Speed of convergence of Lebesgue integrable function as domain goes to zero 
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? 
 A: 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 $p=1$.
A: 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 $f\in L^p[0,1]$.
