# Conditions of parameter $\lambda$ ensuring integral is 0

Let $1 \le p \le \infty$. I am seeking to find the values of the parameter $\lambda$ such that: $$\displaystyle \lim_{\epsilon \to 0^+} \frac{1}{\epsilon^\lambda} \int_{0}^{\epsilon} f = 0 \ \ \forall f \in L^{p}[0,1]$$

Any help will be greatly appreciated!

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• Let's figure out, for which $\lambda$, the limit will for sure go to zero. To see this, rewrite the above integral as $$\frac{1}{\epsilon^{\lambda}} \int_0^{\epsilon} f = \frac{1}{\epsilon^{\lambda}} \int_0^1 f(x) \chi_{[0,\epsilon]} \, dx,$$ where $\chi$ is the indicator function. Now apply a well-known inequality that allows you to relate the above expression to $\|f\|_{L^p}$. You should find that if $\lambda$ is small enough, we're good to go.
• If $\lambda$ is too large, we expect the limit won't go to zero, since we'll be dividing by a large power of a small number. So what remains to be shown is that, when $\lambda$ is larger than the threshold determined in the above step, the limit diverges for at least one $f$. So try to find a particular $f \in L^p$ such that the limit diverges. Hint: For which $\alpha$ is $x^{\alpha} \in L^p[0,1]$?
Using Holder I get that the original expression is $\le \frac{1}{\epsilon^{\lambda}}\|f\|_p$. It seems to me that no matter the value of $\lambda$ this thing will always be infinite. What am I missing? – user44069 Feb 14 '13 at 3:55
You're missing a term. What is the $L^q$ norm of $\chi_{[0,\epsilon]}$? – Christopher A. Wong Feb 14 '13 at 5:47
Would it be $\sqrt[q]{\epsilon}$? – user44069 Feb 14 '13 at 7:29
Yes! So Holder's should give you that your integral is bounded by $\epsilon^{-\lambda} \epsilon^{1/q} \|f\|_p$ And $1/p + 1/q = 1$, so... – Christopher A. Wong Feb 14 '13 at 9:30