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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 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 .

Mohammed M. Zerrak
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    Interesting question. Lebesgue or Riemann integral? – Theo Bendit Aug 30 '18 at 02:22
  • simple riemann integration – Mohammed M. Zerrak Aug 30 '18 at 02:24
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    If $f$ is differentiable at $x$ and $f'(x)\ne 0,$ then near $0$ the integrand looks like $f'(x)/t,$ which is not integrable. This is implies that if $f$ is differentiable everywhere, and satisfies the condition, then $f$ is constant. – zhw. Aug 30 '18 at 02:26
  • no the function is not differentiable , – Mohammed M. Zerrak Aug 30 '18 at 02:31
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    I understand that. That's why it's a comment, not an answer. – zhw. Aug 30 '18 at 02:33
  • The question then becomes when does the series $$\sum n^2\ (f(x+\frac{1}{n})-f(x)) $$ converges – Tutankhamun Aug 30 '18 at 03:05
  • @PanchalShamsundar If you're going to turn it into a series, there's an assumption of monotonicity. You need $t^2\left(f\left(x + \frac{1}{t}\right) - f(x)\right)$ to be monotone, which looks tricky to prove in general. – Theo Bendit Aug 30 '18 at 03:24
  • @MohammedM.Zerrak You said "simple Riemann integration". So are we to assume the integrand is bounded on $[0,1]$ for every $x?$ – zhw. Sep 01 '18 at 21:15
  • I don't think so , riemann integration the way i studied states that for every epsilon there exist a delta such that for all x and y less than delta , the integral between x and y is bounded by epsilon – Mohammed M. Zerrak Sep 02 '18 at 02:07
  • @MohammedM.Zerrak If $g$ is Riemann integrable on $[a,b],$ then $g$ is bounded on $[a,b].$ It's part of the definition. – zhw. Sep 02 '18 at 16:25
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    @zhw: $\dfrac{f(x + t) - f(x)}{t^2}$ is continuous, therefore bounded on segment $[\delta, 1]$ for every $\delta > 0$. Therefore $\int\limits_\delta^1 \dfrac{f(x + t) - f(x)}{t^2} ~dt$ exists for every $\delta > 0$. The statement that $\int\limits_0^1 \dfrac{f(x + t) - f(x)}{t^2} ~dt$ converges is, by definition, the statement that $\lim\limits_{\delta \to 0} \int\limits_\delta^1 \dfrac{f(x + t) - f(x)}{t^2} ~dt$ exists. No boundedness of $\dfrac{f(x + t) - f(x)}{t^2}$ on the whole segment $[0, 1]$ is required. – Kaban-5 Sep 02 '18 at 16:42
  • In other words, it is assumed it is a convergent improper Riemann integral. – zhw. Sep 02 '18 at 17:22
  • @MohammedM.Zerrak Where does this problem come from? – zhw. Sep 02 '18 at 18:18
  • One of the problem to be solved within half an hour in the enyrance exam of the french school ENS Lyon – Mohammed M. Zerrak Sep 02 '18 at 18:23
  • With the assumption that $f$ is differentiable it may be possible to show $f$ is constant via integration by parts. – asd Sep 02 '18 at 19:01
  • @asd the differentiable case is simple; see my comment above – zhw. Sep 03 '18 at 21:07
  • @zhw I agree, I am just offering another option. – asd Sep 04 '18 at 01:01
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    Duplicate here? https://math.stackexchange.com/questions/2870314/finding-a-set-of-continuous-functions-with-a-certain-property – Rigel Sep 05 '18 at 17:04
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    Please do not repost questions. – quid Sep 08 '18 at 16:26

4 Answers4

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Edit: (added details about $h$ being bounded on an interval) Not a complete argument (due to the DCT portion), but a start. Below it is assumed that $\max_{x} \int\limits_{0}^1 \frac{|(f(x+t)-f(x))|}{t^2} dt$ is finite which was not given in the problem. Let $h(x)=\int\limits_{0}^1 \frac{f(x+t)-f(x)}{t^2}dt$ and consider $H_{s}(w)=\int\limits_{s}^w h(x) dx$ for some $s,w$. $h$ is continuous as DCT implies $$|h(x+\delta)-h(x)|\leq\int\limits_{0}^1 \frac{|(f(x+\delta+t)-f(x+\delta))-(f(x+t)-f(x))|}{t^2}dt$$ can be made arbitrarily small by taking $\delta$ sufficiently small. This holds as $\frac{|(f(x+\delta+t)-f(x+\delta))-(f(x+t)-f(x))|}{t^2}\leq \frac{|(f(x+t)-f(x))|}{t^2}+\frac{|(f(y+t)-f(y))|}{t^2}\leq 2\max_{x}\frac{|(f(x+t)-f(x))|}{t^2}$. Since the integrand defining $h(x)$ is absolutely integrable, and $h(x)$ is bounded on any interval $(s,w)$ (by continuity), Fubini's applies and

$$H_{s}(w)=\int\limits_{0}^1 \int\limits_{s}^{w} \frac{f(x+t)-f(x)}{t^2} dx dt = \int\limits_{0}^1 \frac{F(w+t)-F(s+t)-(F(w)-F(s))}{t^2}dt$$ where $F(w)-F(s)=\int\limits_{s}^w f(x) dx$, (noting that $F$ is differentiable by FTC). The above integral is finite only if $$[F(w+t)-F(s+t)-(F(w)-F(s))]'=0,\ \text{at }t=0 \text{ i.e. }F'(w)=F'(s)$$ for all choices of $w,s$. If $h(x)$ is bounded for all choices of $x$ in the interval $(s,w)$ then $H_{s}(w)$ must be finite for all choices of $s<w$ and so $f(s)=f(w)$ for all $s,w$ and $f$ must be constant.

