Question: This is the last part of a 5 part question I am working on. Let $(X,\mu)$ be a possibly infinite measure space. Assume $\exists r < \infty$ with $\|f\|_r < \infty$ and that $\|f \|_{\infty} < \infty$. Show that $\lim_{p \rightarrow \infty} \|f_p\| = \|f\|_{\infty}$.

This is from Real and Complex by Rudin, chapter 3 exercise 14.

Progress: I have shown that $\|f\|_{\infty} \le \lim_{p \rightarrow \infty} \|f\|_p$ as follows,

Fix $\epsilon > 0$. Let $E = \{x : |f(x)| > \|f\|_{\infty} - \epsilon \}$. Then observe $$ \|f\|_p \ge \left( \int_{E} |f|^p d\mu \right)^{1/p} > \left( \int_{E}(\| f \|_{\infty} - \epsilon)^{p} d\mu \right)^{1/p} = \left( \|f\|_{\infty} - \epsilon \right) \mu(E)^{1/p}, $$ thus, $\lim_{p \rightarrow \infty} \|f\|_p \ge \|f\|_{\infty} - \epsilon$ since $\mu(E) < \infty$.

I attempted something similar for the other direction, but could not say the measure of a set was finite like (I think) I need for this argument to work. Here is what I tried:

Since $\|f\|_r < \infty, \exists R$ so that $|x| > R \implies f(x) < \frac{1}{2}$. Let $A = \{ x : |x| \le R \}$ and $B = \{x : |x| > R \}$. Then, $$\|f\|_{p} \le \left( \int_{A} |f|^p d \mu + \int_{B} \frac{1}{2^p} d\mu\right)^{1/p} = \left( \int_{A} |f|^p d\mu + \frac{1}{2^p} \mu( B ) \right)^{1/p}.$$

If $\mu(B) < \infty$ this can easily show the desired result. Moreover, if I could show that there is a family of sets $\{B_p\}$ that act similarly so that $\mu(B_p)$ grows slower than $e^p$, then I can also complete the proof.


  • $\begingroup$ See this. $\endgroup$ Commented Feb 17, 2014 at 0:42
  • $\begingroup$ I was trying to find somewhere on the site where this was already asked, and could not find it. However, now that you found it for me, I don't see how they showed this direction. It reads, $$\|f\|_p \le \left( \int_{X} |f(x)|^{p-q} |f(x)|^{q} d\mu \right)^{1/p} \le \|f\|_{\infty}^{\frac{p-q}{p}}\|f\|_{q}^{q/p}.$$ I think the first inequality should just be an equality OR they intended $|f(x)|^{p-q}$ to be $\|f\|_{\infty}^{p-q}$. Even in this latter case, I do not see how the 2nd inequality follows. $\endgroup$
    – mlg4080
    Commented Feb 17, 2014 at 1:04
  • 2
    $\begingroup$ The second inequality is straightforward: all that is happening is that the first $|f(x)|$ is bounded by $\|f\|_\infty$. $\endgroup$ Commented Feb 17, 2014 at 1:13
  • $\begingroup$ Wow. Of course. Thank you. I had convinced myself it should be via Holder. (As did the answer below... because as I now see it's a very simple case of Holder's.) $\endgroup$
    – mlg4080
    Commented Feb 17, 2014 at 1:16
  • $\begingroup$ But it is Hölder. I answered on the comment to my answer. $\endgroup$ Commented Feb 17, 2014 at 1:17

1 Answer 1


For the first part you have to use $\liminf$, as you don't still know that the limit exists.

For the second part, you are thinking as if $X$ was $\mathbb R^n$, which it might not be. One way of attacking the problem along your line of thought would be to assume $\|f\|_\infty=1$ (i.e., work with $f/\|f\|_\infty$). Then, for $p>r$, $$ \left(\int_X|f|^pd\mu\right)^{1/p}\leq\left(\int_X|f|^rd\mu\right)^{1/p}=\|f\|_r^{r/p} $$ Then $$ \limsup_{p\to\infty}\|f\|_p\leq1. $$ Now you can scale back with $\|f\|_\infty$ to get $$ \limsup_{p\to\infty}\|f\|_p\leq\|f\|_\infty. $$

Another way of doing this second part is to use Hölder's inequality: $$ \|f\|_p^p=\int_X|f|^pd\mu=\int_X|f|^r|f|^{p-r}d\mu\leq \|f\|_\infty^{p-r}\,\|f\|_r^r. $$ So $$ \limsup_{p\to\infty}\|f\|_p\leq\limsup_{p\to\infty}\|f\|_\infty^{(p-r)/p}\,\|f\|_r^{r/p}=\|f\|_\infty. $$

  • $\begingroup$ I like your first way. I feel silly for not thinking of passing to $r$ with how much time I spent on this. Also, for the latter, does Holder's inequality hold for $1 \le p \le \infty?$ I've only seen it for $1 < p < \infty$. $\endgroup$
    – mlg4080
    Commented Feb 17, 2014 at 1:15
  • $\begingroup$ Yes, Hölder holds for $1,\infty$ too. The proof is straightforward: $$ \int |f|\,|g|\leq\int \|f\|_\infty\,|g|=\|f\|_\infty\,\int|g|=\|f\|_\infty\,\|g\|_1.$$ $\endgroup$ Commented Feb 17, 2014 at 1:16
  • $\begingroup$ Thanks for the answer(s) and clarifications. Helps a lot! $\endgroup$
    – mlg4080
    Commented Feb 17, 2014 at 1:18
  • $\begingroup$ You are very welcome. Most of us love users who post thoughtful questions. $\endgroup$ Commented Feb 17, 2014 at 2:01

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