The following is a problem from Royden and Fitzpatrick's Real Analysis book.
Find the values of the parameter $\lambda$ for which $$ \lim\limits_{\epsilon\rightarrow0^{+}} \frac{1}{\epsilon^\lambda}\int_0^\epsilon f(x)dx = 0 \;\;\; \forall f \in L^p[0,1] $$
By Holder's inequality,
$$ \frac{1}{\epsilon^\lambda}\int_0^{\epsilon}f(x)dx \leq \frac{1}{\epsilon^\lambda}||f||_p\epsilon^{1-1/p} = ||f||_p\epsilon^{1-1/p-\lambda} $$
So if $\lambda < 1-1/p$, the limit holds. For the case, $\lambda > 1-1/p$ we can consider the following counterexample, $f(x) = \lambda x^{\lambda-1}$. It is simple to see that $f \in L^p[0,1]$, but, $$\frac{1}{\epsilon^\lambda}\int_0^\epsilon \lambda x^{\lambda-1}dx = 1 $$
The case I'm having difficulty with is the borderline one, $\lambda = 1 - \frac{1}{p}$. Obviously Holder will not work here. The above counterexample doesn't work here either.
Should the limit hold in this case? If not, can someone provide a counterexample?
Holder's inequality also gives the bound $\|fI_{(0,\epsilon)}\|_p \epsilon^{1-1/p-\lambda}$ so the limit is zero if $\lambda =1-\frac 1 p$.