asd
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  • Can i ask why the h function is bounded , and what happens if the integral is not necessarily absolutely convergent ? – Mohammed M. Zerrak Sep 02 '18 at 11:43
  • @asd: I also don't understand how boundendness of $h$ follows. You just claim that it is true. Can you explain it in more detail, please? – Kaban-5 Sep 02 '18 at 16:44
  • @asd: Okay, so you are just using unusual terminology. But in this case $h$ does not have to be absolutely integrable: finiteness does not imply absolute integrability. – Kaban-5 Sep 02 '18 at 16:52
  • @Kaban-5 Read the argument. There is no assumption on $h$ being absolutely integrable. All that is assumed is that the integral defining $h$ is absolutely integrable, which was given as an assumption "You can assume that the integral converges absolutely". – asd Sep 02 '18 at 16:54
  • @asd: Okay, but then how does infiteness of $H_s (w)$ (which is not even well defined even if we extend the possible values with positive infinity, because $h$ may take negative values) contradict anything? The all information we get is that $H_s (w)$ can't be finite. – Kaban-5 Sep 02 '18 at 16:57
  • Let us continue this discussion in chat. – asd Sep 02 '18 at 16:58
  • You are using the absolute integrability of the integrand defining $h$,$ I'm guessing that is what @Kaban-5 meant. Also you are incorrectly using the translation result. (the denominators would have to be translated too) – zhw. Sep 02 '18 at 17:57
  • @zhw It has been updated. There was a problem in the last version and there is still a gap now. – asd Sep 02 '18 at 17:59
  • Briefly, the above argument shows that if $$\int_0^1 \frac {|f(x+t) - f(x)|} {t^2} \ \mathrm dt<C$$ for all $x$ and some constant then $f$ must be constant. – asd Sep 02 '18 at 18:01
  • Edit: Deleted previous comments about $h$ being bounded. – asd Sep 02 '18 at 20:41
  • your answer is incomplete , however you sure helped me solve it – Mohammed M. Zerrak Sep 06 '18 at 16:33
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[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.

Rigel
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Although Rigel’s answer solved the matter brilliantly, I would like to present an alternative solution to this:

Consider the sets $A_{\varepsilon,x} =\{ u > x, \, |f(u)-f(x)| < \varepsilon |u-x|\}.$

Notice that these sets are clearly open, by the continuity of $f$. Also, these sets are nonempty for every $x \in \mathbb{R}$, and $x \in \overline{A}\backslash A$ : indeed, as they are nested in $\varepsilon$, if one of them is empty/does not accumulate around $x$, every other one with $\eta < \varepsilon$ also is. Also, it means that we can assume without loss of generality that all points $y>x$ sufficiently close to $x$ satisfy

$$ f(y) \ge f(x) + \varepsilon(y-x).$$

Plugging this back into the property satisfied by $f$ gives us then a contradiction.

Claim: The set $A_{\varepsilon,x}$ is dense in $(x,+\infty).$

Proof: Suppose its intersection with an interval $(a,b)$ is empty, and consider $ a’ = \sup_{u<b} A_{\varepsilon,x} \le a$. It holds then for this $a’$ that

$$ |f(a’)-f(x)|\le \varepsilon (a’-x).$$

As the set $A_{\delta,a’}, \, \delta < \varepsilon,$ is nonempty and accumulates around $a'$, there is $b’\in (a’,b)$ such that

$$|f(b’)-f(a’)| < \delta(b’-a’).$$

This implies that

$$|f(b’)-f(x)| \le |f(b’)-f(a’)| + |f(a’)-f(x)| < \varepsilon (a’-x) + \delta(b’-a’) < \varepsilon (b’-x),$$

A contradiction to the definition of $a’. \, \square$

Now we finish: as the set $A_{\varepsilon,x}$ is open and dense in $(x,+\infty),$ it means that every point $y \in (x,+\infty)$ satisfies

$$ |f(y)-f(x)| \le \varepsilon(y-x).$$

This implies, in particular, that $f$ is differentiable at $x$ and that $f'(x) = 0.$ As this was valid for all $x \in \mathbb{R},$ we conclude that $f$ is differentiable and $f' =0,$ i.e., $f$ is constant, as desired.

João Ramos
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To see the opposite direction:

See Lebesgue's Differentiation Theorem for Continuous Functions

Hence $\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \int_{0}^{\epsilon} (f(x+t)-f(x)) dt = 0$.

shows that $\int_{0}^{\epsilon} (f(x+t)-f(x)) dt $ drops atleast as fast as $O(\epsilon)$ as $\epsilon \rightarrow 0$.

Balaji sb
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