# Prove that $|f| \geq 1$ a.e.

Let $$f \in L_\infty(\mathbb{R})$$ be a function such that $$\int_{(x-a,x+a)} |t-x|^{-\frac{1}{4}} f(t) dt \geq \sqrt8 a^{\frac{3}{4}}$$ for every $$x \in \mathbb{R}$$ and $$a > 0$$. Prove that $$|f| \geq 1$$ a.e.

I have tried contradiction but I am not able to see how the $$L_\infty(\mathbb{R})$$ condition is being used here. Any help would be appreciated.

• How would you start this problem?Do you think the method of contradiction would work here? Jul 15, 2020 at 15:53
• I have tried this but it's not clear to me how to use the fact that the function is in $L_\infty(\mathbb{R})$.
– Nick
Jul 15, 2020 at 16:03
• I think of this : on each finite interval that $|t-x|^{-\frac 14}$ and $f$ are integrable on, maybe Cauchy Schwarz (You are using $f \in L_{\infty}$ since that is sufficient for it to be integrable over a finite interval)? I can't think of a better way to get the integral of $f$ (or something) from the given condition. Remember, if you show that the integral of $|f|$ over any finite interval is at least the length of that interval, you are done(why?) Jul 15, 2020 at 16:05
• Using absolute value on the integral the proof by reductio ad absurdum seems to be straightforward... Jul 15, 2020 at 16:34
• @ DonAntonio The last equality is obviously incorrect because the integrand is positive (except $t=x$). Jul 15, 2020 at 18:05

Use the Lebesgue differentiation theorem. First, Cauchy-Schwarz gives you $$\int_{x-a}^{x+a} \frac{f(t)}{|t-x|^{1/4}} \, dt \le \left( \int_{x-a}^{x+a} \frac{1}{|t-x|^{1/2}} \, dt \right)^{1/2} \left( \int_{x-a}^{x+a} f(t)^2 \, dt \right)^{1/2}.$$ You can calculate $$\int_{x-a}^{x+a} \frac{1}{|t-x|^{1/2}} \, dt = 4 a^{1/2}$$ so that $$\int_{x-a}^{x+a} \frac{f(t)}{|t-x|^{1/4}} \, dt \le 2a^{1/4} \left( \int_{x-a}^{x+a} f(t)^2 \, dt \right)^{1/2} = \sqrt{8} a^{3/4} \left( \frac{1}{2a}\int_{x-a}^{x+a} f(t)^2 \, dt \right)^{1/2}.$$ In light of the assumption on the integral this gives you $$\frac{1}{2a}\int_{x-a}^{x+a} f(t)^2 \, dt \ge 1$$ for all $$x$$ and for all $$a > 0$$. The differentiation theorem tells you that $$\lim_{a \to 0^+} \frac{1}{2a}\int_{x-a}^{x+a} f(t)^2 \, dt = f(x)^2$$ almost everywhere, and at any point $$x$$ where this limit holds you find $$f(x)^2 \ge 1$$.

Now that the question has been answered let's try to see if an improvement is possible. Consider conjugate indices $$p$$ and $$q$$ with $$1 \le q < 4$$. Holder's inequality gives you $$\int_{x-a}^{x+a} \frac{f(t)}{|t-x|^{1/4}} \, dt \le \left( \int_{x-a}^{x+a} \frac{1}{|t-x|^{q/4}} \, dt \right)^{1/q} \left( \int_{x-a}^{x+a} |f(t)|^p \, dt \right)^{1/p}.$$ Again you can calculate $$\int_{x-a}^{x+a} \frac{1}{|t-x|^{q/4}} \, dt = \frac{2a^{1-\frac q4}}{1 - \frac q4}$$, and in tandem with $$\left(\int_{x-a}^{x+a} f(t)^p \, dt \right)^{1/p} = (2a)^{\frac 1p} \left( \frac 1{2a}\int_{x-a}^{x+a} |f(t)|^p \, dt \right)^{1/p}$$ arrive at the inequality $$\sqrt{8} a^{\frac 34} \le \frac{2a^{3/4}}{(1 - \frac q 4)^{1/q}}\left( \frac 1{2a}\int_{x-a}^{x+a} |f(t)|^p \, dt \right)^{1/p}.$$ The factors of $$a^{\frac 34}$$ cancel, and upon letting $$a \to 0^+$$ you get $$|f(x)| \ge \frac{\sqrt 8 (1 - \frac q4)^{1/q}}{2}$$ almost everywhere. When $$q = 2$$ this is the bound previously obtained. Taking $$q$$ very close to $$1$$ you can get a lower bound slightly larger than $$1.06$$.

• Very clear answer! To use the Lebesgue differentiation theorem, we need $f(t)^2 \in L_{1,\text{loc}}(\mathbb{R})$. Does this follow from the fact that $f$ is bounded?
– Nick
Jul 15, 2020 at 19:04
• It sure does. In fact $f$ belongs to $L_{p,\mathrm{loc}}(\mathbf R)$ for any $0 < p < \infty$. Jul 15, 2020 at 19:06

First take a $$u-$$subsitution of $$u = t-x$$ to rewrite the integral as $$\int_{-a}^a |u|^{-1/4}f(u+x)dx \geq \sqrt{8}a^{3/4} > 0$$ The function $$|u|^{-1/4}$$ is an even function on a symmetric domain, since the above needs to be positive, then the odd the odd part of the function is annihilated. We can proceed on the assumption that $$f$$ must be an even function. We can thus use that $$f$$ is an even function to deduce that for all $$a>0$$ we have $$\int_0^a |u|^{-1/4}f(u+x)dx \geq \sqrt{2}a^{3/4} > 0$$ Notice that the set $$\{(0,a), [0,a), [0,a], (0,a]: a >0\}$$ generates $$\mathcal{B}([0,\infty))$$. Since the integral of $$|u|^{-1/4}f(u+x)$$ is nonnegative on a set which generates the Borel set on $$[0,\infty)$$, we can prove that the integral must be nonnegative on all Borel sets on $$[0,\infty)$$. As a consequence of this, we can realize that $$|u|^{-1/4}f(u+x)\geq 0$$ a.e., then we have that $$f \geq 0$$ a.e.

Now I am stuck :P

Note $$\int_{x-a}^{x+a}|x-t|^{-1/4}f(t)dt=\int_{0}^a\frac{f(x-t)+f(t+x)}{t^{1/4}}dt\geq 2\sqrt{2}a^{3/4},$$ we have \begin{align} \int_{0}^a\frac{f(x+t)+f(t-x)-3\sqrt{2}/2}{t^{1/4}}dt\geq 0. \end{align} Since $$a\geq 0$$, $$\frac{1}{a}\int_{0}^a\left({f(x+t)+f(t-x)-3\sqrt{2}/2}\right)d\mu(t)\geq 0,$$ Then since $$f\in L^\infty(d\mu)$$, by Dominated convergence theorem, take limit $$a\to 0$$, we have $$2f(x)\geq \frac{3\sqrt{2}}{2},~a.e.,$$ thus $$f(x)\geq \sqrt{\frac{9}{8}}>1,a.e.$$.

So of course we have $$|f|>1,a.e.$$.