Are integrable, essentially bounded functions in L^p?

Question

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!

Answer 1

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 $X\setminus A$.

Answer 2

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] $.