Functional equation involving sine function: $ \sin x + f ( x ) = \sqrt 2 f \left( x - \frac \pi 4 \right) $

Question

Let $ f : \mathbb R \to \mathbb R $ be a continuous function such that $$ \sin x + f ( x ) = \sqrt 2 f \left( x - \frac \pi 4 \right) \text . $$ Find $ f $.

I noticed that a solution for $ f $ is the cosine function. I don't know how to continue. Is there a way I could link it to d'Alembert functional equation?

Answer 1

Write the eqn like this. $${1\over \sqrt 2}\sin x+{1\over \sqrt 2}f(x)=f(x-{\pi\over 4})$$ $$\sin {\pi\over 4}\sin x+\cos {\pi\over 4}f(x)=f(x-{\pi\over 4})$$

Does it look familiar? The procedure in solving such problems is to modify values so as to make them look familiar..

Answer 2

Can you assume that $f''$ exist? In this case note that is is suffices to prove that there exist a function $g$ such that $g(x)=\sqrt{2}g(x-\frac{\pi}{4})$ since $$ -\sin x+f''(x)=\sqrt{2}f''\left(x-\frac{\pi}{4}\right) $$ Hence $$ (f+f'')(x)=\sqrt{2}(f+f'')\left(x-\frac{\pi}{4}\right) $$

Answer 3

Let $g(x) = f(x) - \cos x$. We have $$\sin x + \cos x + g(x) = \sqrt 2 \cos \left( x - \frac \pi 4 \right) + \sqrt 2 g \left( x - \frac \pi 4 \right)$$ which yields $$g(x) = \sqrt 2 g \left( x - \frac \pi 4 \right) \tag {*} \label { * }$$ Now note that every function $g$ satisfying \eqref{ * } gives a solution for the original equation by adding $\cos x$. There are a lot of such functions since every function $g_0 : [ 0 , \frac \pi 4 ) \to \mathbb R$ can be extended uniquely to a $g : \mathbb R \to \mathbb R$ satisfying \eqref{ * }. Further assumptions like continuity or differentiability give you some conditions on $g_0$, especially some conditions concerning the value $g_0 (x)$ near $x = 0$ and $x = \frac \pi 4$. But these assumptions are not strong enough to force $g (x) = 0$.