Jtdylktuy

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: $langle f, grangle = int_{0}^{1}overline{f(t)}g(t)dt$. Moreover, let the operator $L_a$ be defined as: $L_a(f(t)) = f(t+a)$ and the space $H_a = {fin V: L(f) = f}$. 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 $ninmathbb{Z}$. 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 $langle f, grangle = 0$ for $fin H_a$ but this lead me nowhere.

Any ideas on how to attack this problem?

If $alpha=frac 12$ then for every $fin V$ and $gin H_alpha$
$$
int^1_0f(x)g(x)dx=int^frac{1}{2}_0f(x)g(x)dx+int^1_frac{1}{2}f(x)g(x)dx=int^frac{1}{2}_0left[f(x)+fleft(x+frac 12right)right]g(x)dx
$$
From the fundamental lemma of calculus of variations if $f$ belongs to orthogonal of $H_alpha$ then
$$
overline{f(x)}+overline{fleft(x+frac 12right)}=0Leftrightarrow f(x)=-fleft(x+frac 12right)
$$
for every $xinmathbb R$.

If $alpha=frac{1}{sqrt 2}$ we have to prove that $H_alpha$ contains only constant functions. If $fin H_alpha$ let $G={kneq 0 : f$ is $k$-periodic$}cup{0}$ it's clear that $(G, +)$ is a group with $0$ as identity element. Not let $k_nin G$ such that $k_nrightarrow kneq 0$ then $k_nneq 0$ for a certain $n$, due to continuity
$$
f(x+k)=lim_{nrightarrow +infty}f(x+k_n)=f(x)
$$
so $kin G$ 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
$$
T={alpha k : kinmathbb Z}
$$
Let $alpha=inf{xin T : x>0}in T$, if $alpha>0$ then the statement follows immediately, else $alpha=0$ and exists $x_nin T$ such that $x_n>0$ and $x_nrightarrow 0$. Let $uinmathbb R$ then exists $k_ninmathbb Z$ such that
$$
k_nx_nleq uleq (k_n+1)x_n
$$
then
$$
lvert u-k_nx_nrvertleq x_nrightarrow 0Rightarrow k_nx_nrightarrow u
$$
due to $k_n x_nin T$ then also $uin T$ ($T$ is closed) so $T=mathbb R$.

Now $G$ is a closed subgroup, if $Gneqmathbb R$ then exists $alpha>0$, $k, k'inmathbb Z$ such that
$$
1=kalpha\
frac{1}{sqrt 2}=k'alpha
$$
but then
$$
frac{1}{sqrt 2}=frac{k'}{k}inmathbb Q
$$
that's absurd so $G=mathbb R$ and $f(x)=f(x+0)=f(0)$ for every $xinmathbb R=G$.

Orthogonal of $H_alpha$ contains all periodic $f$ such that
$$
int^1_0f(x)dx =0
$$
and the proof is concluded.