Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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 $f\in L^p[0,1]$.

share|cite|improve this answer
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
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 $p=1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.