# Convexity of a function with log and det

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!

• Maybe you can look at it from another perspective. Hint hint. Oct 26 '16 at 11:22
• Oh! do I compute the Hessian of the function and see if it is positive semi-definite to show that it is convex? Oct 26 '16 at 11:45
• That was not the perspective I was thinking of. Oct 26 '16 at 12:07
• The question is titled perspective of log determinant. Oct 26 '16 at 13:39
• @JohanLöfberg took his funny pills this morning Oct 26 '16 at 13:48

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.