If $f:\mathbb{R}\to\mathbb{R}$ is infinitely-differentiable, and $f(x+y)-f(y-x)=2xf^\prime(y)$, then it is a polynomial of degree less than $2$

Question

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

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

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.

Answer 1

Hint:

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

Answer 2

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

Proposition. Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions such that $$f(x+y)-f(y-x)=2\,x\,g(y)\tag{*}$$ for all $x,y\in\mathbb{R}$. 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 $y\in\mathbb{R}$.

Now, define $F,G:\mathbb{R}\to\mathbb{R}$ by $F(x):=f(x)-f(0)-cx$ and $G(x):=g(x)-c$ for all $x\in\mathbb{R}$. 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 $x,y\in\mathbb{R}$. This shows that $$F(x)-F(y)=(x-y)\,G\left(\frac{x+y}{2}\right)\tag{#}$$ for each $x,y\in\mathbb{R}$. In particular, we have $$F(x+t)-F(y-t)=(x-y+2t)\,G\left(\frac{x+y}{2}\right)$$ for every $x,y,t\in\mathbb{R}$. Consequently, $$(x-y)\,\big(F(x+t)-F(y-t)\big)=(x-y+2t)\,\big(F(x)-F(y)\big)$$ for all $x,y,t\in\mathbb{R}$.

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 $$xt\,G\left(\frac{x}{2}+t\right)=(x+2t)\,F(x)$$ for all $x,t\in\mathbb{R}$. Plugging in $t:=\dfrac{x}{2}$ in the previous equation, we get $$F(x)=\frac{x}{4}\,G(x)\text{ for every }x\in\mathbb{R}\,.\tag{@}$$ Thus, (#) becomes $$x\,G(x)-y\,G(y)=4\,(x-y)\,G\left(\frac{x+y}{2}\right)\text{ for all }x,y\in\mathbb{R}\,.$$ Plugging in $y:=0$ in the equation above and recalling that $G(0)=0$, we get $$G\left(\frac{x}{2}\right)=\frac{G(x)}{4}\text{ for all }x\in\mathbb{R}\,.$$ Ergo, we have $$x\,G(x)-y\,G(y)=(x-y)\,G(x+y)\text{ for every }x,y\in\mathbb{R}\,.$$ 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 $x,y\in\mathbb{R}$.

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