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My question is inspired by this one : Integral of the multiplication of two funtions

I wanted to find a counter example when $f>0$, but we have all latency we want for $g$.

Because when $f$ annulates or is negative on a small interval then it is easier to find a counter example by making $g^+,g^-,g^0$ disagree with $f$ values.


So here is the precise question :

  • given a function $f$ such that $f>0$ and continuous and $\displaystyle 0<\int_0^\infty f(t)dt<+\infty$
  • can we find $g$ such that $\displaystyle 0<\int_0^\infty g(t)dt<+\infty$ but $\displaystyle \int_0^\infty f(t)g(t)dt\le 0$ ?

There is no special requirements on $g$, it can be piecewise or we can smoothen it later.

My attempt at a solution is given below.

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zwim
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2 Answers2

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Sketch: Since $\int_0^\infty f(t)dt<+\infty$, the measure of $[f\leq 1/n]$ for each $n$ is infinite. So for any $n$ we can find a closed set $S_n$ disjoint of $[0,1]$ with measure 1 i.e. $|S_n|=1$ such that $f\leq 1/n$ on $S_n$.

Let $m>0$ be the minimum of $f$ on $[0,1]$, define $g(x)= -\frac{2}{m},\,\forall x \in [0,1]$. Therefore $\int_{0}^{1}fg \leq -2$.

Now define $g=\sqrt{n}$ on $S_n$ and zero elsewhere.

Claim: We can find $n$ such that $\int_{0}^{\infty}fg \leq -1$ and $0<\int_{0}^{\infty}g <\infty$.

Proof: First notice $\int_{0}^{\infty}fg =\int _{0}^{1}fg +\int_{S_n}fg\leq -2+\frac{1}{\sqrt{n}}$

Also, $\int_{0}^{\infty}g =\int _{0}^{1}g +\int_{S_n}g=-\frac{2}{m}+\sqrt{n}$. Now, it is evident that such an $n$ exists. $\blacksquare$

clark
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  • I read it thouroughly, and I'm convinced. This is a nice idea to work on $[f\le1/n]$ sets, I was wondering how to exploit the condition that the integral is finite in my attempt, and this is indeed an elegant way to do it, because we are not to worry about local behaviour. – zwim Mar 27 '17 at 06:02
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Here is my attempt :

$\displaystyle F(x)=\int_0^x f(t)\;dt$ is a continuous and increasing function between $0$ and $\displaystyle I=\int_0^\infty f(t)dt$.

Suppose $\frac{F(x)}{x}$ is constant on $\mathbb R^+$ then there would exists some $a>0$ such that $F(x)=ax$.

But then $f(x)=F'(x)=a$ and $\displaystyle \int_0^\infty f(x)dx=+\infty$.

Thus there exists $(\alpha,\beta)\in\mathbb (R^{+*})^2$ such that

$\frac{F(\alpha)}{\alpha}\neq\frac{F(\beta)}{\beta}\tag{E}$


Assume $\alpha<\beta$ and let's define $g$ piecewise constant like below :

$\begin{cases} \forall x\in[0,\alpha[, & g(x)=c \\ \forall x\in[\alpha,\beta[, & g(x)=-1 \\ \forall x\in[\beta,+\infty[, & g(x)=0 \end{cases}$

$\begin{array}{l} 0\ge\displaystyle \int_0^\infty f(x)g(x)dx= \displaystyle \int_0^{\alpha}(c)f(x)dx+\int_{\alpha}^{\beta}(-1)f(x)dx+\int_{\beta}^{\infty} (0)f(x)dx \\ \iff cF(\alpha)-(F(\beta)-F(\alpha))+0\le 0 \end{array}$

But we also want $\displaystyle \int_0^\infty g(x)dx=c\alpha-(\beta-\alpha)>0$.

This is possible if $\quad\big(\frac{\beta-\alpha}{\alpha}\big)<c\le\big(\frac{F(\beta)-F(\alpha)}{F(\alpha)}\big)$

The double condition on $c$ requires that $\frac{F(\alpha)}{\alpha}<\frac{F(\beta)}{\beta}$

Instead we can reverse the intervals where $g$ is negative :

$\begin{cases} \forall x\in[0,\alpha[, & g(x)=-c \\ \forall x\in[\alpha,\beta[, & g(x)=1 \\ \forall x\in[\beta,+\infty[, & g(x)=0 \end{cases}$

The condition on $c$ become $\quad\big(\frac{F(\beta)-F(\alpha)}{F(\alpha)}\big)\le c < \big(\frac{\beta-\alpha}{\alpha}\big)$

And if requires that $\frac{F(\alpha)}{\alpha}>\frac{F(\beta)}{\beta}$


But according to $(E)$ one of these two situations always happens, and the choice of a convenient $c$ is not an issue since $c=\frac{u+v}{2}$ always agrees the inequality $u<c<v$.

So we have succeeded in building a function $g$ that answers the question.

zwim
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  • My first attempt was unsuccessful, but I think I have finally found an answer, so I edited it. This comment is for people who may had come earlier and saw the failure, or for people who wonder why I post a question whose answer I know. – zwim Mar 27 '17 at 07:35
  • nice, this looks good to me – clark Mar 27 '17 at 07:44