# Convex function

Convex function on an interval.
A function (in black) is convex if and only if the region above its graph (in green) is a convex set.

In mathematics, a real-valued function f(x) defined on an interval is called convex (or convex downward or concave upward) if the line segment between any two points on the graph of the function lies above the graph, in a Euclidean space (or more generally a vector space) of at least two dimensions. Equivalently, a function is convex if its epigraph (the set of points on or above the graph of the function) is a convex set. Well-known examples of convex functions are the quadratic function ${\displaystyle f(x)=x^{2}}$ and the exponential function ${\displaystyle f(x)=e^{x}}$ for any real number x.

Convex functions play an important role in many areas of mathematics. They are especially important in the study of optimization problems where they are distinguished by a number of convenient properties. For instance, a (strictly) convex function on an open set has no more than one minimum. Even in infinite-dimensional spaces, under suitable additional hypotheses, convex functions continue to satisfy such properties and, as a result, they are the most well-understood functionals in the calculus of variations. In probability theory, a convex function applied to the expected value of a random variable is always less than or equal to the expected value of the convex function of the random variable. This result, known as Jensen's inequality, underlies many important inequalities (including, for instance, the arithmetic–geometric mean inequality and Hölder's inequality).

Exponential growth is a special case of convexity. Exponential growth narrowly means "increasing at a rate proportional to the current value", while convex growth generally means "increasing at an increasing rate (but not necessarily proportionally to current value)".

## Definition

Let X be a convex set in a real vector space and let f : XR be a function.

${\displaystyle \forall x_{1},x_{2}\in X,\forall t\in [0,1]:\qquad f(tx_{1}+(1-t)x_{2})\leq tf(x_{1})+(1-t)f(x_{2}).}$
${\displaystyle \forall x_{1}\neq x_{2}\in X,\forall t\in (0,1):\qquad f(tx_{1}+(1-t)x_{2})

## Properties

Suppose Template:Mvar is a function of one real variable defined on an interval, and let

${\displaystyle R(x_{1},x_{2})={\frac {f(x_{1})-f(x_{2})}{x_{1}-x_{2}}}}$

(note that R(x1, x2) is the slope of the purple line in the above drawing; note also that the function R is symmetric in x1, x2). Template:Mvar is convex if and only if R(x1, x2) is monotonically non-decreasing in x1, for x2 fixed (or vice versa). This characterization of convexity is quite useful to prove the following results.

A convex function Template:Mvar defined on some open interval Template:Mvar is continuous on C and Lipschitz continuous on any closed subinterval. Template:Mvar admits left and right derivatives, and these are monotonically non-decreasing. As a consequence, Template:Mvar is differentiable at all but at most countably many points. If Template:Mvar is closed, then Template:Mvar may fail to be continuous at the endpoints of Template:Mvar (an example is shown in the examples' section).

A function is midpoint convex on an interval Template:Mvar if

${\displaystyle \forall x_{1},x_{2}\in C:\qquad f\left({\frac {x_{1}+x_{2}}{2}}\right)\leq {\frac {f(x_{1})+f(x_{2})}{2}}.}$

This condition is only slightly weaker than convexity. For example, a real valued Lebesgue measurable function that is midpoint convex will be convex by Sierpinski Theorem.[1] In particular, a continuous function that is midpoint convex will be convex.

A differentiable function of one variable is convex on an interval if and only if its derivative is monotonically non-decreasing on that interval. If a function is differentiable and convex then it is also continuously differentiable. For the basic case of a differentiable function from (a subset of) the real numbers to the real numbers, "convex" is equivalent to "increasing at an increasing rate".

A continuously differentiable function of one variable is convex on an interval if and only if the function lies above all of its tangents:[2]:69

${\displaystyle f(x)\geq f(y)+f'(y)(x-y)}$

for all x and y in the interval. In particular, if f′(c) = 0, then Template:Mvar is a global minimum of f(x).

A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative there; this gives a practical test for convexity. Visually, a twice differentiable convex function "curves up", without any bends the other way (inflection points). If its second derivative is positive at all points then the function is strictly convex, but the converse does not hold. For example, the second derivative of f(x) = x4 is f ′′(x) = 12x2, which is zero for x = 0, but x4 is strictly convex.

More generally, a continuous, twice differentiable function of several variables is convex on a convex set if and only if its Hessian matrix is positive semidefinite on the interior of the convex set.

Any local minimum of a convex function is also a global minimum. A strictly convex function will have at most one global minimum.

For a convex function f, the sublevel sets {x | f(x) < a} and {x | f(x) ≤ a} with aR are convex sets. However, a function whose sublevel sets are convex sets may fail to be a convex function. A function whose sublevel sets are convex is called a quasiconvex function.

Jensen's inequality applies to every convex function f. If X is a random variable taking values in the domain of f, then E(f(X)) ≥ f(E(X)). (Here E denotes the mathematical expectation.)

If a function Template:Mvar is convex, and f(0) ≤ 0, then Template:Mvar is superadditive on the positive half-axis.

Proof: Since Template:Mvar is convex, let y = 0, then:

${\displaystyle f(tx)=f(tx+(1-t)\cdot 0)\leq tf(x)+(1-t)f(0)\leq tf(x),\quad \forall t\in [0,1].}$

From this we have:

${\displaystyle f(a)+f(b)=f\left((a+b){\frac {a}{a+b}}\right)+f\left((a+b){\frac {b}{a+b}}\right)\leq {\frac {a}{a+b}}f(a+b)+{\frac {b}{a+b}}f(a+b)=f(a+b)}$

## Strongly convex functions

The concept of strong convexity extends and parametrizes the notion of strict convexity. A strongly convex function is also strictly convex, but not vice versa.

