A question about $f(x)\equiv 0$ If $f(x) \in C(-\infty+\infty)$, $\;g(x)=f(x) \int_0^x f(t)\,dt\,$ and $\;g(x)$ is monotone-decreasing in $(-\infty,+\infty),$
Prove:$f(x)\equiv 0$.
It is easy to get $g(0)=0$,and I'm thinking about proving $g(x)\equiv 0$,then $f(x)\equiv 0$ follows.
But I havn't any ideas to prove $g(x)\equiv 0$.
 A: Let
$$F(x) := \int_0^x f(t)\,dt.$$
Then we have
$$g(x) = \frac{d}{dx} \left(\frac{1}{2} F(x)^2\right).$$
Since $g$ is assumed to be monotonically decreasing (non-strictly), we have $g(x) \geqslant 0$ for $x < 0$ and $g(x) \leqslant 0$ for $x > 0$. Hence we have
$$\frac{1}{2} F(x)^2 \leqslant \frac{1}{2} F(0)^2$$
for $x < 0$ as well as for $x > 0$. But since $F(0)= 0$, it follows that $F\equiv 0$. From that, $f \equiv 0$ follows easily.
A: I am going to prove by contradiction that
if $\;x_0\;$ is a real number and if $\;f\;$ is a continuous function on $\;\left]-\infty,+\infty\right[\;$ such that $\;\int_\limits{x_0}^x f(t)dt=0\;\;\;\forall x\in\mathbb{R}\;,\;\;$ then
$f(x)=0\;\;\;\forall x\in\mathbb{R}$.
Proof:
If there were $\;x^*\in\mathbb{R}\;$ such that $\;f\left(x^*\right)\ne0\;$, without loss of generality, we could suppose that $\;f\left(x^*\right)>0\;$.
By continuity of the function $\;f\;$, it would exist $\;\delta>0\;$ for which $\;f(x)>0\;\;\;\;\forall x\in\left[x^*-\delta,x^*+\delta\right]$.
Hence $\;\;\int_\limits{x^*-\delta}^{x^*+\delta} f(t)dt>0\;.\color{blue}{\quad{(*)}}$
On the other hand we get that
$\int_\limits{x^*-\delta}^{x^*+\delta} f(t)dt=\int_\limits{x_0}^{x^*+\delta} f(t)dt-\int_\limits{x_0}^{x^*-\delta} f(t)dt=0\;,\color{blue}{\quad{(**)}}$
indeed, for hypothesis, it results that
$\;\int_\limits{x_0}^x f(t)dt=0\;\;\;\forall x\in\mathbb{R}$.
But $\;(**)\;$ contradicts $\;(*)\;$.
So it is not possible that there exists some $\;x^*\in\mathbb{R}\;$ for which $\;f\left(x^*\right)\ne0\;$, otherwise it would lead to a contradiction.
Hence,
$f(x)=0\;\;\;\forall x\in\mathbb{R}$.
