$f:\R^n\rightarrow \R$ is convex if $\dom f$ is a convex set and
$$ f\left( \theta x + (1-\theta)y\right) \le \theta f(x) + (1-\theta) f(y) $$for all $x,y\in \dom f$ and $0\le \theta\le 1$.
for all $x,y\in \dom f$, $x\ne y$, and $0\le \theta\le 1$.
Convex functions:
Concave functions:
Convex functions:
Concave functions:
$c(x) = h\left( f_1(x), \dots, f_k(x)\right)$ is convex when $h$ is convex and for each $i$
Rough idea: for $x\in\R$ with differentiable $f$ and $h$, $$ c''(x) = f'(x)^T \nabla^2 h\left(f(x)\right)f'(x) + \nabla h\left(f(x)\right)^Tf''(x) $$
is convex.
is convex.
is convex.