# Convergence of $\Vert f \Vert_p$ with respect to $p$

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!

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Uh, in general $\|f\|_p$ does not converge to $\|f\|_\infty$. This characterisation of $L^\infty$ requires you divide by an appropriate factor of the measure of the total space. (In particular, it is only true as stated if the the total measure is 1.) See Problem 7.1 in Gilbarg and Trudinger's Elliptic Partial Differential Equations of Second Order. Part (iii) of that problem is the correctly stated version of the convergence you are thinking about. – Willie Wong Nov 10 '10 at 10:46
Thanks. Just added under probability measure. But I cannot thing out a way to show it is true... – steveO Nov 10 '10 at 10:52
One can prove a version of the first statement (without using that the space has finite measure) if $f \in L^{p_0} \cap L^\infty$ for some $1 \leq p_0 < \infty$. To see this just write $|f|^p = |f|^{p_0} |f|^{p - p_0}$ and do some syntax juggling. – Jonas Teuwen Nov 10 '10 at 12:33
Just for reference, this is Exercise 5 part d in Chapter 3 on page 71 of Rudin's Real and Complex Analysis, 3rd edition. – Jonas Meyer Nov 10 '10 at 17:37

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 $\text{exp}(\int log |f|)$. – Jonas Teuwen Nov 10 '10 at 12:30