analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
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A convex function is a real-valued function defined on a convex set whose graph is the boundary of a convex set.
There is another context where people say a function is convex if it is a Lipschitz function between metric spaces with Lipschitz constant (or Lipschitz modulus) 1. These are different concepts of convexity, although there are relations between convexity and Lipschitz continuity, as we shall see below.
Let $D$ be a convex space. A function $f \colon D \to \mathbb{R}$ is convex if the set $\{(x, z) \in D \times \mathbb{R}: z \geq f(x)\}$ is a convex subspace of $D \times \mathbb{R}$. Equivalently, $f$ is convex if for all $x, y \in D$,
whenever $0 \leq t \leq 1$.
This definition obviously extends to functions $f \colon D \to I$ where $I$ is an interval of $\mathbb{R}$ (whether open or closed or half-open, it doesn’t matter).
The function $f$ is called concave if it satisfy the reverse inequality to the one given above, or equivalently if $-f$ is convex.
A function $f$ is strictly convex if the inequality holds strictly whenever $0 \lt t \lt 1$. In high-school mathematics, one often says concave upward for strictly convex and concave downward for strictly concave.
A homomorphism of convex sets, i.e. a convex-linear map, $D \to \mathbb{R}$, is of course convex.
The norm function $\mathbb{C} \to \mathbb{R}$ is convex. This follows readily from multiplicativity ${|x y|} = {|x|} \cdot {|y|}$ and the triangle inequality ${|x + y|} \leq {|x|} + {|y|}$.
More generally, for a normed vector space $V$, the norm function ${\|(-)\|}$ is convex, again by the triangle inequality and the scaling axiom ${\|\alpha v\|} = {|\alpha|} \cdot {\|v\|}$ for scalars $\alpha$.
For any twice-differentiable function $f \colon (a, b) \to \mathbb{R}$, the second derivative $f''$ is nonnegative iff $f$ is convex; this may be proven using the mean value theorem. Examples include the exponential function $\exp: (-\infty, \infty) \to \mathbb{R}$ and the $p$-power function $[0, \infty) \to \mathbb{R}: t \mapsto t^p$ if $p \geq 1$.
More generally, for an open convex region $D \subseteq \mathbb{R}^n$ in a Euclidean space, a twice-differentiable function $f: D \to \mathbb{R}$ is convex iff its Hessian is a positive semidefinite bilinear form.
Any positive $\mathbb{R}$-linear combination of convex functions on $D$ is again convex.
The pointwise maximum of two convex functions on $D$ is again convex.
If $g: V \to W$ is a homomorphism or convex-linear map between convex spaces, and if $f: W \to \mathbb{R}$ is convex, then $f \circ g: V \to \mathbb{R}$ is convex.
In the next two examples, $I \subseteq \mathbb{R}$ is an interval.
It is not generally true that a composition $D \stackrel{g}{\to} I \stackrel{f}{\to} \mathbb{R}$ of convex functions is convex. For example, this fails for the case $D = \mathbb{R}$ and $g(x) = x^2 + 1$ and $f: [1, \infty) \to \mathbb{R}$ given by $f(x) = x^{-1}$.
However, if $f: I \to \mathbb{R}$ is both monotone increasing and convex, then for any convex function $g: D \to I$, the composite $f \circ g: D \to \mathbb{R}$ is convex (as is easily shown).
A special case of the last class of examples are functions of the form $e^f = \exp \circ f$ where $f$ is convex. We say a function $f: D \to (0, \infty)$ is log-convex if $\log f$ is convex. We see then that log-convex functions are also convex. The identity function on $\mathbb{R}$ is an example of a convex function that is not log-convex (or use $x \mapsto x^2$ for a strict example).
If $f: (a, b) \to \mathbb{R}$ is convex and $u \lt v \lt w \in (a, b)$, then
Consequently, the function $x \mapsto \frac{f(x) - f(v)}{x - v}$ on the domain $u \lt x \lt v$ is increasing and is bounded above by $\frac{f(v)-f(w)}{v-w}$; similarly, this function on the domain $v \lt x \lt w$ is increasing and bounded below by $\frac{f(u)-f(v)}{u-v}$.
The proof is virtually trivial; just write $v$ as a convex combination $t u + (1-t)w$ and use the definition of convexity.
A convex function $f: (a, b) \to \mathbb{R}$ is continuous.
For any point $x_0 \in (a, b)$ there are $s, t, u, v \in (a, b)$ with $s \lt t \lt x_0 \lt u \lt v$. Then for any $x \in (t, x_0) \cup (x_0, u)$ we have by Lemma
so that $K = \max\{\left\vert \frac{f(s)-f(t)}{s-t}\right\vert, \left\vert \frac{f(u)-f(v)}{u-v}\right\vert\}$ serves as a Lipschitz constant, i.e., we have
for all $x \in (t, u)$. This is enough to force continuity at the point $x_0$.
In the converse direction, we have the following result which frequently arises in practice.
If $D \subseteq \mathbb{R}^n$ is a convex set and $f: D \to \mathbb{R}$ is a continuous function such that
then $f$ is convex.
For any fixed $x, y \in D$, the function $h: [0, 1] \to \mathbb{R}$ defined by
is continuous. The inverse image $h^{-1}([0, \infty))$ is closed in $[0, 1]$ and, by an easy induction argument, contains the set of dyadic rational numbers $t \in [0, 1]$, which is dense in $[0, 1]$. Being closed and dense, $h^{-1}([0, \infty))$ is all of $[0, 1]$, i.e., $h(t) \geq 0$ for all $t \in [0, 1]$, but this is precisely the condition that $f$ is convex.
Another easy consequence of Lemma is
For a convex function $f: (a, b) \to \mathbb{R}$, the right-hand and left-hand derivatives
of $f$ exist at every point $x_0 \in (a, b)$, and the left-hand derivative at $x_0$ is less than or equal to the right-hand derivative at $x_0$.
It further follows from Lemma that if $f$ is convex and we define $(D f)(x_0)$ to be the average of the right-hand and left-hand derivatives, then $D f$ is monotone increasing and hence is discontinuous at at most countable many points of jump discontinuity (whence $D f$ is Riemann-integrable, for instance).
See also
Last revised on December 17, 2023 at 01:17:04. See the history of this page for a list of all contributions to it.