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Given an arbitrary measure space (of possibly infinite measure), if $f \in L^1 \cap L^\infty$, then by Hölder's inequality, $f^2 \in L^1$, so $f \in L^2$.

Intuition suggests that $f \in L^p$ even for any $1 \le p \le \infty$ (since we have eliminated the only two things that can go wrong for $f$ to be in $L^p$; blow-up & non-decay).

This does not seem to follow from the common inequalities, hence my question: Is it true that

$L^1 \cap L^\infty \subset L^p$

in general, and if so how can I prove it? Many thanks in advance for any hints!

Davide Giraudo
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Henry Wegener
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    Note, the only interesting case is indeed if the measure is not finite. Otherwise $L^p\subset L^q$ for $p\ge q$ – math Nov 10 '13 at 16:04

2 Answers2

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Yes, it's true. Let us define $A = \{x : \lvert f(x)\rvert > 1\}$. Then $\mu(A) \leqslant \int_A \lvert f(x)\rvert\,d\mu \leqslant \lVert f\rVert_1$, and hence

$$\int_X \lvert f(x)\rvert^p\,d\mu = \int_A \lvert f(x)\rvert^p\,d\mu + \int_{X\setminus A} \lvert f(x)\rvert^p\,d\mu \leqslant \lVert f\rVert_\infty^p\cdot \lVert f\rVert_1 + \lVert f\rVert_1 < \infty$$

since $\lvert f(x)\rvert^p \leqslant \lvert f(x)\rvert$ on $X\setminus A$.

Daniel Fischer
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More generally we have a family of interpolation inequalities: assume $1\leq q \leq r \leq p \leq \infty$ with $$ \frac1r = \frac\theta q + \frac{1-\theta}{p},\qquad \theta\in(0,1). $$ Then, if $f \in L^q \cap L^p$ we have $$ f\in L^r \qquad\text{and}\qquad \| f\|_r \leq \| f \|_q^{\theta} \, \| f\|_p^{1-\theta}. $$

To prove this, just write $$ \int_X |f(x)|^r \, d\mu = \int_X |f(x)|^{\theta r} \, |f(x)|^{(1-\theta)r}\, d\mu $$ and then use Holder's inequality with exponents $a = q/(\theta r)$ and $a' = p/[(1-\theta)r] $.

user109420
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  • We know that $f\in L^q\cap L^p$. When we write what you suggest, aren't we implicitly assuming that $f^\theta\in L^q$ and $f^{1-\theta}\in L^p$ in order to use Holder's inequality? How do we know this? – Laars Helenius Dec 09 '13 at 04:23
  • I may have answered my own comment: $$|f|^\theta|f|^{1-\theta}\le \theta |f|+(1-\theta)|f|=|f|.$$ This is just a generalized arithmetic/geometric mean inequality, right? – Laars Helenius Dec 09 '13 at 04:28