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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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    if it were true it would imply that a$n$y bou$n$ded fu$n$ction is integrable, which is only true on finite measure spaces.2012-04-24

2 Answers 2

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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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    @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.2012-04-21
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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 $\|f\|_{p-1} \le \|f\|_p$.