1
$\begingroup$

Let $f$ be an integrable function on $\mathbb{R}$ such that for every continuous $g$ on $\mathbb{R}$, $\displaystyle\int_\mathbb{R} fg \ dm=0$. Show that $f$ is zero a.e.

This is a qualifier sample problem and my idea is to proceed by contradiction. My problem here is that $f$ is not necessarily nonnegative so I'm having difficulties on establishing inequalities to arrive at a contradiction. Thanks!

$\endgroup$
7
  • 1
    $\begingroup$ Hint: You can always write a measurable function $f$ as $f =f^++f^-$ where $f^+(x)=\max\{0,f(x)\}$ and $f^-(x)=-\max\{0,-f(x)\}$. So $f\neq 0$ a.e. implies $f^+$ or $f^-$ must be non-zero on some set of positive measure. $\endgroup$
    – Dayton
    Apr 26, 2019 at 17:45
  • $\begingroup$ If we write $f$ in that form, I don't see why that works. $\endgroup$ Apr 26, 2019 at 18:06
  • $\begingroup$ $\int_\mathbb R fg~dm=0$ for every continuous $g$ implies that $\int_I f~dm=0$ over every interval $I$. Let $g$ be a bump with height 1 on $I$ and very steep slopes. If $f$ were to have some oscillatory/cancellation behavior, then it will need to oscillate infinitely fast to do so on every interval. Furthermore there must be a simple function that approximates $f$ on $I$. Such a simple function must be very small on any set of positive measure. Not rigorous, just some thoughts. $\endgroup$
    – jdods
    Apr 26, 2019 at 18:20
  • 1
    $\begingroup$ You can cheat; the assumptions imply that the Fourier transform $\hat{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx$ vanishes, and since the Fourier transform is injective, $f=0$. The cheating part is that you should prove that the Fourier transform is injective, of course. $\endgroup$ Apr 26, 2019 at 18:58
  • 1
    $\begingroup$ @GiuseppeNegro LOL. Now we are cheating, we could also say that $f$ is weakly differentiable with weak derivative $0$, so it must be constant a.e.. But then the only integrable constant is zero. Here we did all kind of cheating (and circle reasoning)... $\endgroup$
    – Shashi
    Apr 26, 2019 at 19:03

2 Answers 2

0
$\begingroup$

Hint: If $b\in L^\infty(\mathbb R),$ then there exists a sequence $g_n$ of bounded continuous functions on $\mathbb R$ such that $\|g_n\|_\infty\le \|b\|_\infty$ for all $n,$ and $g_n \to b$ pointwise a.e.

$\endgroup$
0
$\begingroup$

$\newcommand{\sign}{\operatorname{sign}}$By taking continuous functions $g$ with support in $(0,1)$, we can conclude that the statement is true for all functions $g\in C_c((0,1))$, i.e. functions with compact support in $(0,1)$, right? So let us focus on the function $f|_{(0,1)}$ on $(0,1)$ and forget about $\mathbb R$ for the moment.

Since $(0,1)$ equipped with the Lebesgue measure $m$ is a finite measure space, Lusin tells us that for every $\varepsilon>0$ there exists a compact set $K_\varepsilon\subset (0,1)$ such that $m((0,1)\setminus K_\varepsilon)<\varepsilon$ and a continuous function $h_\varepsilon$ with compact support such that $h_\varepsilon=\sign(f)$ on $K_\varepsilon$ and $\sup_{x\in (0,1)}|h_\varepsilon(x) |\leq 1$. This means for instance that we can take $g=h_\varepsilon$ and get $$0=\int_{(0,1)}fg\,dm=\int_{K_\varepsilon} |f|\,dm + \int_{(0,1)\setminus K_\varepsilon}fh_{\varepsilon}\,dm$$ Now we take a sequence of $\varepsilon$s, say $\varepsilon_n=1/n$, and apply DCT (HOW?) to get that $$\int_{(0,1)}|f|\,dm=0$$ which implies $f=0$ a.e. on $(0,1)$. But we could also take $(a,b)$ instead of $(0,1)$, right? Of course, try it yourself and see if you understand it.

$\endgroup$
1
  • $\begingroup$ Edited. I have added a crucial thing about $h_\varepsilon$. $\endgroup$
    – Shashi
    May 5, 2019 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.