Finding a set of continuous functions with a certain property 2

Question

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 .

Answer 1

[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.

Answer 2

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.

Answer 3

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 $\epsilon \rightarrow 0$.

Answer 4

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.