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.
 A: 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 $1.06$.
A: 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
A: 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 $|f|>1,a.e.$.
