Show that the function $f: S \to \mathbb R$ given by $$f(x,s,t):=-\ln(st - ||x||^2)$$ is convex on $$S := \left\{(x,s,t) \in \mathbb R^n \times \mathbb R \times \mathbb R: \frac{\|x\|^2}{s}<t, s>0, t>0 \right\}$$

$-\ln(x)$ is a convex monotonic decreasing function, hence we can use composition by showing the inner part is concave. Also we know $-||x||^2$ is concave and $st$ is neither convex nor concave on $S$. Also in $S$ we have $st>||x||^2$ but I was unable to show the convexity of the function as a whole

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    $\begingroup$ What are your thoughts on this? What have you tried already (perhaps, what do you know about the convexity of $x^2$ and $-\ln x$?) Questions without visible effort or context tend to get downvoted. $\endgroup$ – postmortes Feb 15 at 10:43

I think you can try the following:

$-\ln(st-\| x\|^2)=-\ln(s(t-\frac{| x\|}{s}^2))=-\ln(s)-\ln(t-\frac{| x\|}{s}^2)$

Then you have a sum of two convex functions.


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