$. Take $m = (1,0,0)$ and $f = (1,1,1)$. It follows that $$\langle f -m , m \rangle = \langle(0,1,1), (1,0,0) \rangle = 0,$$ but for $m' = (0,1,0) \in M$ we have that $$\langle f - m , m' \rangle = \langle (0,1,1), (0,1,0) \rangle = 1\not= 0. $$

Answer 2

Counterexample: $m=(0,1)\in M=\mathcal H=\Bbb R^2$ (with its standard inner product) and $m'=f-m=(1,0)$.