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I'm preparing for an exam in measure theory and found this exercise:

For $p \in [1, \infty]$ find values of $\lambda$ such that $\lim\limits_{\varepsilon \to 0^{+}} \frac{1}{\varepsilon^{\lambda}} \int_{0}^{\varepsilon} f = 0$ for all $f \in L^{p}[0, 1]$.

By Hölder's inequality the limit is 0 if $\lambda$ < $\frac{p-1}{p}$ and $f(x)=x^{-\alpha}$, $\alpha$ < $\frac{1}{p}$ is a counterexample for the case $\lambda$ > $\frac{p-1}{p}$.

But I'm not sure about $\lambda =\frac{p-1}{p}$, can anyone help? 

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For $p=1$ we have $|\int_0^\epsilon f(x)\,dx| \le \int_{[0,\epsilon]} |f|\, dx \to 0$, so $\lambda = 0$ works for $p=1$. –  martini Apr 24 '12 at 7:05
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For $p=\infty$, if we take $f(x) = 1$, then the limit is $1$, so $\lambda = 1$ does not work for $p = \infty$. –  copper.hat Apr 24 '12 at 7:26
    
If $f$ is a "bit" more regular, then $$\lim_{\varepsilon \to 0} \varepsilon^{-1} \int_0^\varepsilon f=f(0).$$ I believe that $1-\lambda > 0$ is a necessary condition. –  Siminore Apr 24 '12 at 8:34

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up vote 3 down vote accepted

Let $E_p:=\{\lambda\in\mathbb R: \forall f\in L^p[0,1], \lim_{\varepsilon\to 0^+}\frac 1{\varepsilon^{\lambda}}\int_{[0,\varepsilon]}f(x)dx=0\}$.

  • First case: $p=1$: solved by martini and thomas.en, we have $E_1=(-\infty,0]$.
  • Second case: $1< p<\infty$: Indeed, using maps $x\mapsto x^{-\alpha}$ with $\alpha<\frac 1p$ we can see that $E_p\subset (-\infty,\frac{p-1}p]$, and Hölder's inequality shows that $(-\infty,\frac{p-1}p)\subset E_p$, so we have to determine whether $\frac{p-1}p\in E_p$.

Fix a function $f\in L^p[0,1]$, and put $f_n:=f\chi_{\{-n\leq f(x)\leq n\}}\in L^p[0,1]$. We can write \begin{align} \left|\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}f(x)dx\right|&\leq \frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}\left|f(x)-f_n(x)\right|dx+ \frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}\left|f_n(x)\right|dx\\\ &\leq \frac 1{\varepsilon^{\frac{p-1}p}}\lVert f-f_n\rVert_{L^p}\varepsilon^{\frac{p-1}p}+n\varepsilon^{\frac 1p}\\\ &= \lVert f-f_n\rVert_{L^p}+n\varepsilon^{\frac 1p} \end{align} so for each integer $n$ $$\limsup_{\varepsilon\to 0^+}\left|\frac 1{\varepsilon^{\frac{p-1}p}}\int_{[0,\varepsilon]}f(x)dx\right|\leq \lVert f-f_n\rVert_{L^p}.$$ Since by the monotone convergence theorem we have $$\lVert f-f_n\rVert_{L^p}^p=\int_{[0,1]}\chi_{\{|f|\geq n\}}|f|^p\to 0,$$ we conclude that $\frac{p-1}p$ works so for $1< p<\infty$ we have $E_p=\left(-\infty,\frac{p-1}p\right]$.

  • Third case: $p=+\infty$. As siminore showed, with $f=1$ we can see that if $\lambda\in E_{\infty}$ then $\lambda<1$, and if $\lambda<1$ then for $f\in L^{\infty}[0,1]$ we have $$\left|\frac 1{\varepsilon^{\lambda}}\int_{[0,\varepsilon]}f(x)dx\right|\leq \lVert f\rVert_{\infty}\varepsilon^{1-\lambda},$$ which converges to $0$ as $\varepsilon\to 0$. So $E_{\infty}=(-\infty,1)$.
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