I read in these notes (and a few other places) that, for a non-negative function, the logarithm of the Laplace transform is convex and lower-semicontinous. For the convexity part I can see intuitively that it should be true, but the proof doesn't seem obvious, and I'd like to know what step I'm missing.
More precisely, let $V$ be a finite-dimensional vector space $V$ with dual $V^\star$, and let $f$ be a non-negative function $f\colon V\to [0,+\infty]$.
Then let the log-Laplace transform of $f$ be $ \bar f\colon V^\star\to \mathbb{R}\cup \{+\infty,-\infty\}$, given by $$ \bar f(x^\star) = \log\mathscr{L}[f](x^\star) = \int_V e^{-\langle x,x^\star\rangle} f(x)\,\mathrm{d}x. $$ I want to show that $\bar f$ is a lower-semicontinuous convex function of $x^\star$.
It seems like it should be a simple case of substituting into the definition of a convex function. That is, to show it's convex we should be able to show that $\bar f\big((1-\lambda)y + \lambda y'\big)\le (1-\lambda)\bar f(y) + \lambda \bar f(y')$. We have $$ \bar f\big((1-\lambda)y + \lambda y'\big) = \log \int_V e^{-(1-\lambda)\langle x, y\rangle -\lambda\langle x, y'\rangle } f(x)\,\mathrm{d}x = \\ \log \int_V \Big(e^{-\langle x, y\rangle}f(x)\Big)^{1-\lambda} \Big(e^{-\langle x, y'\rangle}f(x)\Big)^{\lambda} \,\mathrm{d}x, $$ and on the other hand, $$ (1-\lambda)\bar f(y) + \lambda \bar f(y') = \log \left( \int_V e^{-\langle x,y\rangle} f(x)\,\mathrm{d}x \right)^{1-\lambda} \left( \int_V e^{-\langle x,y'\rangle} f(x)\,\mathrm{d}x \right)^{\lambda}. $$ But I'm missing the next step, which must be to somehow show, using the non-negativity of $f$, that this product of integrals is greater than the integral of products. Can anybody see what that step is and how to show it?
Note: I realise that another way to prove it would be to differentiate $\bar f$ by its argument and show that the resulting matrix is nonnegative definite. However, I feel like the method sketched above is more likely to be illuminating to me, especially in terms of issues around $\pm \infty$ and the effective domain of $\bar f$, so I'd prefer to complete the proof this way if possible.
I believe this is related to Cramér's theorem in large deviations theory, so I'm adding that as a tag as well.
This is just the Hölder's inequality with $p=1/\lambda$ and $q=1/(1-\lambda)$. The $\log$ on both side can be removed since $\exp$ is strictly monotone and bijective.