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Question: Find all "tame" solutions of $f(x+y)+f(x-y)=2(f(x)+f(y))$.

This is how I've tried to begin— Plug in $y=0$ to get $f(0)=0$.

$y\to x$ gives $f(2x)=2^2f(x)$. Also, I've noted that $f$ is an even function by $y\to -x$ substitution.

Now, I somewhat think that $f(kx)=k^2f(x)$ but I'm not sure how to land onto there. And at the end of the day, I believe $f(x)=ax^2$ but I cannot proceed into that stage.

Mathejunior
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    Inductively, suppose that we know $f(kx)=k^2f(x)$ for $k≤n$. Then let $y=nx$. We have $f((n+1)x+f((n-1)x)=2\left(f(x)+f(nx)\right)\implies f((n+1)x+(n-1)^2f(x)=2f(x)+2n^2f(x)$ and the desired result follows at least for integers $k$. – lulu Dec 24 '17 at 18:53
  • Fine, I get it. – Mathejunior Dec 24 '17 at 18:59
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    Well, what I wrote is far from a complete solution. Next step is to show it holds for rational $k$. Then, I suppose, invoke continuity (unstated but I figure that is what "tame" means) to show it works for all $k$. – lulu Dec 24 '17 at 19:02
  • @lulu Yes I got that. – Mathejunior Dec 24 '17 at 19:04
  • If "tame" includes continuous differentiability, then taking the derivative in $x$ on both sides gives $f'(x+y)+f'(x-y)=2f'(x),$, which after a little more work implies that $f'$ is linear. – dxiv Dec 24 '17 at 19:54

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You have made a good start, but not sure where to go from there. My approach may help you.

Let $$e(x,y) := f(x+y)+f(x-y)-2(f(x)+f(y)). $$

We are given that $\;e(x,y)=0\;$ for all $\;x,y.\;$

Since $\;e(0,0) = -2f(0),\;$ then $\;f(0)=0.\;$ Since $\;e(0,x) = -2f(0) + f(-x) -f(x),\;$ then $\;f(-x) = f(x).\;$ Let

$$ g(x,y) := f(x+y)-f(x)-f(y). $$

Now evaluate to get

$$ 2(g(x_1,y)+g(x_2,y)-g(x_1+x_2,y)) = \\ e(x_1,x_2)+e(x_1+x_2+y,y)-e(x_1+y,x_2+y), $$

but this is $0$ by property of $\;e(x,y).\;$ Thus, $\;h(x):=g(x,y)\;$ satisfies Cauchy's functional equation, and assuming $h(x)$ is "tame", then $\;h(x) = g(x,y) = x\, c(y)$ for some function $c(y).$ Using the same reasoning with $\;h(y) := g(x,y)\;$ we have $\;g(x,y)=y\,d(x)\;$ for some function $d(x).$ Thus, $\;g(x,y) = t\,xy\;$ for some constant $t.\;$ Now $\;\lim_{y\to0} g(x,y)/y = t\,x $ and, assuming $f'(0)$ exists, by definition, also $\lim_{y\to0} g(x,y)/y = f'(x)-f'(0).\;$ But $f(-x)=f(x)$ so $f'(0)=0$ and thus $f'(x)=t\,x\;$ and $f(x)=t\,x^2/2.$

Somos
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  • could you prove that the only nonnegative functions $f:\mathbb R\to\mathbb R_0^+$ which satisfy the quadratic functional equation of the OP are $f(x)=kx^2;;\forall;x\in\mathbb R;?$ $f(rx)=r^2f(x);;\forall;r\in\mathbb Q;$ implies that $f(r)=kr^2;;\forall;r\in\mathbb Q;$ where $k=f(1).;$ Could you prove that $f(x)=kx^2;$ for any $;x\in\mathbb R;,;$ not only for any $x\in\mathbb Q;,;$ by using the fact the $f$ is nonnnegative without using the continuity ? – Angelo Aug 28 '22 at 10:28