Orthogonal subspace on a Hilbert space

Question

Let $X$ be a Hilbert space and $\varphi \in X' \setminus \left\{0\right\}$. If $$C=\left\{x \in X: \varphi(x)=1\right\},$$ how can I find $C^{\perp}$?

Answer 1

Think about it geometrically. $C$ is a hyperplane that doesn't pass through the origin, so $C^\perp$ probably should be $\lbrace 0 \rbrace$. There are a few ways to think about this, but I like to think about $\operatorname{span} C$.

Note that $C - C$, the subspace of all vectors between points in $C$, is a subspace of codimension $1$, and will be contained in the span of $C$. Therefore, $\operatorname{span} C$ is of codimension at most $1$; either $1$ or $0$. But it can't be $1$, otherwise it would equal $C - C$. Thus, $\operatorname{span} C = X$, and hence $C^\perp = \lbrace 0 \rbrace$.

Or we can go less geometrically. By the Riesz Representation Theorem, there exists some $u \in X \setminus \lbrace 0 \rbrace$ such that $\langle \cdot, u \rangle = \phi(\cdot)$. Suppose $v \in C^\perp$. Note that $\frac{u}{\|u\|^2} \in C$, so $\langle v, u \rangle = 0$. But then $v + \frac{u}{\|u\|^2} \in C$, so $$0 =\left\langle v, v + \frac{u}{\|u\|^2} \right\rangle = \langle v, v \rangle + \frac{\langle v, u \rangle}{\|u\|^2} = \|v\|^2,$$ hence $v = 0$.

Answer 2

It is $C^\perp = \{0\}$. Indeed, take $x \in C^\perp$.

Assume $\phi(x) \ne 0$. Write it in polar form $\phi(x) = |\phi(x)|e^{i\psi}$.

We have:

$$\phi\left(x\cdot\frac{e^{-i\psi}}{|\phi(x)|}\right) = \frac{\phi(x)e^{-i\psi}}{|\phi(x)|} = \frac{|\phi(x)|}{|\phi(x)|} = 1$$

so $x\cdot\frac{e^{-i\psi}}{|\phi(x)|} \in C$.

Therefore $x \perp x\cdot\frac{e^{-i\psi}}{|\phi(x)|}$ so:

$$0 = \left\langle x, x\cdot\frac{e^{-i\psi}}{|\phi(x)|}\right\rangle = \overbrace{\frac{e^{i\psi}}{|\phi(x)|}}^{\ne 0} \cdot \|x\|^2 \implies x= 0$$

Otherwise, if $\phi(x) = 0$, take any $y \in C$. Notice that $\phi(x+y) = \phi(x) + \phi(y) = 1$ so $x + y\in C$.

Therefore $x \perp x + y$ and $x \perp y$ so:

$$0 = \langle x, x + y\rangle = \langle x, x, \rangle + \langle x, y \rangle = \|x\|^2 \implies x = 0$$

Hence $C^\perp = \{0\}$.