Show that: $f(x)$ is a constant function 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 use the following lemma? But I can't it, thanks.
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 $h&gt;0$.
$$
\left|g(x)-\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds\right|=\frac{1}{2\,h}\left|\int_{x-h}^{x+h}(g(x)-g(s))\,ds\right|\le\frac{1}{2\,h}\int_{x-h}^{x+h}\left|g(x)-g(s)\right|\,ds.$$
Use that $g$ is continuous at $x$ to show that the last expression converges to $0$ as $h\to0$.
Define $F (x):= \int_{0}^x f(t)dt$
By integration by parts, we observe that:
$$\begin{align}\frac{1}{h}\int_{-h}^h f(x+t)tdt &amp;= \frac{1}{h}\int_{0}^h \big[ f(x+t)-f(x-t)]tdt\\&amp; = F(x+h)+F(x-h)  -  \frac{1}{h}\int_{0}^h \big[ F(x+t)+F(x-t) \big]dt
\end{align}$$
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:
$$T(G)(c)=0+0+\frac{4}{3}\epsilon&gt;0 $$
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 $\epsilon&gt;0$ 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.
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 $\xi=\xi(y)$: $0&lt;\xi(y)&lt;y$ 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 $y\rightarrow0^{-}$