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Let $f$ be a $L^{p}$ function on $\mathbb{R}$. If $p>4/3$, prove that:

$\lim_{t\rightarrow0^{+}}\int_{0}^{t}x^{-1/4}f(x)dx=0.$

The natural procedure here is to bound the integral by using Hölder's inequality; after using the Hölder inequality, we get that as $t\rightarrow0^{+}$ the integral from $0$ to $t$ of $\left|x^{-1/4}f(x)\right|$ goes to zero. Does the desired result follow immediately from the Dominated Convergence Theorem?

2 Answers 2

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If $pq=p+q$, then $q<4$, so $x^{-1/4}$ is in $L^q$ on $(0,t)$ for all $t>0$. Hence you can apply Hölder's inequality as you indicated to conclude that $x^{-1/4}f(x)$ is integrable on each $(0,t)$.

Yes, you can apply dominated convergence. For example, the integrable function $x^{-1/4}|f(x)|\chi_{(0,1)}(x)$ dominates the functions $x^{-1/4}|f(x)|\chi_{(0,t)}(x)$ for $t<1$, which converge pointwise to $0$ as $t\to 0$.

But you could also apply the fact that if $g$ is integrable, then for all $\varepsilon>0$ there exists $\delta>0$ such that $m(A)<\delta$ implies $|\int_A g|<\varepsilon$. In other words, you could apply absolute continuity of the integral.

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To use the dominated convergence theorem, you have to show that the integrands $x^{-1/4}|f(x)|\chi_{(0,t)}$ are bounded by an integrable function, and this essentially means that you have to show $\int_0 ^\varepsilon x^{-1/4} |f(x)|dx < \infty$ for an $\varepsilon>0$ of your choice. This appears slightly weaker than your claim, but you still need Hölder's inequality to show this.