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For a Lebesgue measurable function $f$, $\Vert f \Vert_p$ converges to $\Vert f \Vert_\inf$ as $p$ goes to infinity.

I was wondering if $\Vert f \Vert_p$ converges to $\exp(\int \log |f|)$ as $p$ goes to $0$ for probability measure. How to prove this?

Thanks!

  • 0
    Just for reference, this is Exercise 5 part d in Chapter 3 on page 71 of Rudin's *Real and Complex Analysis*, 3rd edition.2010-11-10

2 Answers 2

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As my approach is different to that of Willie Wong, I give it here:

Assume $f$ is in some $L^{p_0}$.

Let $(p_n)$ be such that $0 < p_n < p_0$ and $p_n \downarrow 0$. Define:

$g_n(x) = \frac{1}{p_0} (|f(x)|^{p_0} - 1) - \frac{1}{p_n} (|f(x)|^{p_n} - 1).$

Recall that $\frac{1}{p} (x^p - 1) \downarrow \log x$ as $p \downarrow 0$ for all x > 0. So let the $g_n \to g$. Then by the Monotone Convergence Theorem we have that $\int g_n \uparrow \int g.$

This implies that $\int \frac{1}{p_n} (|f|^{p_n} -1) \downarrow \int \log |f|.$

Now, by Jensen's inequality we have

$\text{exp} \left ( \int \log |f| \right ) \leq \left ( \int |f|^{p_n} \right )^\frac{1}{p_n}$

and by the inequality $|t| \leq \text{exp}(|t| - 1)$ we get

$\left ( \int |f|^{p_n} \right )^\frac{1}{p_n} \leq \text{exp} \left ( \int \frac{1}{p_n} (|f|^{p_n} - 1) \right ).$

So, this together with our previous claim proves the theorem.

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HINT

Approximate by non-negative step functions (if $f$ vanishes on a set of positive measure, using a Hölder inequality you can argue that $\|f\|_p \leq C |\mathrm{supp}(f)|^{\frac1p - 1} \searrow 0$). For step functions, that $\|f\|_p\geq \exp \int \log |f|$ is a consequence of the arimetic-geometric mean inequality.

The AM-GM inequality becomes an equality when all its terms are equal. For a non-negative step function $f$, as $p\to 0$, you have that $f \to 1$. By continuity you can quantify the difference between AM-GM. So you have that the convergence is true for step functions.

To finish off you use the usual diagonal argument to extract a subsequence from the approximating sequence of step functions, together with a suitable sequence of $p$'s.

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    I would suggest taking a sequence $(p_n)$ such that 0 < p_n < p_0 with $p_n \downarrow 0$ and then define something like $f_n(x) = 1/p_0 (|f(x)|^{p_0} - 1) - 1/p_n (|f(x)|^{p_n} - 1)$ and use that $1/p (t^p - 1)$ decreases to $\log t$ as $p \to 0$ for all t > 0. Then note that $\int f_n \uparrow \int f$. Now we can bound $(\int |f|^{p_n})^{1/p_n}$ in both directions by things that go to the required $\text{exp}(\int log |f|)$.2010-11-10