Let $H$ be a Hilbert space and let $X \leq H$ where $X$ is a closed subspace of $H$. Suppose that there is another (closed?) subspace $Y$ with $X \oplus Y = H$ and $X \perp Y$. Is it true that $Y = X^\perp?$
Attempt:
Let $y \in Y$. Then $\langle y,x \rangle = 0$ for all $x \in X$, so $y \in X^\perp$. Hence $Y \subseteq X^\perp$.
Conversely, assume $y \in X^\perp$. Then $\langle y,x \rangle =0$ for all $x \in X$. Write $y = x_1 + y_1$ with $x_1 \in X, y_1 \in Y$. We wish to show that $x_1 = 0$. Then I'm stuck.
Is the statement even true?
Best Answer
Let $y \in X^{\perp}$. We can write $y=u+v$ with $u \in X$ and $v \in Y$. Note that $v \in X^{\perp}$. Now $u=y-v \in X^{\perp}$. In particular $u \perp u$ so $u=0$. Hence $y=v \in Y$.