# Problem involving a functional equation.

$$\mathbf {The \ Problem \ is}:$$Find all possible functions $$f \colon \Bbb R→\Bbb R$$ such that $$f$$ is infinitely differentiable on $$\Bbb R$$ and $$f$$ satisfies an equation: $$f(y+x)-f(y-x)= 2xf'(y)\text{ for all }x,y\in\Bbb R .$$

$$\mathbf {My \ attempt}:$$ Then, obviously $$(f(b) - f(a)) = (b-a) f'((a+b)/2)$$ for all $$a,b$$ and we know that :
$$f''(x) = \lim_{h→0} (f(c+h) + f(c-h)-2f(c))/h^2$$

The answer provided is that $$f$$ must be polynomial of order at most $$2$$ , then I tried putting values of $$f(c+h)$$ in the above limit but after that I can't approach properly .

• Can someone please edit the text using Latex (as I am completely unable of doing it) ??? – Rabi Kumar Chakraborty Jul 20 at 19:53

We need only assume that $$f$$ is once differentiable and the existence of higher derivatives follows because with $$x=\frac12$$, $$f'(y)=f(y+\tfrac12)-f(y-\tfrac12)$$ and the right hand side is differentiable so that by induction $$f^{(n+1)}(x)=f^{(n)}(x+\tfrac12)- f^{(n)}(x-\tfrac12).$$
By differentiating the functional equation with respect to $$x$$, we find $$f'(y+x)+f'(y-x)=2f'(y)$$ and hence if we fix $$\epsilon>0$$ and let $$g(y):=f'(y)-f'(y-\epsilon)$$, we have $$g(y+\epsilon)=g(y).$$ So $$g$$ is periodic with period $$\epsilon$$, and is smooth, hence attains local extrema inside each interval of length $$>\epsilon$$, hence $$g'$$ has zeroes in such intervals, i.e., $$f''$$ has equal values a distinct points, and finally by Rolle, $$f'''$$ has a root in each interval of length $$>\epsilon$$. As $$\epsilon$$ was arbitrary, the roots of $$f'''$$ are dense in $$\Bbb R$$, and as $$f'''$$ is continuous, it follows that $$f'''$$ is identically zero.