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 $x$

Question

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 $h\to0$.

Answer 1

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.

Answer 2

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*}

Answer 3

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.

Answer 4

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 $\xi=\xi(y)$: $0<\xi(y)<y$ 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 $y\rightarrow0^{-}$