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This question is related to Exercise 3.3.7(b) in Cohn, Measure Theory, 2nd edition, which reads as follows:

Let $(X, \mathcal A, \mu)$ be a finite measure space, and let $f$ be an $\mathcal A$-measurable real- or complex-valued function on $X$. Show that if $f \in \mathcal L^{\infty}(X, \mathcal A, \mu)$, then $\|f\|_{\infty} = \lim_{p \to \infty}\|f\|_p$.

I was able to solve this. I was then curious to know whether convergence is necessarily monotonic.

Using $f = 1$ (the constant function), I observed that $\|f\|_p = \mu(X)^{1/p}$, so $\|f\|_p \uparrow \|f\|_{\infty}$ if $0 < \mu(X) \leq 1$, whereas $\|f\|_p \downarrow \|f\|_{\infty}$ if $\mu(X) \geq 1$. Therefore monotonic convergence in either direction is certainly possible.

By experimenting with step functions of the form $a\chi_{[0, \alpha]} + b \chi_{(\alpha, \beta]}$, I was able to find an example where convergence is not monotonic: for this example, as $p$ increased, initially $\|f\|_p$ decreased, then subsequently increased to converge to $\|f\|_{\infty}$. However, I was only able to find such an example when the sum of the interval lengths exceeded $1$ (i.e. $\mu(X) > 1$)

I have not found a nonmonotonic example where $X$ is an interval of length $1$ or smaller, even by writing a Matlab script to generate random step functions on this interval. In all cases, $\|f\|_p \uparrow \|f\|_{\infty}$ when $X = [0,1]$.

This leads me to speculate that

If $\mu(X) \leq 1$, then $\|f\|_p \uparrow \|f\|_{\infty}$ as $p \to \infty$.

To prove this, it would suffice to show that $1 \leq p < q < \infty$ implies $\|f\|_p \leq \|f\|_q$ when $\mu(X) \leq 1$. But I haven't been able to show this. Is it true?

I know that for any finite measure space, $p \leq q$ implies $\mathcal L^q \subset \mathcal L^p$, but that doesn't necessarily mean that $\|f\|_p \leq \|f\|_q$ (at least if $\mu(X) > 1$), as my examples have shown.

My guess is that there is some clever Hölder inequality manipulation which will give me what I want, but I haven't spotted it yet.

1 Answers1

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Yes; apply Holder's inequality (you can also use Jensen if you like): \begin{align*} \|f\|_{L^p}=\|f\cdot1\|_{L^p}\leq\|f\|_{L^q}\|1\|_{1/(1/p-1/q)}=\|f\|_{L^q}\mu(X)^{1/(1/p-1/q)}\leq\|f\|_{L^q} \end{align*} since we assumed $\mu(X)<1$, and $1/(1/p-1/q)\geq0$ from $p\leq q$.

Edit: Please note I used a slightly generalized version of the Holder inequality for the case where a product is being estimated in $L^p$ for $p$ not necessarily $1$. This is described here.

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Funktorality
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  • My Hölder manipulation skills aren't what they should be. How do you get a $p$-norm on the left hand side of the Hölder inequality? i.e. how do I start with $|fg|_1 \leq |f|_r |g|_s$ (where $1/r + 1/s = 1$) to obtain $|f|_p$ on the LHS? –  Jul 09 '16 at 06:20
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    @Bungo \begin{align} \int f^p &\leq \left(\int (f^p)^{q/p}\right)^{p/q}\left(\int 1^{1/(1-p/q)}\right)^{1-p/q} \ &= |f|_q^p\mu(X)^{1-p/q} \end{align} – Ningxin Jul 09 '16 at 06:21
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    There's a more general version of the Holder inequality that says $|fg|{L^r}\leq|f|{L^p}|g|_{L^q}$ when $1/r=1/p+1/q$. – Funktorality Jul 09 '16 at 06:23
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    @Bungo I edited my answer to include a MSE link where they derive the more general inequality from the basic one. – Funktorality Jul 09 '16 at 06:29
  • @QiyuWen Ah, that's what I was missing. Thanks! –  Jul 09 '16 at 06:34
  • @StanCoreyCarter I think I've seen that more general version before, but I had forgotten about it. Thanks for the help! –  Jul 09 '16 at 06:34