# Convergence of $L^p$ means and/or norms

Dear all,

I have a question about convergence of $L^p$-means. It can be shown (Inequalities, Theorem 193, Hardy, Littlewood, Polya) that

$\forall f \in L^p(D,\mu)\cap L^\infty(D,\mu), M_p(f) = \left( \frac{\int_D|f|^pd\mu}{\mu(D)} \right)^{1/p} \to \|f\|_\infty, \textrm{ for } p \to \infty$

Translating this in terms of convergence, we have the follwings:

$\forall \epsilon>0, \forall f, p_0 = p_0(f,\epsilon): \forall p\geq p_0, |M_p(f)-\|f\|_\infty| < \epsilon$

or equivalently, $\forall \epsilon>0, \forall f \neq 0, p_0 = p_0(f,\epsilon): \forall p\geq p_0, |\frac{M_p(f)}{\|f\|_\infty} - 1| < \epsilon$

Now, here is my question: given a suitable $f$ and an $\epsilon$, is there any way to computes/estimate this $p_0$ coming in the convergence property? If not, is there a way to add some constraints to make it possible? (for instance, $f$ should be defined on a bounded domain, or $f$ should have 0 mean: $\int f = 0$ or any other things)

Thanks a lot for any clue/references

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Have you tried working through the proof? It seems to me that if you go through it carefully, you can derive a rather explicit estimate of $p_0$ for a given $f$ and $\epsilon$. –  Deane Yang Mar 26 '12 at 9:06
I had another look at the proof, and what is said, is that $M_p(f) \geq (\|f\|_\infty - \epsilon)(\mu(\{x:f(x) \geq \|f\|_\infty-\epsilon\}))^{1/p}$, which is not exactly done... or did I miss something? Thanks anyway! –  Jean-Luc Bouchot Mar 26 '12 at 14:13
Yes, that formula looks pretty good. You should be able to estimate $p_0$ using it. –  Deane Yang Mar 26 '12 at 19:51
Oh yes... I feel really stupid, I was focusing on something else, and didn't see that... Thanks for pointing this out! Helped me a lot. –  Jean-Luc Bouchot Mar 27 '12 at 10:34