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

Can someone help do this?

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2 Answers 2

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By Holder's inequality, we have $$\int_0^\epsilon f\leq\big(\int_0^1f^p\big)^{\frac{1}{p}}\big(\int_0^\epsilon\big)^{\frac{1}{q}}=\epsilon^{{\frac{1}{q}}}\big(\int_0^1f^p\big)^{\frac{1}{p}}$$ where $\frac{1}{q}=1-\frac{1}{p}=\frac{p-1}{p}.$ Hence, if $\lambda<\frac{1}{q}=\frac{p-1}{p},$ then $$0\leq\lim\limits_{\epsilon \to 0^{+}} \frac{1}{\epsilon^{\lambda}} \int_{0}^{\epsilon} f\leq\lim\limits_{\epsilon \to 0^{+}}\epsilon^{\frac{1}{q}-\lambda}\big(\int_0^1f^p\big)^{\frac{1}{p}}= 0$$ since $f\in L^p[0,1]$.

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Does the convergence hold for other values though? My instinct is no but I'm having a hard time figuring out how to prove it... –  Red Rover Nov 20 '11 at 7:21
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To complete the proof, consider $f(x)=x^{-a}$ with $a\lt1/p$. Then the integral is of the order of $\epsilon^{1-a}$ hence you need that $1-a\gt\lambda$ for every $a\lt1/p$, that is, that $\lambda\leqslant1/q$. –  Did Nov 20 '11 at 9:15
    
Thanks Didier Piau! –  Red Rover Nov 20 '11 at 11:05
    
Red: A previous comment was marred by misprints. I turned it into an answer. –  Did Nov 20 '11 at 13:23

Let $H_\epsilon(f,\lambda)=\epsilon^{-\lambda}\int\limits_0^\epsilon f$. For every $a\gt0$, let $f_a(x)=1/x^{a}$ and $g_a(x)=1/(x^{a}\log(x))$.

Let us first consider the case $p\gt1$.

As mentioned and proved by Paul, if $\lambda\lt1-1/p$, $H_\epsilon(f,\lambda)\to0$ for every $f$ in $L^p$.

On the other hand, if $\lambda\gt1-1/p$, $f_a$ is in $L^p$ for every $a\lt1/p$ but $H_\epsilon(f_a,\lambda)$ is of the order of $\epsilon^{-\lambda}\times\epsilon^{1-a}$ hence $H_\epsilon(f_a,\lambda)\to\infty$ if $a\gt1-\lambda$. Since $1-\lambda\lt1/p$, both conditions can be met by the same parameter $a$.

If $\lambda=1-1/p$, $g_{1/p}$ is in $L^p$ but $H_\epsilon(g_{1/p},\lambda)\to\infty$.

Turning to the case $p=1$, note that $H_\epsilon(f,0)\to0$ for every $f$ in $L^1$ but for every $\lambda\gt0$ and $1-\lambda\lt a\lt1$, $H_\epsilon(f_a,\lambda)\to\infty$.

Finally, the suitable values of $\lambda$ are $\lambda\lt1-1/p$ for $p\gt1$ and $\lambda=0$ if $p=1$.

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