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Problem: For $1\le p\le +\infty$, find the values of the parameter $\lambda$ for which $\mathop{lim}\limits_{\epsilon\to 0+} \frac1{\epsilon^\lambda} \int_0^\epsilon f = 0$ for all $f \in L^p[0, 1]$.

I have tried to use $H\ddot{o}lder$'s inequality to prove that every $\lambda<1/q,\ q$ is the conjugate of $p$, satisfies the equation:

$$\forall f \in L^p[0,1], \int_0^\epsilon f dm=\int_0^1 f\chi_{[0,\epsilon]}dm\le\Vert f\Vert_p\epsilon^{1/q}.$$

As for $\lambda>1/q$, let $f(x)=x^{-\frac1 {kp}},k>1$, it is easy to find that $\forall\lambda>1/q,$ there exists a $k_\star$ s.t. $\mathop{lim}\limits_{\epsilon\to 0+} \frac1{\epsilon^\lambda} \int_0^\epsilon f = +\infty.$

My problem is how to solve the condition of $\lambda=1/q.$ I 've thought it for several hours but have no idea. Could anyone help out? Thanks a lot in advance!

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$\lambda =\frac1q$ might not work, take for example $f\equiv 1$ and $p=\infty$, then $q=1$. We have $$\frac{1}{\epsilon} \int^\epsilon_0 f= 1$$

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  • $\begingroup$ I agree with you, but in my view $p$ seems no need to be $\infty$. Then how about other conditions? $\endgroup$
    – LonnerT
    May 23, 2020 at 5:51
  • $\begingroup$ But, you require the result to hold for all $p \in [1,\infty]$ and other conditions are satisfied. $\endgroup$
    – Mathronaut
    May 23, 2020 at 6:03
  • $\begingroup$ I think $p$ is a number given in advanced. If it holds for every $p$, there'd be no need to say $f \in L^p[0,1]$.(btw it is a problem from Royden's Real Analysis.) $\endgroup$
    – LonnerT
    May 23, 2020 at 6:09
  • $\begingroup$ I'm not saying $f \in \cap_{p\ge 1} L^p$, You are required to fix a $p$ in $[0,\infty]$ and then find all $\lambda$'s for which the limit is zero. So, to construct a counterexample for the critical case $\lambda =\frac1q$, you choose $p= \infty$, and then prove that $lambda$ is strictly less than $q=1$. $\endgroup$
    – Mathronaut
    May 23, 2020 at 6:27

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