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?

  • $\begingroup$ 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. $\endgroup$ – Eric Nov 20 '18 at 21:28
  • $\begingroup$ 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. $\endgroup$ – Unknown Nov 20 '18 at 21:32
  • $\begingroup$ I meant only constant functions specifically in the case of $a= \frac{1}{\sqrt{2}}$. Sorry for the lack of clarity. $\endgroup$ – Eric Nov 20 '18 at 21:36
  • $\begingroup$ Almost a duplicate of math.stackexchange.com/questions/3006189/… $\endgroup$ – 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.

  • $\begingroup$ Awesome thank you!! $\endgroup$ – Unknown Nov 21 '18 at 7:38

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