If $f(t)$ is periodic and $f(t) + C \cdot f(t + t_1)$ has the same shape of $f(t)$ for each value of $C$ and $t_1$, then $f(t)$ has the shape of a sine wave.
Is there a simple proof?
Is there an intuitive explanation? I mean without using, for example, the Fourier Transform.
Thanks in advance!
One way to describe what makes translating sine waves special is the following: sine waves $f(t)$ have the property that the subspace of the space of functions spanned by their translates $f(t + t_0), t_0 \in \mathbb{R}$ is $2$-dimensional, spanned by $\cos t, \sin t$. What other functions have this property? (I'm not going to require periodicity yet; this condition is already very restrictive.)
It turns out that the answer is the following. Let $V$ be the subspace spanned by translates, and let me assume that $f(t)$ is smooth. Because $V$ is finite-dimensional, it is in particular closed, so if $g(t) \in V$ is differentiable then $g'(t)$, which is a limit of functions in $V$, is also in $V$. Hence the functions $f(t), f'(t), f''(t) \in V$ must be linearly dependent, which means $f(t)$ satisfies a differential equation of the form
$$a f''(t) + b f'(t) + c f(t) = 0.$$
Solutions to differential equations of this form are very restricted. They are linear combinations of functions of the following forms:
In particular, the only ones that are also periodic are sine waves.