Need a bit of help with a parameter integral problem.
We have, $X$ is a finite measure space with measure $\mu$ and $f:X\rightarrow [0 , \infty)$ is a measurable function.
The parameter integral is defined as
$$\psi (t) := \int_{X} \exp(-tf(x)).$$
I have already shown that this is a continuous function of $t \in[0 , \infty)$.
I have also shown that if $ f\in L^{m}(x) $ for some $m\ \epsilon \ \mathbb{N}$ then the $k$-th derivative is
$(-1)^{k}\left \| f \right \|_{k}^{k}$ where $k = 1 , 2 ,\dots , m$.
My question is, assuming there exists $\varepsilon > 0 $ such that $\psi(t) < \infty$ for all
$t \in(-\varepsilon ,\varepsilon )$.
How do I show $\left \| f \right \|_{p}< \infty $ for all $1 < p < \infty$?
Thanks
With $t:=-\varepsilon/2$, we get $$\int_X\exp\left(\frac\varepsilon 2f(x)\right)\mathrm d\mu(x)<\infty.$$ Since $s^k/k!\leqslant \sum_{j=0}^\infty s^j/j!=e^s$ for each positive $s$, this implies that $f$ belongs to $\mathbb L^k$ for each positive integer $k$. Indeed, $$\left(\frac{\varepsilon}2\right)^k\frac 1{k!}\int_Xf(x)^k\mathrm d\mu(x)<\infty.$$ If $K\int |g(x)|\mathrm d\mu(x)$ is finite for some positive constant $K$, then $\int |g(x)|\mathrm d\mu(x)$ is finite.