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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 .

Davide Morgante
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Mohammed M. Zerrak
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  • This is really off-topic without your efforts, but a good problem to me! – Nosrati Aug 02 '18 at 17:35
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    I think I can prove that if the function is differentiable, then it must be constant. – Kenny Lau Aug 02 '18 at 17:40
  • What's reaultinv function? – Nosrati Aug 02 '18 at 17:47
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    Converges in what sense? As a Lebesgue integral, or just an improper Riemann integral? – David C. Ullrich Aug 02 '18 at 18:00
  • @user108128 I think we can safely assume the OP means "resulting" when he used the text "reaultinv." – Mason Aug 02 '18 at 18:02
  • I couldn't find such function so you are right ;)) – Nosrati Aug 02 '18 at 18:08
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    Can somebody who is also interested set up a bounty on this? – Kaban-5 Aug 19 '18 at 21:31
  • I tried putting a bounty but i think that i cant – Mohammed M. Zerrak Aug 20 '18 at 16:40
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    Okay, I started a bounty on this question. I hope it will help. – Kaban-5 Sep 01 '18 at 09:29
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    you probably wan't a more specific name in the title like "determine all continuous functions such that operator $\int_{0}^1 \ldots$ converges". It may be an obsession of my own but think it really helps future readers to find an specific answer. – user1868607 Sep 01 '18 at 09:40
  • Please answer the question of how the integral is supposed to converge. As a Riemann integral, an improper Riemann integral, or a Lebesgue integral? – zhw. Sep 01 '18 at 16:34
  • @zhw.@David C. Ulrich. Y. Yes. If f is the characteristic function of the irrationals then the Riemann integral does not exist but the Lebesgue integral is 0. – DanielWainfleet Sep 02 '18 at 02:53
  • @DanielWainfleet: $f$ is required to be continuos in the original post, making this distinction moot. The real question which zhw and David C. Ulrich asked is most probably about whether the convergence is required to be absolute or not and for some reason they phrased that statement as Lebesgue vs improper Riemann integration (Lebesgue integration corresponds to absolute convergence and improper Riemann corresponds to possibly conditional convergence), which may be slightly confusing at first (it was to me). – Kaban-5 Sep 02 '18 at 09:41
  • @Kaban-5. I thought of my error of mine this A.M. before I even got to the computer – DanielWainfleet Sep 02 '18 at 14:30

2 Answers2

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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.

Rigel
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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.

Kenny Lau
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    I figured that as well , but the heart of the problem lies in the weak hypotesis of the function being only continuous . And the condition for every x is very essential because i can demonstrate A LOT of functions that are continuous but only verify the condition for infinitely many numbers , and don't do the same for all real numbers. – Mohammed M. Zerrak Aug 02 '18 at 18:26