Convexity of a function with log and det

Question

I am currently working with "Additional Exercises for Convex Optimization" by Boyd and Vandenberghe. Problem 2.6 asks to show that $$ f(X,t) = nt \log t - t \log \det X $$ with $\textbf{dom} \ f = \textbf{S}_{++}^n \times \textbf{R}_{++}$, is convex in $(X,t)$. I know that a function is convex if for $\theta \in (0,1)$, we have $$ f(\theta x_1 + (1-\theta) x_2) \leq \theta f(x_1) + (1-\theta) f(x_2) $$ But I was not really sure how to prove that a function is convex if there are two arguments. Could you please give me a hint on how to approach this problem? Thanks in advance for your help!

Answer 1

1, $t \log t$ is convex because its hessian is positive: $(t \log t)'=\log t + 1$; $(t \log t)'' = (\log t + 1)' = 1/t > 0, t \in {R}_{++}$

2, $\log \det X$ is concave (refer to Boyd page 74 'Log-determinant') hence its negative is convex;

Thus $f(X,t) = nt \log t - t \log \det X$ is convex.

Answer 2

I think the function is convex over $\mathcal{S}_{++}^n\times\mathcal{R}_{++}$. The proof has three steps:

1. Show the function $g(X)=-\log{\det{X}}$ is a convex function in $X\in \mathcal{S}_{++}^n$. (Chapter 3.1.5 Log-determinant in Convex Optimization by Boyd and Vandenberghe)
2. Show the function $f(X,t) = nt \log t - t \log \det X$ is the perspective of $g(X)$. $$ f(X,t) = -t \log{\frac{\det X}{t^n}} = -t \log{\det{\frac{X}{t}}}=t\cdot g(\frac{X}{t})$$
3. By Chapter 3.2.6 in Convex Optimization by Boyd and Vandenberghe, the perspective function $f(X,t)$ is a convex function since $g(X)$ is a convex function.