Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is this true? $$ \|f \|_{L^{p-1} }\leq \|f\|_{L^{p}}\;\; $$

Specifically I know $\;\;\|f\|_{L^{2}} \leq \|f\|_{L^{\infty}}$ $\;$ but I can't figure out why?

share|cite|improve this question
what you say is true on a finite measure space, where you just pull out the bound, but not on say $\mathbb R$ with lebesgue measure. – mike Apr 21 '12 at 12:25
Are you sure this is not true on $\mathbb{R}$? My lecture notes say its true on the circle, but only for $f \in C(\mathbb{T})$ ? – rk101 Apr 21 '12 at 12:33
if it were true it would imply that any bounded function is integrable, which is only true on finite measure spaces. – mike Apr 24 '12 at 18:56
up vote 1 down vote accepted

Your first inequality is not necessarily true. With $X=[0,1]$, $p=2$ and $f(x)=x$ we have $$ \| f \|_1 = \int^1_0 x dx = 1/2 $$ while $$ \| f \|_2 = \left( \int^1_0 x^2 dx \right)^{1/2}=1/9 .$$

However your second one is correct: $$ \| f \|_p \leq \| f \|_{\infty}. $$ This follows so quickly from the definition of essential supremum and basic estimates, I recommend you try to prove this again.

Whilst we don't have you first inequality, we do have the follow inclusion: $ L^p \subseteq L^q $ for a finite measure space and $ p \leq q .$ This follows from an application of Holder's inequality. I suggest you give that exercise a try as well.

In general measure spaces however, there are not any guaranteed inclusions. This great question shows that.

share|cite|improve this answer
Your counterexample has a mistake: the $1/9$ should be $1/\sqrt{3}$, and it is not a counterexample. Actually the statement is true for a space of measure 1 (a probability space), and as you suggest it can be proved with Holder's inequality. – Nate Eldredge Apr 21 '12 at 12:50
I still can't show $\|f\|_p \leq \|f\|_{\infty}$. Best I can do is $\|f\|_p \leq \int |f(x)|^p\!\ \mathrm{d}x \leq \|f\|_{\infty}^p \int \!\mathrm{d}x$. Which still looks wrong! – rk101 Apr 21 '12 at 13:05
Use $f(t)=1$ for $t \in [0,10]$ and $0$ elsewhere to get a counterexample for $\|f\|_p \leq \|f\|_{\infty}$. Like the other one, this one is true when the measure space has total measure ${}\le 1$. – GEdgar Apr 21 '12 at 13:17
@rk01: You know $\lVert f \rVert_p^p = \int |f(x)|^p\,dx \le \lVert f \rVert_\infty^p \int 1 dx = \lVert f \rVert_\infty^p$ (when your space has measure 1). Now use the fact that $t^{1/p}$ is an increasing function, so you can apply it to both sides of the inequality. – Nate Eldredge Apr 21 '12 at 14:36

To prove the stated inequality (for normalized measure spaces): Let $X$ be a finite measure space and $q \le p$. We have using Hölder for $f \in L^p$ \begin{align*} \|f\|_q &= \bigl\||f|^q\bigr\|_1^{1/q}\\\ &= \bigl\|1 \cdot |f|^q\bigr\|_1^{1/q}\\\ &= \|1\|_{p/(p-q)} \bigl\||f|^q\bigr\|_{p/q}^{1/q}\\\ &= \mu(X)^{(p-q)/p} \|f\|_p \end{align*} If $\mu$ is normalized, i. e. $\mu(X) = 1$ (for example in $[0,1]$ or $\mathbb T$ with normalized arclength), then $\|f\|_q \le \|f\|_p$ for every $q \le p$. especially $\|f\|_{p-1} \le \|f\|_p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.