I need help finding the set of continuous functions $f : \Bbb R \to \Bbb R$ such that for all $x \in \Bbb R$, the following integral converges:
$$\int_0^1 \frac {f(x+t) – f(x)} {t^2} \ \mathrm dt$$
I think it might be the set of constant functions but i havent been able to prove it 🙁
I was thinking that you can use the stone weiestrass theorem considering the set of continuous functions on a closed interval(non trivial) ,and a subset which contains the set of continuous functions whose integral above diverges in some point in that interval along with with the set of constant functions.
So in order to solve the problem i need only to prove that if two functions do not meet the condition of the problem then their product does not as well .
I hope you can provide some insight and thank you .
Best Answer
[Duplicate here.]
Let us prove that $f$ is constant.
Assume by contradiction that there exist $x_0 < x_1$ such that $f(x_0)\neq f(x_1)$. W.l.o.g. we can assume $f(x_1) > f(x_0)$ (otherwise it is enough to change $f$ with $-f$), so that $$ m := \frac{f(x_1) - f(x_0)}{x_1 - x_0} > 0. $$ Let us consider the continuous function $$ g(x) := f(x) - m(x-x_0). $$ By Weierstrass' theorem, $g$ admits a minimum point $c$ in the interval $[x_0, x_1]$. Since $g(x_0) = g(x_1)$, it is not restrictive to assume that $c\in [x_0, x_1)$.
Let $\delta := \min\{1, x_1 - c\}$. We have that $$ 0 \leq \int_0^\delta \frac{g(c+t) - g(c)}{t^2}\, dt = \int_0^\delta \left( \frac{f(c+t) - f(c)}{t^2} - \frac{m}{t}\right)\, dt = -\infty, $$ a contradiction.