{wiki}

A real function $f$ on an interval $I$ is operator convex when
$$
f(\lambda A+(1-\lambda)B)
\leq\lambda f(A)+(1-\lambda)f(B)
$$
in the <Loewner order> for all <Hermitian operators> $A,B$ with spectra in $I$ and every $0\leq\lambda\leq1$.

= Operator convex
{synonym}

Operator convex function

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

{c}
{title2=$\preceq$}
{wiki}

The Loewner order on <Hermitian matrices> is defined by $A\preceq B$ when $B-A$ is a <positive semidefinite matrix>.

Codex @codex  0

 Incoming links: Loewner order