Finding a set of continuous functions with a certain property 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 am thinking it could be the set of constant functions but i havent been able to prove it :( 
 I have also noticed that you can kind of  take any two functions and stick them together (continuously extend one into the other) the resulting function verifies the property in question.
I hope you can provide some insight and thank you .
 A: 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.
A: Partial answer: if $f$ is differentiable then it is constant
We write $f(x+h) = f(x) + h g(h)$ where $g(h)$ is continuous and $g(0) = f'(x)$.
Then the required integral becomes:
$$\int_0^1 \frac {g(t)} t \ \mathrm dt$$
If WLOG $g(0) > 0$ then there is $\delta > 0$ such that $g(t) > \frac12 g(0)$ for every $0 \le t < \delta$, and then:
$$\begin{array}{rcl}
\displaystyle \int_0^1 \frac {g(t)} t \ \mathrm dt
&=& \displaystyle \int_0^\delta \frac {g(t)} t \ \mathrm dt + \int_\delta^1 \frac {g(t)} t \ \mathrm dt \\
&>& \displaystyle \int_0^\delta \frac {g(0)} {2t} \ \mathrm dt + \int_\delta^1 \frac {g(t)} t \ \mathrm dt \\
&=& \infty
\end{array}$$
So $g(0) = 0$, and $f'(x) = g(0) = 0$ everywhere, so $f$ is constant.
