$\int_{X}e^{c|f|}\, d\mu < \infty$ implies$\|f\|_{L^{p}(X)} \leq Cp$

Question

Suppose $(X, \mu)$ is a measure space with finite measure and $f: X \rightarrow \mathbb{R}$ a measurable function and there exists a $c > 0$ such that $\int_{X}e^{c|f|}\, d\mu < \infty$. Why does there exist a constant $C > 0$ such that $\|f\|_{L^{p}(X)} \leq Cp$ for all $1 \leq p < \infty$?

Using the fact that for $n = 1, 2, 3, \ldots$, $c^{n}|f(x)|^{n}/n! \leq e^{c|f(x)|}$, I can show for these $n$, $\|f\|_{L^{n}(X)} \leq Cn$ for some $C > 0$. However, I can't seem to be able to prove this for, say $p \in (n, n + 1)$.

How do they come up with these problems!?? – Christian Chapman Jun 22 '14 at 21:39 — Christian Chapman, Jun 22 '14 at 21:39

score 3 · Accepted Answer · edited Apr 13 '17 at 12:19

3

Let $a$ be a real number and $n$ such that $n\leqslant a\lt n+1$. Using Lyapunov's inequality with $p=n$ and $q=n+1$, we have with the notations of the link that $\lambda=n+1-a$, and the conclusion follows.

edited Apr 13 '17 at 12:19

Community

1

answered Jun 22 '14 at 21:37

Davide Giraudo

181,608

$\int_{X}e^{c|f|}\, d\mu < \infty$ implies$\|f\|_{L^{p}(X)} \leq Cp$

1 Answers1