Ytukyg

$S$ is set of family of infinite differentiable function from $mathbb R to mathbb R$ with $forall x,yin R$

$$f(x+y)-f(y-x)=2xf^prime(y)$$

then I have to prove that $S$ only contain all polynomials of degree less than $2$.

My attempt:

I can show that all polynomial of degree less than 2 satisfies that property

From given equation, I think it uses Mean value theorem, but I am not able to show that is only function.

Please only provide me hint. I wanted to solve this problem.

Hint:

Differentiate both sides w.r.t. $x$ twice.

Here is a generalization. I do not assume differentiability of $f$ or $g$ in the proposition below.

Proposition. Let $f,g:mathbb{R}tomathbb{R}$ be functions such that
$$f(x+y)-f(y-x)=2,x,g(y)tag{*}$$
for all $x,yinmathbb{R}$. Then, $f$ is a linear function and $g$ is a constant function such that the (first) derivative of $f$ equals $g$.

Plugging in $y:=0$ in (*), we obtain $f(x)-f(-x)=2cx$, where $c:=g(0)$. That is,
$$begin{align}2x,g(y)+2x,g(-y)&=big(f(x+y)-f(y-x)big)-big(f(-x-y)-f(x-y)big)\&=big(f(x+y)-f(-x-y)big)+big(f(x-y)-f(y-x)big)\&=2c(x+y)+2c(x-y)=4cx,.end{align}$$
Therefore,
$$g(y)+g(-y)=2c$$
for all $yinmathbb{R}$.

Now, define $F,G:mathbb{R}tomathbb{R}$ by $F(x):=f(x)-f(0)-cx$ and $G(x):=g(x)-c$ for all $xinmathbb{R}$. Then, we see that $F$ is an even function with $F(0)=0$, $G$ is an odd function (whence $G(0)=0$), and
$$F(x+y)-F(y-x)=2,x,G(y)$$
for all $x,yinmathbb{R}$. This shows that
$$F(x)-F(y)=(x-y),Gleft(frac{x+y}{2}right)tag{#}$$
for each $x,yinmathbb{R}$. In particular, we have
$$F(x+t)-F(y-t)=(x-y+2t),Gleft(frac{x+y}{2}right)$$
for every $x,y,tinmathbb{R}$. Consequently,
$$(x-y),big(F(x+t)-F(y-t)big)=(x-y+2t),big(F(x)-F(y)big)$$
for all $x,y,tinmathbb{R}$.

Taking $y:=0$ in the equation above and using the fact that $F$ is even with $F(0)=0$, we have
$$x,big(F(x+t)-F(t)big)=(x+2t),F(x),.$$
From (#), we conclude that
$$xt,Gleft(frac{x}{2}+tright)=(x+2t),F(x)$$
for all $x,tinmathbb{R}$. Plugging in $t:=dfrac{x}{2}$ in the previous equation, we get
$$F(x)=frac{x}{4},G(x)text{ for every }xinmathbb{R},.tag{@}$$
Thus, (#) becomes
$$x,G(x)-y,G(y)=4,(x-y),Gleft(frac{x+y}{2}right)text{ for all }x,yinmathbb{R},.$$
Plugging in $y:=0$ in the equation above and recalling that $G(0)=0$, we get
$$Gleft(frac{x}{2}right)=frac{G(x)}{4}text{ for all }xinmathbb{R},.$$
Ergo, we have
$$x,G(x)-y,G(y)=(x-y),G(x+y)text{ for every }x,yinmathbb{R},.$$
Replacing $y$ by $-y$ in the equation above and noting that $G$ is odd, we get
$$(x+y),G(x-y)=x,G(x)+y,G(-y)=x,G(x)-y,G(y)=(x-y),G(x+y)$$
for all $x,yinmathbb{R}$.

This shows that there exists $kinmathbb{R}$ such that $G(x)=kx$ for all $xinmathbb{R}$. Since $Gleft(dfrac{x}{2}right)=dfrac{G(x)}{4}$ for all $xinmathbb{R}$, we deduce that $k=0$, making $Gequiv 0$. By (@), we then see that $Fequiv 0$. Consequently, $g$ is a constant function, and $f$ is a linear function such that $f'=g$.