# 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 $2$

$$S$$ 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.

Hint:

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

• thanks a lot. I differentiate twice I got 0 that means it is polynomial of degree of order at most 2.Is this argument? – MathLover Dec 5 '18 at 16:04
• It yields $f''(y+x)-f''(y-x)=0$ for all $x,y\in \Bbb R$, so $f''$ is constant. Then $f$ is a polynomial and its degree is $\le 2$. – ajotatxe Dec 5 '18 at 16:06

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 $$f'=g$$.