I am reading through a set of notes about concentration of Gaussian measure, and on page 56, they make the following claim that I am failing to see the proof of:
Now we estimate the second summand at the right hand side. Using
Young's inequality with $y = e^{\lambda \tilde Z} \geq 0$ and $x = \Sigma^2 – (e-1)E\Sigma^2$, we get
\begin{align*}E(\Sigma^2e^{\lambda\tilde Z} &= (e-1)(E\Sigma^2)Ee^{\lambda \tilde Z}+E\left[(\Sigma^2 – (e-1)E\Sigma^2)e^{\lambda \tilde Z}\right]\\
& \leq (e-1)(E\Sigma^2)Ee^{\lambda \tilde Z} +\lambda E(\tilde Ze^{\lambda \tilde Z})-Ee^{\lambda \tilde Z} + E e^{\Sigma^2 – (e-1)E\Sigma^2}\end{align*}
This inequality is baffling to me. I'm assuming they mean Young's product inequality, but stated in its usual form, Young's product inequality does not allow us to put "$x$" in an exponent. The best I could think of was using the fact that $\require{enclose} Where are the authors getting this inequality from?
\enclose{horizontalstrike}{\frac{x^p}{p} = \frac{e^{p\log x}}{p} = \frac{x e^p}{p}}$ so that if we chose $p = x$, we could get from Young's inequality:
$$\require{enclose}
\enclose{horizontalstrike}{x\cdot y \leq e^{x} +y^{\frac{x}{x-1}}\cdot \frac{x-1}{x}}$$
for $\require{enclose}
\enclose{horizontalstrike}{x,y \geq 0}$, but even then, it seems to me that the second term becomes quite intractable.
Note: A lot of variables in the above quote are defined in the preceding parts of the proof, so I may have left out important context for this problem that can be found in the link.
Best Answer
They use a generalized version of Young's inequality (see here, the last example under "Generalization using Fenchel-Legendre transform"). In particular, given two real numbers $x > 0$, $y$, one gets:
$$xy \leq x \ln x - x +e^y.$$
This version of Young's inequality was already used in the document you linked, at the bottom of page 43.
Note by the way that there is a mistake in your computation ($e^{p \ln x} \neq x e^p$).