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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?

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  • $\begingroup$ 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. $\endgroup$ – mike Apr 21 '12 at 12:25
  • $\begingroup$ Are you sure this is not true on $\mathbb{R}$? My lecture notes say its true on the circle, but only for $f \in C(\mathbb{T})$ ? $\endgroup$ – rk101 Apr 21 '12 at 12:33
  • $\begingroup$ if it were true it would imply that any bounded function is integrable, which is only true on finite measure spaces. $\endgroup$ – mike Apr 24 '12 at 18:56
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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.

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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 21 '12 at 12:50
  • $\begingroup$ 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! $\endgroup$ – rk101 Apr 21 '12 at 13:05
  • $\begingroup$ 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 ${}\le 1$. $\endgroup$ – GEdgar Apr 21 '12 at 13:17
  • $\begingroup$ @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. $\endgroup$ – Nate Eldredge Apr 21 '12 at 14:36
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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 $\|f\|_{p-1} \le \|f\|_p$.

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