Suppose $Z$ is standard normal and $U$ is another random variable. Let $f(Z,U) \ge 0$ also $ E[f^p(Z,U)] <\infty$ for $1 \le p <\infty$> we want to find a bound on the following expectation \begin{align} E\left[f(Z,U)\exp \left(\frac{tZ^2}{2}\right) \right] \end{align} for $ t \in (0,1)$.
The assumption is that \begin{align} E\left[f(Z,U)\exp \left(\frac{tZ^2}{2}\right) \right] < \infty \end{align} fo all $t \in (0,1)$
The bound I have is due to Holder \begin{align} E\left[f(Z,U)e^{\frac{tZ^2}{2}} \right] &\le \left(E\left[f(Z,U)^q \right] \right)^{1/q} \left(E\left[\exp \left(\frac{tpZ^2}{2}\right) \right] \right)^{1/p}\\ &= \left(E\left[f(Z,U)^q \right] \right)^{1/q} \left(\frac{1}{1-tp} \right)^{\frac{1}{2p}} \end{align} the last using by moment generation function of $Z^2$ which is valid for $tp < 1$.
However, my problem is that the above bound is only finite for $tp < 1$ or in other words for $t < \frac{1}{p} \le 1$ and does not hold for all values of $t \in (0,1)$.
My question is this the best bound we can do? Or is there a better bound?
This question is related to something I asked before here but not exactly
Thank you