So I am for the moment studying the properties of the Boltzmann-Gibbs statistical entropy
\begin{equation} S_{BG}=-k_B \sum_{i}p_i\ln p_i, \end{equation} where of course $k_B$ is the Boltzmann constant. Now, among the properties it is stated that the entropy has an extremum for equal probabilities and then it mentions that $S_{BG}$ is concave, therefore this particular extremum is a maximum.
Now, I can define: \begin{equation} p^{''}_i=\lambda p_i+(1-\lambda)p_i^{'} \end{equation} for two distinct probability sets $\{ p_i\}$, $\{ p_i^{'} \}$ (associated with the system states W) and then use the definition of a concave functional to prove: \begin{equation} S_{BG}(\{ p_i^{''}\})>\lambda S_{BG}(\{ p_i^{}\})+(1-\lambda)S_{BG}(\{ p_i^{'}\}) \end{equation}
That's as far as I can get. I cannot prove the concavity inequality. Even though I do know that it has something to do with the one dimension function $f(x)=x\ln (x), x>0$ which is convex, therefore $-f(x)$ will be concave. But that is considered for a real function of 1-dimension. Anyway, I am sure it's simple but I can't see it.
Any ideas?? Thanks!
If $f(x)$ is concave, then so is $g(x_1, \ldots, x_n) = f(x_1)$.
A sum of concave functions is concave.