solve the functional equation $f(x+t)-f(x-t)=4xt$

Question

I think this question might be related with arbitrary functions, but I’m not sure. I also tried to set $t$ to different values but couldn’t get it to work. I tried to set $t=x$ and end up with $f(2x)=f(0)+4x^2$, $f(x)=f(0)+x^2$.

Answer 1

Set $x=t$ and we get $$f(2x)-f(0)= 4x^2$$ so $f(x) =x^2+a$ where $a= f(0)$.

Check: If we now put this in to starting equation we get: $$ (x+t)^2+a-(x-t)^2-a = 4xt$$ which is always true. So $\boxed{f(x) =x^2+a}$ for all real $a$.

Answer 2

Let $f(q) = q^2$ + const., that solves it directly. Is your question for a solution or for all solutions?

Answer 3

For arbitrary reals $a$ and $b$ plug $x=\frac{a+b}{2}$ and $t=\frac{a-b}{2}$. Then, we obtain that $$ f(a)-f(b)=4\cdot\frac{(a+b)(a-b)}{4} $$ or $$ f(a)-f(b)=a^2-b^2. $$ Hence, $f(x)-x^2\equiv c$ for some constant $c$. It's easy to see that function $f(x)=x^2+c$ is satysfying the condition.

Answer 4

As long as $f(x)$ is continuous in the Lipschitz sense, we have

$$ 2\lim_{t\to 0}\left(\frac{f(x+t)-f(x-t)}{2t}\right) = 2f'(x) = 4x\Rightarrow f(x) = x^2+C $$