Bound on integrable nonnegative function $F$ given inequality with compactly supported continuous functions.

Question

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.

Answer 1

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*}

Answer 2

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.