2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ How would you start this problem?Do you think the method of contradiction would work here? $\endgroup$ Jul 15, 2020 at 15:53
  • $\begingroup$ I have tried this but it's not clear to me how to use the fact that the function is in $L_\infty(\mathbb{R})$. $\endgroup$
    – Nick
    Jul 15, 2020 at 16:03
  • $\begingroup$ 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?) $\endgroup$ Jul 15, 2020 at 16:05
  • $\begingroup$ Using absolute value on the integral the proof by reductio ad absurdum seems to be straightforward... $\endgroup$
    – DonAntonio
    Jul 15, 2020 at 16:34
  • 2
    $\begingroup$ @ DonAntonio The last equality is obviously incorrect because the integrand is positive (except $t=x$). $\endgroup$ Jul 15, 2020 at 18:05

3 Answers 3

2
$\begingroup$

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 $1.06$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Nick
    Jul 15, 2020 at 19:04
  • 1
    $\begingroup$ It sure does. In fact $f$ belongs to $L_{p,\mathrm{loc}}(\mathbf R)$ for any $0 < p < \infty$. $\endgroup$
    – Umberto P.
    Jul 15, 2020 at 19:06
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $|f|>1,a.e.$.

$\endgroup$

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.