A differentiable function Template:Mvar is called strongly convex with parameter m > 0 if the following inequality holds for all points x, y in its domain:[3]

${\displaystyle (\nabla f(x)-\nabla f(y))^{T}(x-y)\geq m\|x-y\|_{2}^{2}}$

or, more generally,

${\displaystyle \langle \nabla f(x)-\nabla f(y),(x-y)\rangle \geq m\|x-y\|^{2}}$

where ${\displaystyle \|\cdot \|}$ is any norm. Some authors, such as [4] refer to functions satisfying this inequality as elliptic functions.

An equivalent condition is the following:[5]

${\displaystyle f(y)\geq f(x)+\nabla f(x)^{T}(y-x)+{\frac {m}{2}}\|y-x\|_{2}^{2}}$

It is not necessary for a function to be differentiable in order to be strongly convex. A third definition[5] for a strongly convex function, with parameter m, is that, for all x, y in the domain and ${\displaystyle t\in [0,1]}$,

${\displaystyle f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)-{\frac {1}{2}}mt(1-t)\|x-y\|_{2}^{2}}$

Notice that this definition approaches the definition for strict convexity as m → 0, and is identical to the definition of a convex function when m = 0. Despite this, functions exist that are strictly convex but are not strongly convex for any m > 0 (see example below).

If the function Template:Mvar is twice continuously differentiable, then Template:Mvar is strongly convex with parameter m if and only if ${\displaystyle \nabla ^{2}f(x)\succeq mI}$ for all x in the domain, where I is the identity and ${\displaystyle \nabla ^{2}f}$ is the Hessian matrix, and the inequality ${\displaystyle \succeq }$ means that ${\displaystyle \nabla ^{2}f(x)-mI}$ is positive semi-definite. This is equivalent to requiring that the minimum eigenvalue of ${\displaystyle \nabla ^{2}f(x)}$ be at least m for all x. If the domain is just the real line, then ${\displaystyle \nabla ^{2}f(x)}$ is just the second derivative ${\displaystyle f''(x)}$, so the condition becomes ${\displaystyle f''(x)\geq m}$. If m = 0, then this means the Hessian is positive semidefinite (or if the domain is the real line, it means that ${\displaystyle f''(x)\geq 0}$), which implies the function is convex, and perhaps strictly convex, but not strongly convex.

Assuming still that the function is twice continuously differentiable, one can show that the lower bound of ${\displaystyle \nabla ^{2}f(x)}$ implies that it is strongly convex. Start by using Taylor's Theorem:

${\displaystyle f(y)=f(x)+\nabla f(x)^{T}(y-x)+{\frac {1}{2}}(y-x)^{T}\nabla ^{2}f(z)(y-x)}$

for some (unknown) ${\displaystyle z\in \{tx+(1-t)y:t\in [0,1]\}}$. Then

${\displaystyle (y-x)^{T}\nabla ^{2}f(z)(y-x)\geq m(y-x)^{T}(y-x)}$

by the assumption about the eigenvalues, and hence we recover the second strong convexity equation above.

A function Template:Mvar is strongly convex with parameter m if and only if the function ${\displaystyle x\mapsto f(x)-{\frac {m}{2}}\|x\|^{2}}$ is convex.

The distinction between convex, strictly convex, and strongly convex can be subtle at first glimpse. If Template:Mvaris twice continuously differentiable and the domain is the real line, then we can characterize it as follows:

Template:Mvar convex if and only if ${\displaystyle f''(x)\geq 0}$ for all Template:Mvar.
Template:Mvar strictly convex if ${\displaystyle f''(x)>0}$ for all Template:Mvar (note: this is sufficient, but not necessary).
Template:Mvar strongly convex if and only if ${\displaystyle f''(x)\geq m>0}$ for all Template:Mvar.

For example, consider a function Template:Mvar that is strictly convex, and suppose there is a sequence of points ${\displaystyle (x_{n})}$ such that ${\displaystyle f''(x_{n})={\frac {1}{n}}}$. Even though ${\displaystyle f''(x_{n})>0,}$ the function is not strongly convex because ${\displaystyle f''(x)}$ will become arbitrarily small.

A twice continuously differentiable function Template:Mvaron a compact domain ${\displaystyle X}$ that satisfies ${\displaystyle f''(x)>0}$ for all ${\displaystyle x\in X}$ is strongly convex. The proof of this statement follows from the extreme value theorem, which states that a continuous function on a compact set has a maximum and minimum.

Strongly convex functions are in general easier to work with than convex or strictly convex functions, since they are a smaller class. Like strictly convex functions, strongly convex functions have unique minima.

### Uniformly convex functions

A uniformly convex function,[6][7] with modulus ${\displaystyle \phi }$, is a function Template:Mvarthat, for all x, y in the domain and t ∈ [0, 1], satisfies

${\displaystyle f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)-t(1-t)\phi (\|x-y\|)}$

where ${\displaystyle \phi }$ is a function that is increasing and vanishes only at 0. This is a generalization of the concept of strongly convex function; by taking ${\displaystyle \phi (\alpha )={\frac {m}{2}}\alpha ^{2}}$ we recover the definition of strong convexity.

## Notes

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## References

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• Borwein, Jonathan, and Lewis, Adrian. (2000). Convex Analysis and Nonlinear Optimization. Springer.
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• Hiriart-Urruty, Jean-Baptiste, and Lemaréchal, Claude. (2004). Fundamentals of Convex analysis. Berlin: Springer.
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