Finding a set of continuous functions with a certain property [duplicate]

Solution 1:

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.

Solution 2:

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.