Show that $f(x)$ is a constant function if $\lim\limits_{h\to 0}\frac{1}{h^3}\int_{-h}^{h}f(x+t)\cdot t\,dt=0$ for all $x$

Let $$f(x)$$ be a continuous function on $$\mathbb{R}$$, such that for any real number $$x$$ we have: $$\lim_{h\to 0}\dfrac{1}{h^3}\int_{-h}^{h}f(x+t)\cdot t\,dt=0.$$ Show that $$f(x)$$ is a constant function.

Maybe we can use the following lemma?

Lemma. If $$g$$ is a continuous function, then $$\lim_{h\to0}\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds=g(x).$$

Proof. We may assume $$h>0$$. $$\left|g(x)-\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds\right|=\frac{1}{2\,h}\left|\int_{x-h}^{x+h}(g(x)-g(s))\,ds\right|\le\frac{1}{2\,h}\int_{x-h}^{x+h}\left|g(x)-g(s)\right|\,ds.$$ Use that $$g$$ is continuous at $$x$$ to show that the last expression converges to $$0$$ as $$h\to0$$.

• Please don't change the statement of the question after users have already thought about your problem. It's okay to make a separate question for the case $h^3$. The question with $h^2$ is very nice, and you got a good answer showing why it is false. Commented May 18, 2021 at 16:26
• @AlexOrtiz: If there is a simple counter example such as $f(x) = x$ then I would suspect a typo, and ask for clarification in a comment. But there are surely different opinions on that. Commented May 18, 2021 at 19:26
• I think that Paresseux Nguyen's answer, which is a wonderful piece of analysis, should be accepted. Commented May 28, 2021 at 20:12
• It is also a pity that the question author did not care to award the bounty. Commented May 28, 2021 at 20:20
• @NN2: No, the bounty was for 50 points. But inequality did not award the bounty (or accept an answer). In that case, half of the amount is automatically assigned to the highest voted answer that was posted during the bounty period. Commented Jun 6, 2021 at 16:22

Define $$F (x):= \int_{0}^x f(t)dt$$

By integration by parts, we observe that: \begin{align}\frac{1}{h}\int_{-h}^h f(x+t)tdt &= \frac{1}{h}\int_{0}^h \big[ f(x+t)-f(x-t)]tdt\\& = F(x+h)+F(x-h) - \frac{1}{h}\int_{0}^h \big[ F(x+t)+F(x-t) \big]dt \end{align} So the provided limit for f implies that for all $$x \in \mathbb{R}$$ $$T(F)(x)=0$$ Where $$T$$ is an operator defined as $$T(g)(x)=\lim_{h \rightarrow 0^+} \frac{1}{h^2} \left\{ g(x+h)+g(x-h)-\frac{1}{h}\int_{0}^h \big[ g(x+t)+g(x-t)\big]dt \right\}$$ for any continuous function $$g$$. Note that $$T(g)$$ is not necessarily defined on all points in $$\mathbb{R}$$.

Let $$\epsilon$$ be any positive number , $$(a,b)$$ be any interval in $$\mathbb{R}$$. Define the functions $$G(x)= F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a)+\epsilon(x-a)(x-b)$$ and $$H(x)= F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a)\boldsymbol{-}\epsilon(x-a)(x-b)$$

We prove that $$G(x) \le 0$$ everywhere on $$[a,b]$$. If not then, because $$G(a)=G(b)=0$$, there is a point $$c \in (a,b)$$ at which $$G$$ attains its positive maximum. Observe that: $$T(G)(c)= \lim_{h \rightarrow 0^+} \frac{1}{h^2} \left\{ G(c+h)+G(c-h)-\frac{1}{h}\int_{0}^h \big[ G(c+t)+G(c-t)\big]dt \right\}$$ Because $$c$$ is a maxima of $$G$$, the function $$u(t):=G(c+t)+G(c-t)$$ attains its local maximum at $$t=0$$. So we can choose a strictly decreasing positive sequence $$(h_n, n \in \mathbb{N})$$ such that $$h_n$$ converge to $$0$$ and $$h_n \in \text{argmin}_{ t \in [0,h_n]} u(t)$$ for all $$n$$ . Thus, $$G(c+h_n)+G(c-h_n)-\frac{1}{h_n}\int_{0}^{h_n} \big[ G(c+t)+G(c-t)\big]dt \le 0$$ Thus $$T(G)(c) \le 0$$. Note that the limit $$T(G)(c)$$ exists because $$T(F)$$ exists and in fact, according to the definition, we see that: $$T(G)(c)=0+0+\frac{4}{3}\epsilon>0$$ Which is a contradiction to what we have just proven.

