# Orthogonal complement of a vector space

I am working on a problem where I have to find the orthogonal complement of a specific vector space. Let $$V = \{f: \mathbb{R} \rightarrow \mathbb{C}$$, such that f is continuous and periodic with period $$1\}$$. We define the inner product on $$V$$ as: $$\langle f, g\rangle = \int_{0}^{1}\overline{f(t)}g(t)dt$$. Moreover, let the operator $$L_a$$ be defined as: $$L_a(f(t)) = f(t+a)$$ and the space $$H_a = \{f\in V: L(f) = f\}$$. Now, I want to find the orthogonal complement of $$H_a$$ for $$a=\frac{1}{2}$$ and $$a=\frac{1}{\sqrt2}$$.

I have worked out that $$L_a$$ will be unitary for all $$a$$ and it will be self adjoint for all $$a=\frac{n}{2}$$ where $$n\in\mathbb{Z}$$. My initial thought was to use the fact that if an operator is self adjoint its eigenspaces are orthogonal however i realized that $$H_a$$ is infinite dimensional and therefore this fact does not apply. I also tried finding functions $$g$$ such that $$\langle f, g\rangle = 0$$ for $$f\in H_a$$ but this lead me nowhere.

Any ideas on how to attack this problem?

• I may be mistaken but isn't any function in $H_a$ also $a$ periodic? This would mean that $H_{\frac{1}{2}}$ is the space of continuous $\frac{1}{2}$ periodic functions and $H_{\frac{1}{\sqrt{2}}}$ is the space of continuous functions which are both $1$ periodic and $\frac{1}{\sqrt{2}}$ periodic, which I believe would just be constant functions. – Eric Nov 20 '18 at 21:28
• Yes, $H_a$ is the space where $f$ is both $1$ and $a$ periodic however this is not true only for constant functions. If we consider $a=\frac{1}{2}$, for example $f(x) = \sin(2\pi x)$ has this property. – Unknown Nov 20 '18 at 21:32
• I meant only constant functions specifically in the case of $a= \frac{1}{\sqrt{2}}$. Sorry for the lack of clarity. – Eric Nov 20 '18 at 21:36
• Almost a duplicate of math.stackexchange.com/questions/3006189/… – Kavi Rama Murthy Nov 20 '18 at 23:43

If $$\alpha=\frac 12$$ then for every $$f\in V$$ and $$g\in H_\alpha$$ $$\int^1_0f(x)g(x)dx=\int^\frac{1}{2}_0f(x)g(x)dx+\int^1_\frac{1}{2}f(x)g(x)dx=\int^\frac{1}{2}_0\left[f(x)+f\left(x+\frac 12\right)\right]g(x)dx$$ From the fundamental lemma of calculus of variations if $$f$$ belongs to orthogonal of $$H_\alpha$$ then $$\overline{f(x)}+\overline{f\left(x+\frac 12\right)}=0\Leftrightarrow f(x)=-f\left(x+\frac 12\right)$$ for every $$x\in\mathbb R$$.
If $$\alpha=\frac{1}{\sqrt 2}$$ we have to prove that $$H_\alpha$$ contains only constant functions. If $$f\in H_\alpha$$ let $$G=\{k\neq 0 : f$$ is $$k$$-periodic$$\}\cup\{0\}$$ it's clear that $$(G, +)$$ is a group with $$0$$ as identity element. Not let $$k_n\in G$$ such that $$k_n\rightarrow k\neq 0$$ then $$k_n\neq 0$$ for a certain $$n$$, due to continuity $$f(x+k)=\lim_{n\rightarrow +\infty}f(x+k_n)=f(x)$$ so $$k\in G$$ and $$G$$ is closed.
We now prove that if $$(T, +)$$ is a closed subgroup of $$\mathbb R$$ then or $$T=\mathbb R$$ or exists $$\alpha>0$$ such that $$T=\{\alpha k : k\in\mathbb Z\}$$ Let $$\alpha=\inf\{x\in T : x>0\}\in T$$, if $$\alpha>0$$ then the statement follows immediately, else $$\alpha=0$$ and exists $$x_n\in T$$ such that $$x_n>0$$ and $$x_n\rightarrow 0$$. Let $$u\in\mathbb R$$ then exists $$k_n\in\mathbb Z$$ such that $$k_nx_n\leq u\leq (k_n+1)x_n$$ then $$\lvert u-k_nx_n\rvert\leq x_n\rightarrow 0\Rightarrow k_nx_n\rightarrow u$$ due to $$k_n x_n\in T$$ then also $$u\in T$$ ($$T$$ is closed) so $$T=\mathbb R$$.
Now $$G$$ is a closed subgroup, if $$G\neq\mathbb R$$ then exists $$\alpha>0$$, $$k, k'\in\mathbb Z$$ such that $$1=k\alpha\\ \frac{1}{\sqrt 2}=k'\alpha$$ but then $$\frac{1}{\sqrt 2}=\frac{k'}{k}\in\mathbb Q$$ that's absurd so $$G=\mathbb R$$ and $$f(x)=f(x+0)=f(0)$$ for every $$x\in\mathbb R=G$$.
Orthogonal of $$H_\alpha$$ contains all periodic $$f$$ such that $$\int^1_0f(x)dx =0$$ and the proof is concluded.