Recall for continuous (e.g., smooth) functions $f, g:[0,L] \to \R$, their dot product is
defined to be
\begin{equation}
\langle f, g \rangle := \frac{1}{L} \int_0^L f(x) g(x) \, dx,
\label{innerproduct}
\end{equation}
(This is not denoted as $f \cdot g$, to avoid confusion
with usual multiplication of functions.)
When $\langle f, g \rangle = 0$ we say $f$ and $g$ are orthogonal. We write $\langle f, g \rangle_{[0,L]}$ rather than $\langle f, g \rangle$ if we want to emphasize $L$. (One can similarly define $\langle f, g \rangle_{[a,b]} = \frac{1}{b-a}\int_a^b f(x) g(x)\,dx$ for any $a < b$ and $f, g: [a,b] \to \R$.)
One thing to keep in mind is that the dot product of two functions is not easily visualized in terms of their graphs, even for the orthogonality condition.
What actually matters about dot products of functions that they does satisfy the usual properties of the dot product that
we are familiar with in $\R^n$, and so allow us to apply to functions
some of the visual intuition we have for geometry with $n$-vectors. For continuous functions $X(x)$ and $Y(x)$ from $[0,L]$ to $\R$, the following hold:
$\langle c_1 X_1 + c_2 X_2, Y \rangle = c_1 \langle X_1, Y \rangle + c_2 \langle X_2, Y \rangle$,
$\langle X, X \rangle = \int_0^L X(x)^2 \, dx \geq 0$, and $\langle X, X \rangle = 0$ precisely when $X$ is the zero function.
Moreover, if $f_1, f_2, \dots, f_m$ are continuous functions on $[0,L]$ that are not identically zero and are pairwise
orthogonal (i.e., $\langle f_i, f_j \rangle = 0$ when $i \ne j$) then they are linearly independent.
Example 1.
For example, consider the functions $f_1(x) = \sin(x)/\sqrt{2}$ and $f_2(x) = \cos(x)/\sqrt{2}$ on the interval $[0,2\pi]$. We find
$\langle f_1, f_1 \rangle = 1$,
$\langle f_2, f_2 \rangle = 1$,
$\langle f_1, f_2 \rangle = 0$.
So $f_1$ and $f_2$ are orthogonal. What's more, if we define $F_\theta (x): = \cos(\theta)f_1(x) + \sin(\theta)f_2(x)$ and $F_{\theta + \pi/2}(x): = -\sin(\theta)f_1(x) + \cos(\theta)f_2(x)$, then using the linearity of the dot product we find
So $F_\theta$ and $F_{\theta + \pi/2}$ are orthogonal as well.
Below is a visualization of $F_\theta$ and $F_{\theta + \pi/2}$. Choose $\theta$ by dragging $P$ on the unit circle, and see how $F_\theta$ and $F_{\theta + \pi/2}$ change.
Example 2.
Here is another exmaple. Consider the functions $g_1(x) = \sqrt{3/52}(5x-8)$
and $g_2(x) = \dfrac{\sqrt{7}}{8}x^3$ on the interval $[0,2]$. The ugly coefficients
make the $g_j$'s have length 1: we find
$\langle g_1, g_1 \rangle = 1$,
$\langle g_2, g_2 \rangle = 1$,
$\langle g_1, g_2 \rangle = 0$.
So $g_1$ and $g_2$ are orthogonal. What's more, if we define $G_\theta (x): = \cos(\theta)g_1(x) + \sin(\theta)g_2(x)$ and $G_{\theta + \pi/2}(x): = -\sin(\theta)g_1(x) + \cos(\theta)g_2(x)$, then using the linearity of the dot product, again we find
so $G_\theta$ and $G_{\theta + \pi/2}$ are orthogonal as well.
Below is a visualization of $G_\theta$ and $G_{\theta + \pi/2}$. Choose $\theta$ by dragging $P$ on the unit circle, and see how $G_\theta$ and $G_{\theta + \pi/2}$ change. Notice that it is hard to see orthogonality just by looking at the graphs.