So $$G(x) \le 0$$ on $$[a,b]$$. Similarly, we can prove that $$H(x) \ge 0$$ everywhere on $$[a,b]$$. These give: $$\left| F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a) \right| \le \epsilon (b-a)^2$$ for all $$\epsilon>0$$ and $$x \in [a,b]$$. Hence forth the conclusion.

Side note: The hypothesis is fairly tight as we can show that the Cantor function satisfies the condition on an almost everywhere set of $$\mathbb{R}$$ even though it is not constant.

• Your argument is not entirely clear to me. $u$ has a maximum at $t=0$, but why does that imply that $u(h_n) - \frac 1 {h_n} \int_0^{h_n} u(t) dt \le 0$ on a sequence $h_n \searrow 0$? – And $h_n \in \text{argmin}_{ t \in [0,h_n]} u(t)$ looks like a cyclic definition to me. Commented May 24, 2021 at 13:32
• @MartinR The latter part is where I started. That existence of the sequence $(h_n)$ is can be shown by the following construction. \begin{align} h_1 &\in \text{argmin}_{t \in [0,b-a]}\\{0\} u(t) \\ h_{n+1} &\in \text{argmin}_{ t \in [0, \text{min}(h_n,1/n)] } u(t) \end{align} \{0\} Commented May 24, 2021 at 13:39
• That is not exactly what you wrote in the answer. – And how do you get from that definition to $u(h_n) - \frac 1 {h_n} \int_0^{h_n} u(t) dt \le 0$? Perhaps I am overlooking something simple ... Commented May 24, 2021 at 13:44
• That is beause $u(h_n)=\text{min}_{ t \in [0,h_n]} u(t)$, in other words, $u(h_n) \le u(t) \forall 0 \le t \le h_n$. So $u(h_n) \le \frac{1}{h_n} \int_{0}^{h_n} u(t)dt$ Commented May 24, 2021 at 13:46
• @ParesseuxNguyen: Now I see it, excellent! Commented May 24, 2021 at 14:42

Prove me wrong, but I think the statement is false. Consider the example $$f(x)=x$$, which is clearly continuous and fulfills. For any $$x\in\mathbb{R}$$, $$\begin{gather*} \lim_{h\to0}\frac1{h^2}\int_{-h}^hf(x+t)t\,\mathrm dt=\lim_{h\to0}\frac1{h^2}\int_{-h}^h(x+t)t\,\mathrm dt=x\lim_{h\to0}\frac1{h^2}\int_{-h}^ht\,\mathrm dt+\lim_{h\to0}\frac1{h^2}\int_{-h}^ht^2\,\mathrm dt\\ = x \lim_{h\to0} \frac{1}{h^2}\left(\frac{h^2}{2}-\frac{h^2}{2}\right) + \lim_{h\to 0}\frac{1}{h^2}\left(\frac{h^3}{3} - \frac{-h^3}{3}\right) = \lim_{h\to 0 } \frac{2}{3} h = 0. \end{gather*}$$

• I'm very very sorry,it's $h^3$, I have edit Commented May 18, 2021 at 7:03

I can provide a straightforward answer, if $$f$$ is additionally differentiable. Then, there holds

$$\int_{-h}^hf(x+t)tdt = \int_{-h}^h (f(x)+f'(x)t+o(t))t dt = f(x)\left(\frac{h^2}{2}-\frac{h^2}{2}\right) + f'(x)\left(\frac{h^3}{3}-\frac{-h^3}{3}\right) + o(h^3) = f'(x) \frac{2}{3} h^3 + o(h^3).$$

Consequently,

$$0=\lim_{h\to0}\frac{1}{h^3}\int_{-h}^hf(x+t)tdt = \frac{2}{3} f'(x),$$

thus $$f$$ is constant. The general case can maybe treated with an argumentation similar to that of Marco Cantarini.

• I have the feeling that it is considerably more difficult if $f$ is not assumed to be differentiable. Marco Cantarini's argument had a flaw, he has deleted his answer in the meantime. Commented May 23, 2021 at 10:04
• polynomials are dense in the space of continuous functions $C([-1,1])$, and they are smooth, so using this probably you can prove the original result Commented Oct 23, 2022 at 15:33

