letumsnemesislh

2022-07-01

I'm trying to solve the following problem.

Let $f$ be an integrable function in (0,1). Suppose that

${\int}_{0}^{1}fg\ge 0$

for any non negative, continuous $g:(0,1)\to \mathbb{R}$. Prove that $f\ge 0$ a.e. in (0,1).

I'm a little unsure on what it is that I must prove in order to conclude that $f\ge 0$. I tried to show that ${\int}_{0}^{1}{f}^{2}\ge 0$ but I couldn't get very far.

I'm seeking hints on how to solve this. Thanks.

Let $f$ be an integrable function in (0,1). Suppose that

${\int}_{0}^{1}fg\ge 0$

for any non negative, continuous $g:(0,1)\to \mathbb{R}$. Prove that $f\ge 0$ a.e. in (0,1).

I'm a little unsure on what it is that I must prove in order to conclude that $f\ge 0$. I tried to show that ${\int}_{0}^{1}{f}^{2}\ge 0$ but I couldn't get very far.

I'm seeking hints on how to solve this. Thanks.

Nirdaciw3

Beginner2022-07-02Added 20 answers

Suppose that $A\subset (0,1)$ is measurable, is of positive measure and $f<0$ on $A$. The idea is that we want to construct a continuous function $g$ such that

${\int}_{0}^{1}fg\phantom{\rule{thinmathspace}{0ex}}dx<0.$

A logical way to do this would be to choose $g$ such that $g\u2a7e0$ in $A$ and $g=0$ on $(0,1)\setminus A$. However, since $A$ is only a measurable set, in general $g$ will be discontinuous.

There exists a (relatively) closed set $F\subset A$ such that $|A\setminus F|<\u03f5$. Choosing $\u03f5=|A|/2>0$, we have that

$|F|=|A|-|A\setminus F|=|A|/2>0.$

Since $|F|>0$, the interior of $F$ is nonempty. Thus, there exists an open set $U$ compactly contained in the interior of $F$ (just take a small ball for example). Define $g$ such that $g$ is continuous, nonnegative, $g=0$ in $A\setminus F$, and $g=1$ in $U$. Then

$\begin{array}{rl}{\int}_{0}^{1}fg\phantom{\rule{thinmathspace}{0ex}}dx& ={\int}_{F}fg\phantom{\rule{thinmathspace}{0ex}}dx\\ & ={\int}_{U}f\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{F\setminus U}fg\phantom{\rule{thinmathspace}{0ex}}dx\\ & \u2a7d{\int}_{U}f\phantom{\rule{thinmathspace}{0ex}}dx\\ & <0.\end{array}$

This completes the proof.

${\int}_{0}^{1}fg\phantom{\rule{thinmathspace}{0ex}}dx<0.$

A logical way to do this would be to choose $g$ such that $g\u2a7e0$ in $A$ and $g=0$ on $(0,1)\setminus A$. However, since $A$ is only a measurable set, in general $g$ will be discontinuous.

There exists a (relatively) closed set $F\subset A$ such that $|A\setminus F|<\u03f5$. Choosing $\u03f5=|A|/2>0$, we have that

$|F|=|A|-|A\setminus F|=|A|/2>0.$

Since $|F|>0$, the interior of $F$ is nonempty. Thus, there exists an open set $U$ compactly contained in the interior of $F$ (just take a small ball for example). Define $g$ such that $g$ is continuous, nonnegative, $g=0$ in $A\setminus F$, and $g=1$ in $U$. Then

$\begin{array}{rl}{\int}_{0}^{1}fg\phantom{\rule{thinmathspace}{0ex}}dx& ={\int}_{F}fg\phantom{\rule{thinmathspace}{0ex}}dx\\ & ={\int}_{U}f\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{F\setminus U}fg\phantom{\rule{thinmathspace}{0ex}}dx\\ & \u2a7d{\int}_{U}f\phantom{\rule{thinmathspace}{0ex}}dx\\ & <0.\end{array}$

This completes the proof.

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