# $f(x+y)+f(x-y)=2[f(x)+f(y)]$

Here is the problem: Find all continuous $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy $$f(x+y)+f(x-y)=2[f(x)+f(y)]\;\;\;(1).$$

Here is my attempt:

Fix $\delta>0$ and let $C=\int_{0}^{\delta}2f(y)dy.$

Then $$\begin{align*} 2\delta f(x)+C&=\int_{0}^{\delta}2[f(x)+f(y)]dy\\ &=\int_{0}^{\delta}f(x+y)+f(x-y)dy\\ &=\int_{x}^{x+\delta}f(y)dy+\int_{x-\delta}^{x}f(y)dy\\ &=\int_{x-\delta}^{x+\delta}f(y)dy. \end{align*}$$

Now since $f$ is continuous, the last expression is a differentiable function of x and thus the first expression must also be differentiable; hence $f$ is differentiable. By induction, f is infinitely differentiable.

Differentiating (1) first with respect to y, we arrive at: $$f'(x+y)-f'(x-y)=2f'(y)\;\;\;(2).$$

Differentiating once more with respect to x, we have: $f''(x+y)=f''(x-y),$ so $f''$ is constant. It follows that $f(x)= ax^2+bx+c$ are the only potential solutions.

Substituting $x=y=0$ in (1) and (2) implies $f(0)=f'(0)=0$; hence $f(x)=ax^2\;\;\;(a\in\mathbb{R})$.

It is easy to check that all such $f$ are indeed solutions. $\square$

Is my proof correct?

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3
Very nice approach. – Andres Caicedo Apr 19 '11 at 5:46
Nice indeed! I was thinking more along the lines of when y = 0, we get f(y) = f(y) = 0... – The Chaz 2.0 Apr 19 '11 at 5:49
How do you get $f(-y)$ in $\int_{x}^{\delta+x}f(y)+f(-y)dy$? It seems wrong, since $f(-y)$ has arguments near $-x$ instead of $x$. – joriki Apr 19 '11 at 5:53
@joriki: You are correct; I have fixed the error. – bobobinks Apr 19 '11 at 6:03
@Mike: I am aware of the approach you suggested; in fact, in most cases it is the better approach as it allows one to reach the same conclusion with weaker hypotheses, i.e, monotonicity away from 0 or boundedness on a closed interval. I was simply looking for an alternate method. In any case, thank you for your suggestion. – bobobinks Apr 19 '11 at 6:19
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## 1 Answer

You might be interested in problem 12 in "functional equations" by Marko Radovanović

The problem states: Find all functions $f,g,h\colon \mathbb R\to\mathbb R$ that satisfy $f(x+y)+g(x-y)= 2h(x)+2h(y)$. The solution is quite short and beautiful (as is typical with that kind of problems.)

Turns out the only solutions are $$h(x) = \alpha x^2 + \beta x + b$$ $$f(x) = \alpha x^2 + 2\beta x + 4 b- a$$ $$g(x) = \alpha x^2 + a$$

For some choice of constants. Specializing to the case where $f=g=h$ yields $\beta = 0$ and $a=b=4b-a$, so $f(x)=g(x)=h(x)=\alpha x^2$.

Edit: I just noted that the author apparently reduces this problem to something very similar to the given problem (the sign on the left hand side is different) and then states it is trivial with the given technique. I think it's not that difficult, something like this might work on his problem --- maybe it provides a shortcut here too (?):

• Show that $f(0) = 0$ and $f(-x) = f(x)$
• Assume that $f(1)=f(-1) = \alpha$, find $f(2)=f(1+1)=4\alpha$
• Find that $f(n) = f((n-1)+1) = n^2 \alpha$ with induction.
• Proceed to deduce that $f(\tfrac{n}{m}) = \frac{n^2 \alpha}{m^2}$.
• Use continuity.
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+1 Very nice reference. – Adrián Barquero Apr 19 '11 at 7:16
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