TITLE: Are integrable, essentially bounded functions in L^p?
QUESTION [7 upvotes]: 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!

REPLY [5 votes]: 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] $.