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Problem

Find the functional $f: \mathbb{R} \to \mathbb{R}$ with $f(x+y)+f(x-y)=2f(x)cosy$

Attempts

Now as I know, the functional equation can be solved more generally as follow:

General problem Find the function$f: \mathbb{R} \to \mathbb{R}$ \begin{align} f(x+y)+f(x-y)=2f(x)h(y) \end{align}

with $h$ being an even function.

General solution

Choose r such that $h(x − 2r) ≡ h(x)$. Let $f_p$ be a particular solution of the problem and $f$ a general solution.

Choose constants $A$ and $B$, not both zero, such that $Af_p(r) + Bf(r) = 0$. Define $F(x) = Af_p(x) + Bf(x)$ so that $F(r) = 0$. Interchanging $x$ and $y$ in the original equation, we see that $f(x+y)-f(x-y)=2f(y)h(x)$

Thus, \begin{align} f(x-y)=f(x)h(y)-f(y)h(x) \end{align}

Now as I see, this is great for odd $f$, but my original problem does not contain that kind of condition, and as I suspect, the solution is $f(x)= acosx+bsinx$ is also not odd.

Is there any way this can be done?

Any help is appreciated.

Nikola Tolzsek
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    Odditivity? What's that? – Gerry Myerson Oct 07 '21 at 10:53
  • 1. Is the "$ 6 $" in "$ h ( x - 2 r ) 6 \equiv h ( x ) $" a typo? If so, that means you're assuming $ h $ is periodic, which was not given (we only knew $ h $ is even). 2. D'Alembert functional equation is of the form $ f ( x + y ) + f ( x - y ) = 2 f ( x ) f ( y ) $, which is different from what you're trying to solve. I suppose "d'Alembert-like functional equation" is a better title. For d'Alembert functional equation itself, take a look at this post. – Mohsen Shahriari Oct 08 '21 at 20:28
  • For the original problem before generalization, see Solving functional equation $f(x+y)+f(x-y)=2f(x)\cos y$?. – Mohsen Shahriari Oct 08 '21 at 20:34
  • 3. What you get by "interchanging $ x $ and $ y $ in the original equation" is $$ f ( x + y ) + f ( x - y ) = 2 f ( y ) h ( x ) \text , $$ not the one you've claimed. You seem to have assumed that $ f $ is odd to get that. That assumption is not valid, as we know that the equation can have solutions which are not odd (when $ h ( x ) = \cos x $, one solutions is $ f ( x ) = \cos x $, which is not odd). 4. The part about "particular" and "general" solutions seem to be irrelevant, as you've made no use of the function $ F $ you've defined. – Mohsen Shahriari Oct 09 '21 at 10:33

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