Let $ X=C[-1,1]$ be inner product space with definition $$\langle f,g\rangle =\int_{-1}^1 f \overline{g} dt .$$ Let $M$ be the subspace defined by $$ M= \left\{f \in X\mid f(t)=0 , -1 \leq t \leq 0 \right\}. $$
What is $M^{\perp}$?
Let $N := \{\, g \in X \mid \text{$g(t) = 0$ for all $t \in (0, 1)$} \,\}$. I'll show that $M^\perp = N$.
From the definition of inner product, $N \subset M^\perp$ is obvious. So assume $g \in M^\perp$ and prove $g \in N$ by contradiction.
Suppose $g(\tau) \neq 0$ for some $\tau \in (0, 1)$. Without loss of generality, we may assume that $g(\tau) > 0$. Since $g$ is a continuous function, there exist a (small) $\delta > 0$ such that $g(t) > 0$ for all $\lvert t - \tau \rvert < \delta$. Then take a function $f \in M$ defined by
$$ f(t) = \begin{cases} 0 & \text{if $\lvert t - \tau \rvert \geq \delta$} \\ 1 + (t - \tau)/\delta & \text{if $\tau - \delta \leq t \leq \tau$} \\ 1 - (t - \tau)/\delta & \text{if $\tau \leq t \leq \tau + \delta $}. \end{cases}$$ (Its graph is something like triangle wave which has peek at $t = \tau$ and its amplitude is $1$.) Since $f \in M$ and $g \in M^\perp$, it should be $\langle f, g \rangle = 0$. However, from the choice of $\delta$ and $f$, it's also $\langle f, g \rangle > 0$. Contradiction.
%% Though I show in case $X$ is a set of continuous real-valued functions, the same argument would work in complex-valued case by separation of integral to real part and imaginary part.