Adjoints and orthogonal projections The proof of problem 10A.2 of Axler's measure theory text seemed fairly straightforward to me. But I found something which appears to be a counterexample, but I know must be untrue (I'm misunderstanding some property of the adjoint or inner product).

I can see that $T^* = P_U$, the orthogonal projection of U. This follows because $v - T^*v$ is orthogonal to $u$ for each $v \in V$ and $u \in U$.
But also, for any $v \in V$ and $u \in U$,
$$
\langle u, v \rangle = \langle Tu, v \rangle = \langle u, T^* v \rangle
$$
Because $0 \in U$, we have
$$\langle 0, v \rangle = \langle 0, T^*v \rangle  $$
which implies that $v = T^*v$. But then $v = P_U(v)$ even when $v \notin U$, which is impossible.
Where is my error?


Because $0\in U$...

That's not how these arguments work. When you have that $\langle u,y\rangle=0$ for all $u$ and you conclude that $y$ is zero, you need to have that $y\in u$ and the way to conclude if that you then get $\|y\|^2=0$. But, in general, from $$\langle u,v-T^*v\rangle=0$$ for all $u\in U$, what you prove is that $v-T^*v\in U^\perp$. Nothing else.
