Convexity: Examples and Counterexamples

The function $f(x)=\mathrm{\mid }x\mathrm{\mid }$ is convex on $\mathbb{R}$. Indeed, for $0\le \lambda \le 1$,

$\mathrm{\mid }\lambda x+(1-\lambda )y\mathrm{\mid }\le \mathrm{\mid }\lambda x\mathrm{\mid }+\mathrm{\mid }(1-\lambda )y\mathrm{\mid }=\lambda \mathrm{\mid }x\mathrm{\mid }+(1-\lambda )\mathrm{\mid }y\mathrm{\mid },$

by the triangle inequality (The triangle inequality) and absolute homogeneity (Basic properties of the absolute value), which is precisely the convexity inequality (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

Convexity does not entail differentiability at every point; the nondifferentiability of absolute value at zero is recorded in $x\mapsto \mathrm{\mid }x\mathrm{\mid }$ is continuous everywhere and not differentiable at $0$: the difference quotient equals $1$ on the right and $-1$ on the left, so the two one-sided limits differ but is not a dependency of this example.

For nonnegative weights ${\lambda }_i$ with ${\sum }_i{\lambda }_i=1$ and real $x_i$,

${\left({\sum }_i{\lambda }_i{x}_i\right)}^2\le {\sum }_i{\lambda }_i{x}_i^2\mathrm{.}$

The function $q(t)=t^2$ is convex because

$\lambda x^2+(1-\lambda )y^2-(\lambda x+(1-\lambda )y{)}^2=\lambda (1-\lambda )(x-y{)}^2\ge 0$

for $0\le \lambda \le 1$, using nonnegativity of squares (Squares of nonzero elements are positive). Applying finite Jensen (Finite Jensen inequality for a convex function and nonnegative weights summing to one) to $q$ gives the displayed inequality.

Every midpoint-convex function $\mathbb{R}\to \mathbb{R}$ is continuous, and hence convex.

For $f(x)=x^3$, the power rule gives $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$ (For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term). It is negative on $(-\mathrm{\infty },0)$ and positive on $(0,\mathrm{\infty })$, while $f$ is continuous at $0$. Therefore the sign-change criterion (A continuous function whose second derivative has opposite signs on the two sides of a point has an inflection point there) makes $0$ an inflection point (An inflection point as a point of continuity where convexity changes to concavity or conversely).