Let $f:\mathbb {R}\to \Bbb R$ be a continuous function. Suppose $f(x)=\frac 1t\int_0^t(f(x+y)-f(y))dy$ for all $x\in\Bbb R$ and all $t>0.$ Then show that $\exists c$ such that $f(x)=cx$ for all $x.$
My solution goes like this:
We have, $f(x)=\frac 1t\int_0^t(f(x+y)-f(y))dy\implies tf(x)=\int_0^t(f(x+y)-f(y))dy$ for all $x\in\Bbb R$ and all $t>0.$ We assume, $F(t)=tf(x)$ and $h(y)=f(x+y)-f(y),$ thus, we have, $F(t)=\int_0^th(y)dy.$ From Fundamental Theorem Of Calculus, we have, $F'(t)=h(t)\implies f(x)=f(x+t)-f(t).$ Hence, the function $f$ satisfies $f(x+t)=f(x)+f(t),\forall x\in \Bbb R, t\in \Bbb R^+.$ This is indeed Cauchy's additive functional equation and since, $f$ is continuous everywhere, so the only solution of this functional equation is $f(x)=cx,$ where $c\in\Bbb R.$
Is the above solution valid? If not, where is it going wrong?