Question: I am aware that the common continuous distributions (like Gaussian, Uniform, Gamma) are log-concave. I am wondering if Bernoulli distributions (a discrete distribution) is log-concave? If so, then does this extend to the Categorical distribution?
My attempt: For a Bernoulli($p$) distribution $$ f(k)=p^k(1-p)^{1-k} \text{ for } k\in\{0,1\}, $$ we need to show that for some $\theta\in(0,1)$, we have $$ f\Big(\theta k_1+(1-\theta) k_2\Big)\geq f(k_1)^\theta f(k_2)^{1-\theta}. $$ I can show this by considering two cases (i) $\theta k_1+(1-\theta) k_2=0$ and $\theta k_1+(1-\theta) k_2=1$. Case (i) leads to $k_1,k_2=0$ which leads to equality in the log-concave inequality. Case (ii) leads to either $k_1,k_2=1$ which leads to equality in the log-concave inequality, or $k_1,k_2\notin\{0,1\}$ which leads to strict inequality in the log-concave inequality.
There are two uses of the phrase "log concave." The continuous version is the one you stated, that $\log f(\theta x+(1-\theta)y)\ge \theta \log f(x)+(1-\theta)\log f(y)$ for all $\theta\in [0,1]$.
There is a discrete analogue as well. We say that a sequence of nonnegative numbers $\{a_n\}_{n\in \mathbb Z}$ is log-concave if it satisfies $$ a_n^2\ge a_{n-1}a_{n+1} $$ for all $n\in \mathbb Z$. If the support of the sequence is a proper subset of $\mathbb Z$, then you can first extend the support to all of $\mathbb Z$ by appending and prepending zeroes. This further implies that $\log a_{r}\ge \theta \log a_n+(1-\theta)\log a_m$, whenever $r=\theta n+(1-\theta)m$ is an integer.
Since the support of the Bernoulli distribution is discrete, we must use the discrete version of log concavity. The probability distribution is the finite sequence defined by $a_0=1-p$ and $a_1=p$. We extend this by defining $a_n=0$ whenever $n\in\mathbb Z\setminus \{0,1\}$. You can then confirm that $a_n^2\ge a_{n-1}a_{n+1}$ always holds for this extended sequence, so $a_n$ is log concave.
It is useful to consider the log-concavity of discrete sequences. For example, the discrete convolution of two log concave sequences is still log concave. Since the binomial distribution is a convolution of several Bernoulli distributions, the fact that the Bernoulli distribution is log concave leads to a quick proof that the binomial distribution is log concave as well.