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

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

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