If $\;\;V=\{ f:\mathbb{R}\rightarrow \mathbb{C} |\; f \text{ is continuous and has period }1\}$, $\;\; \langle f | g \rangle$ is defined as $ \displaystyle \langle f | g \rangle = \int_0^1 \overline{f(t)}g(t)dt$, $\forall f,g \in V\;\;$ and $\displaystyle \;\;H_{\frac{1}{\sqrt{2}}} =\left\{g\in V: g\left(t+\frac{1}{\sqrt{2}}\right)=g(t) \right\}$.
($f \text{ has period } 1$)
($g \text{ has period } 1 \text{ and } \frac{1}{\sqrt{2}}$)
What can be said about $H_{\frac{1}{\sqrt{2}}}^\perp$?
$H_{\frac{1}{\sqrt{2}}}^\perp = \{f\in V: \langle f | g \rangle=0, \forall g\in H_a \} = \left\{f\in V: \langle f | g \rangle=0, \forall g: g\left(t+\frac{1}{\sqrt{2}}\right)=g(t) \right\}$
$\displaystyle 0 = \langle f | g \rangle = \int_0^1 \overline{f(t)}g(t)dt = \int_0^1 \overline{f(t)}g\left(t+n+\frac{m}{\sqrt{2}}\right)dt, \;\; n,m \in \mathbb{Z}, \;\; f\in H_{\frac{1}{\sqrt{2}}}^\perp, \;\; g\in H_{\frac{1}{\sqrt{2}}}$.
Related post: Orthogonal complement of $H_a =\left\{g \in V: g\left(t+\frac{1}{2}\right)=g(t) \right\}$