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!

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


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