Prove that $f\left(x\right)=\sum_{i=1}^{n}x_{i}\ln x_{i}-\left(\sum_{i=1}^{n}x_{i}\right)\ln\left(\sum_{i=1}^{n}x_{i}\right)$ is convex

Question

Im trying to prove that $f\colon\mathbb{R}_{++}^{n}\to\mathbb{R}$ defined by $$ f\left(x\right)=\sum_{i=1}^{n}x_{i}\ln x_{i}-\left(\sum_{i=1}^{n}x_{i}\right)\ln\left(\sum_{i=1}^{n}x_{i}\right) $$ is a convex function. What I thought about is using some of the ln properties which can give us \begin{align*} f\left(x\right) & =\sum_{i=1}^{n}x_{i}\ln x_{i}-\left(\sum_{i=1}^{n}x_{i}\right)\ln\left(\sum_{i=1}^{n}x_{i}\right)=\\ & =\sum_{i=1}^{n}x_{i}\ln x_{i}-\sum_{i=1}^{n}x_{i}\ln\left(\sum_{i=1}^{n}x_{i}\right)=\\ & =\sum_{i=1}^{n}x_{i}\left(\ln x_{i}-\ln\left(\sum_{i=1}^{n}x_{i}\right)\right)=\\ & =\sum_{i=1}^{n}x_{i}\ln\left(\frac{x_{i}}{\sum_{j=1}^{n}x_{j}}\right) \end{align*} and it suffices to show that $f_{i}\left(x\right)=x_{i}\ln\left(\frac{x_{i}}{\sum_{j=1}^{n}x_{j}}\right)$ is convex. But using the Hessian criterion for this function seemed to me like an overkill. Any suggestions?

Answer 1

it suffices to show that $f_{i}\left(x\right)=x_{i}\ln\left(\frac{x_{i}}{\sum_{j=1}^{n}x_{j}}\right)$ is convex

Since, $g(x) = x\log x$ is convex, the perspective $y g(x/y) = x\log (x/y)$ is also convex. Taking $y=\sum_{j=1}^{n}x_{j}$ does not affect convexity.