Convexity of combination of natural logs with sums function

Question

Let $$f(x)=\sum_{i=1}^n x_i \ln x_i - (\sum_{i=1}^n x_i) \ln (\sum_{i=1}^n x_i)$$ be defined over $(0,\infty)^n$, $f:\mathbb{R}^n_{>0}\to \mathbb{R}$. I need to prove that $f$ is convex. I know that $t\mapsto t\ln t$ is convex over $(0,\infty)$. However, I can't use this due to the minus sign in the expression. I tried a lot of things but I'm stuck. Any help would be appreciated.

Answer 1

If I am not mistaken, your function is the convex conjugate of: $$g(y) = \begin{cases} \log\left( \sum_{i=1}^n \exp(y_i)\right) & \text{ if } \sum_{i=1}^n \exp(y_i)\leq1 \\ \infty & \text{ otherwise.} \end{cases} $$ In other words, $$f(x)=\sup_y\left\{x^Ty - g(y)\right\}.$$ Since $f$ is the supremum of affine functions (in $x$), $f$ itself is convex.

Answer 2

I have a solution now. I used induction on $n$. For $n=2$, I directly use a Hessian test. Assuming $n>2$ and the result works for $n-1$, denoting $f$ as $f_n$, we get

$$f_n(x) = f_{n-1}(Ax)+f_2(x_{n-1}, x_n)$$ where

$$A=\begin{bmatrix} 1&0&\dots&0&0\\ 0&1&\dots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\dots&1&1 \end{bmatrix} = [I|e_n]\in \mathbb{R}^{(n-1)\times n}.$$ Since convexity is preserved under an affine change of variables, the result follows. $\square$