Full Question:

Suppose that $F$ is a nonnegative function that is integrable on $\mathbb R$ and there is a constant $C$ such that

$\int_\mathbb R Ff \leq C\int_\mathbb R f$

whenever $f$ is a nonnegative continuous function on $\mathbb R$ having compact support. Prove that $F(x) \leq C$ for almost all $x$.

My thoughts:

I feel like I should use the density of compactly supported continuous functions in $L_1$, but I'm having a tough time getting on the right track. Any help would be appreciated.


This is cheap but here you go:

It's easy to see that indicator functions of compact intervals can be approximated pointwise by nonnegative compactly supported continuous functions$^*$. Thus (by dominated convergence for instance) your assumed inequality extends to such functions. That is, for any interval $I$, \begin{align*} \int\chi_IF\leq C\int\chi_I \end{align*} which implies \begin{align*} \frac{1}{|I|}\int_IF\leq C \end{align*} where $\chi$ is the indicator function. You can conclude by the Lebesgue differentiation theorem.

*If this isn't obvious, what you should do is take $\epsilon\to0$ in the continuous function \begin{align*} \chi_{[a,b],\epsilon}(x)=\left\{\begin{array}{ll}0&x<a-\epsilon\\\text{interpolation}&a-\epsilon\leq x<a+\epsilon\\ 1&a+\epsilon\leq x<b-\epsilon\\\text{interpolation}&b-\epsilon\leq x<b+\epsilon\\ 0&b+\epsilon<x\end{array}\right.. \end{align*}

  • 1
    $\begingroup$ "can be approximated pointwise by nonnegative continuous functions" You should probably add "with compact support". $\endgroup$ – zhw. Jun 17 '16 at 19:07
  • $\begingroup$ Oops you're right. $\endgroup$ – Funktorality Jun 17 '16 at 19:41

We can get the result, with a little more work, without the Lebesgue differentiation theorem.

Suppose $K\subset \mathbb R$ is compact. Choose a bounded open interval $I$ containing $K.$ Then there are continuous functions $f_n:\mathbb R\to [0,1]$ with support in $I$ such that $f_n \to \chi_K$ pointwise everwhere. Therefore

$$\tag 1 \int_{\mathbb R} F\chi_K = \lim \int_{\mathbb R} F\cdot f_n \le \lim C\int_{\mathbb R} f_n = C \int_{\mathbb R} \chi_K = C m(K).$$

We have used the dominated convergence theorem twice here; once for the first equality, where the dominating function is $F,$ and again on the second equality, where the dominating function is $\chi_I.$

Now suppose $m(\{F > C\}) > 0.$ Since $\{F > C\} = \cup_k \{F > C+1/k\},$ there exists $C'>C$ such that $m(\{F > C'\}) > 0.$ By inner regularity, there is a compact $K\subset \{F > C'\}$ of positive measure. We then get

$$\int_{\mathbb R} F\chi_K \ge C'm(K),$$

violating $(1),$ contradiction. Thus $F\le C$ a.e.


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.