# Functional Equation: $f \left(x+\cos(2017y) \right)=f(x)+2017\cos\left(f(y)\right)$

Find all functions $f: \mathbb R \rightarrow \mathbb R$, such that for all $x,y \in \mathbb R$ satisfies the equation: $$f \left(x+\cos(2017y) \right)=f(x)+2017\cos\left(f(y)\right)$$

### My work so far:

Let $f(0)=c$.

1) $x=0 \Rightarrow f\left(\cos(2017y)\right)=c+2017\cos\left(f(y)\right)$

2) $y=0\Rightarrow f(x+1)=f(x)+2017 \cos c$

• Can we assume that $f$ is continuous? Can we assume $f$ is differentiable? – flawr Jul 8 '16 at 8:09
• @flawr: We can't assume that $f$ is continuous – Roman83 Jul 8 '16 at 11:41

For simplicity, let $$a := 2017$$. I claim that the only solutions to the functional equation $$f \big( x + \cos ( a y ) \big) = f ( x ) + a \cos f ( y ) \tag 0 \label 0$$ are the functions of the form $$f ( x ) = n \pi + ( - 1 ) ^ n a x$$, where $$n$$ is a constant integer. It's easy to check that these functions indeed satisfy \eqref{0} and we only need to prove that every solution is of this form.
First, defining $$c := f ( 0 )$$ and letting $$x = 0$$ in \eqref{0} we have: $$f \big( \cos ( a y ) \big) = c + a \cos f ( y ) \tag 1 \label 1$$ So, combining \eqref{0} and \eqref{1} and substituting $$\frac y a$$ for $$y$$, we get: $$f ( x + \cos y ) = f ( x ) + f ( \cos y ) - c \tag 2 \label 2$$ For simplicity, we define $$g ( x ) := f ( x ) - c$$ and \eqref{2} gives us: $$g ( x + \cos y ) = g ( x ) + g ( \cos y ) \tag 3 \label 3$$ By \eqref{3}, we can inductively show that for every positive integer $$m$$: $$g ( m \cos y ) = m g ( \cos y ) \tag 4 \label 4$$ Now, for arbitrary real number $$x$$ and $$y$$, we choose the positive integer $$m$$ in a way that $$\big| \frac x m \big| < 1$$, $$\big| \frac y m \big| < 1$$ and $$\big| \frac { x + y } m \big| < 1$$. So there are real numbers $$\alpha$$, $$\beta$$ and $$\gamma$$ such that $$\frac x m = \cos \alpha$$, $$\frac y m = \cos \beta$$ and $$\frac { x + y } m = \cos \gamma$$. Now by \eqref{3} and \eqref{4} we have: $$g ( x + y ) = g ( m \cos \gamma ) = m g ( \cos \gamma ) = m g \Big( \frac { x + y } m \Big) = m g ( \cos \alpha + \cos \beta ) \\ = m g ( \cos \alpha ) + m g ( \cos \beta ) = g ( m \cos \alpha ) + g ( m \cos \beta ) = g ( x ) + g ( y )$$ So $$g$$ satisfies Cauch's functional equation. Also by \eqref{1} we have $$g \big( \cos ( a y ) \big) = a \cos \big( c + g ( y ) \big)$$. Thus $$g$$ is a bounded function on the interval $$[ -1 , 1 ]$$, since $$\cos$$ is a bounded function. This shows that $$g$$ must be of the form $$g ( x ) = b x$$ for some constant real number $$b$$ (see here). Hence we have: $$b \cos ( a y ) = a \cos ( b y + c )$$ It takes some effort to show that this leads to $$c = n \pi$$ for some constant integer $$n$$, and $$b = ( - 1 ) ^ n a$$ for the same $$n$$.