A real function $f$ on $I=[0,1]$ is an <operator convex function> when, for all <Hermitian operators> $X,Y$ whose spectra lie in $I$ and every $0\leq\lambda\leq1$,
$$
\boxed{
f(\lambda X+(1-\lambda)Y)
\preceq\lambda f(X)+(1-\lambda)f(Y)},
$$
where $\preceq$ is the <Loewner order>. Reversing the inequality defines an <operator concave function>, equivalently $-f$ is operator convex.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 3 / ii / a / Solution

A function is operator concave when its negative is <operator convex>. Equivalently, the defining operator-convexity inequality is reversed.

= Operator concave
{synonym}

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