If $$f(x)$$ is continuous in $$\textbf{R}$$, then we can write $$\int^{h}_{0}f(x+t)tdt-\int^{-h}_{0}f(x+t)tdt=\int^{h}_{0}(f(x+t)-f(x-t))tdt.$$ Hence from remark below (and for all $$x$$ real) we have $$0=\lim_{h\rightarrow 0}\frac{1}{h^3}\int^{h}_{0}(f(x+t)-f(x-t))tdt=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{3h}.\tag 1$$ Clearly about (1) we have: $$\lim_{h\rightarrow 0}h^{-1}\int^{h}_{0}(f(x+t)-f(x-t))tdt=0$$ Set $$F(y):=y^{-1}\int^{y}_{0}(f(x+t)-f(x-t))tdt,$$ then $$F(y)$$ is continuous and differentiatable in $$\textbf{R}$$ with $$F(0)=0$$ and $$F'(0)=0$$. This last follows from the fact that $$F(y)$$ is differentiable in all $$\textbf{R}-\{0\}$$ and from problem condition: $$\lim_{y\rightarrow 0}\frac{F(y)}{y}=0.$$ Also: $$\lim_{y\rightarrow 0}F'(y)=\lim_{y\rightarrow 0}\left(f(x+y)-f(x-y)-\frac{F(y)}{y}\right)=0.$$ All that since $$\lim_{h\rightarrow 0}\frac{1}{h^3}\int^{h}_{0}\left(f(x+t)-f(x-t)\right)tdt=0.\tag 2$$ Using the condition of the problem and Cauchy's mean value theorem (see below) we have $$0=\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=\lim_{y\rightarrow 0^{+}}\frac{F'(y)}{2y}=$$ $$=\lim_{y\rightarrow 0^{+}}\left(\frac{f(x+y)-f(x-y)}{2y}-\frac{1}{2y^3}\int^{y}_{0}(f(x+t)-f(x-t))tdt\right)=$$ $$=\lim_{y\rightarrow 0^{+}}\frac{f(x+y)-f(x-y)}{2y}=(f_{s})'_{+}(x).$$ Also in the same way $$\lim_{y\rightarrow 0^{-}}\frac{f(x+y)-f(x-y)}{2y}=(f_{s})'_{-}(x).$$ and $$(f_s)'_{-}(x)=(f_s)'_{+}(x)=f_s'(x)=\lim_{y\rightarrow 0}\frac{f(x+y)-f(x-y)}{2y}$$ exists. However there exists the

Theorem.

I) If for a function $$f(x)$$ the left $$f'_{-}(x)$$ and right $$f'_{+}(x)$$ derivatives exists, then $$f'_s(x)$$ exist and $$f'_s(x)=\frac{1}{2}(f'_{-}(x)+f'_{+}(x))$$.

II) If $$f_s'(x)$$ exist then it is not necessary that $$f'(x)$$ exist.

Hence we arrive to the result that $$f'_s(x)=0\textrm{, }\forall x\in\textbf{R}.\tag 3$$ However (3) is not so well established since (5) is not applied to functions that change sign infinite times in a neighborhood of $$0$$. For example $$F(y)=y^3\sin\left(\frac{1}{y}\right).$$ The true argument L'Hospital is that $$\textrm{liminf}_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}\leq 0\textrm{ and }\textrm{limsup}_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}\geq 0.\tag 4$$

Notes.

From Cauchy mean value theorem in $$[0,y]$$, we have that exists always $$\xi=\xi(y)$$: $$0<\xi(y) such that $$\frac{F(y)-F(0)}{y^2-0}=\frac{F'(\xi)}{2\xi}.$$ Hence when $$y\rightarrow 0^{+}$$, from the continuity of $$\frac{F'(t)}{2t}=\frac{f(x+t)-f(x-t)}{2t}-\frac{1}{2t^3}\int^{t}_{0}(f(x+t_1)-f(x-t_1))t_1dt_1,$$ in $$(0,y)$$ and knowing that $$\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=0$$, $$\lim_{y\rightarrow 0^{+}}\xi(y)=0$$ we have $$0=\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=\lim_{y\rightarrow 0^{+}}\frac{F'(\xi(y))}{2\xi(y)}=\lim_{t\rightarrow 0^{+}}\frac{F'(t)}{2t}.\tag 5$$ The same way is used when $$y\rightarrow0^{-}$$

• It seems to me that you are applying L'Hospital's rule in the wrong direction: You argue that because the limit $\lim_{h \to 0} \frac{F(h)}{G(h)}$ exists, the limit $\lim_{h \to 0} \frac{F'(h)}{G'(h)}$ exists as well. But L'Hospital's rule is the other way around. Commented May 23, 2021 at 16:54
• I think this is ok since it is given from the question. Commented May 23, 2021 at 16:59
• Please explain how you get $\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h} = 0$. Commented May 23, 2021 at 17:23
• ... and even if that were true, the next step is wrong because suddenly you assume that the limit is zero uniformly in $x$. Commented May 23, 2021 at 17:29
• @NikosBagis: The use of L'Hospital is not correct here, as Martin R pointed out, unless one assumes that $f$ is symmetrically differentiable everywhere. In that case, it is a deep result that $f$ found be constant (Theorem 5.2). In the generality of the OP ($f$ is simply continuous) some other argument may need to be used. Commented May 23, 2021 at 21:37