$\mathbf {The \ Problem \ is}:$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 .

  • $\begingroup$ Can someone please edit the text using Latex (as I am completely unable of doing it) ??? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.