New answers tagged convexity
2
Here is a complete proof.
One proves easily that if $S\in{\mathcal S}^n_{++}$ and if $zz^T\prec S$ (in the order between symmetric matrices), then
$$\frac12 y^TX^{-1}y\ge z\cdot y-\frac12 {\rm Tr}(SX).$$
Hint: start with the obvious inequality
$$\frac12\left( y^TX^{-1}y+z^TXz\right)\ge z\cdot y.$$
On the other hand, the equality is achieved by taking $z=X^{-...
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This is shown in example 3.4 "Matrix Fractional Function" of Convex Optimization – Boyd and Vandenberghe. The epigraph is transformed via Schur Complement (to handle the matrix inverse) into a convex LMI.
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$\newcommand\C{\mathcal C}\newcommand{\R}{\mathbb{R}}\newcommand{\tf}{\tilde f}
\newcommand{\tih}{\tilde h}$The answer is still no.
Indeed, suppose that we have a function $f\colon[0,\infty)\to\R$ such that $f(0)=0$ and $f$ is continuous, nonnegative, nondecreasing, and subadditive, but not superhomogeneous.
Then, for $\C:=[0,\infty)^n$, the function $h\...
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The answer is no. E.g., let $f(0):=0$ and $f(x):=1+m_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$; that is, $f(x)=1+(1+\lfloor\log_2|x|\rfloor)_+$ for all $x\in\mathcal C\setminus\{0\}$. Here, $m_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.
Details: Clearly, $f$ is nonnegative and nondecreasing in each of the ...
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