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In mathematics, a real-valued function ƒ defined on an interval (or on any convex subset of some vector space) is called convex, concave upwards, concave up or convex cup, if for any two points x and y in its domain C and any t in [0, 1], we have

$f(tx+(1-t)y)\leq t f(x)+(1-t)f(y).$

Convex function on an interval.

In other words, a function is convex if and only if its epigraph (the set of points lying on or above the graph) is a convex set.

Pictorially, a function is called 'convex' if the function lies below or on the straight line segment connecting two points, for any two points in the interval.

A function is called strictly convex if

$f(tx+(1-t)y) < t f(x)+(1-t)f(y)\,$

for any t in (0, 1) and x ≠ y.

A function ƒ is said to be concave if −ƒ is convex.

1 Properties
2 Convex function calculus
3 Strongly convex functions
4 Examples
5 See also
6 References
7 External links

Properties

A function (in blue) is convex if and only if the region above its graph (in green) is a convex set.

A convex function ƒ defined on some open interval C is continuous on C and differentiable at all but at most countably many points. If C is closed, then ƒ may fail to be continuous at the endpoints of C.

A function is midpoint convex on an interval C if

$f\left( \frac{x+y}{2} \right) \le \frac{f(x)+f(y)}{2}$

for all x and y in C. This condition is only slightly weaker than convexity. For example, a real valued Lebesgue measurable function that is midpoint convex will be convex.^[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.

A continuously differentiable function of one variable is convex on an interval if and only if the function lies above all of its tangents: ƒ(y) ≥ ƒ(x) + ƒ '(x) (y − x) for all x and y in the interval. In particular, if ƒ '(c) = 0, then c is a global minimum of ƒ(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. If its second derivative is positive then it is strictly convex, but the converse does not hold. For example, the second derivative of ƒ(x) = x⁴ is ƒ "(x) = 12 x², 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 ƒ, the sublevel sets {x | ƒ(x) < a} and {x | ƒ(x) ≤ a} with a ∈ R are convex sets. However, a function whose sublevel sets are convex sets may fail to be a convex function; such a function is called a quasiconvex function.

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

Convex function calculus

If $f$ and $g$ are convex functions, then so are $m (x) = max{f (x), g (x)}$ and $h (x) = f (x) + g (x).$
If $f$ and $g$ are convex functions and if $g$ is non-decreasing, then $h (x) = g (f (x))$ is convex.
Convexity is invariant under affine maps: that is, if $f (x)$ is convex with $x\in\mathbb{R}^n$ , then so is $g (y) = f (A y + b)$ with $y\in\mathbb{R}^m$ , where $A\in\mathbb{R}^{n \times m},\; b\in\mathbb{R}^n.$
If $f (x, y)$ is convex in $(x, y)$ and $C$ is a convex nonempty set, then $g(x) = \inf_{y\in C} f(x,y)$ is convex in $x,$ provided $g(x) > -\infty$ for some $x .$

Strongly convex functions

The concept of strong convexity extends the notion of strict convexity, and a strongly convex function is also strictly convex, but not vice-versa. A function $f$ is called strongly convex with parameter $m > 0$ if it is differentiable and the following equation holds for all points $x, y$ in its domain:

$( \nabla f(x) - \nabla f(y) )^T (x-y) \ge m \|x-y\|^2$

This is equivalent to the following

$f(y) \ge f(x) + \nabla f(x)^T (y-x) + \frac{m}{2} \|y-x\|_2^2$

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

$f(tx+(1-t)y) \le t f(x)+(1-t)f(y) - \frac{1}{2} m t(1-t) \|x-y\|^2 \,$

(note that $t(1-t) \ge 0$ for $t\in[0,1]$ .)

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

Assuming still that the function is twice continuously differentiable, we show that bounding the minimum eigenvalue of $\nabla^2 f(x)$ implies that it is strongly convex. Start by using Taylor's Theorem:

$f(y) = f(x) + \nabla f(x)^T (y-x) + 1/2 (y-x)^T \nabla^2f(z) (y-x)$

for some (unknown) $z \in [x,y]$ . Then $(y-x)^T \nabla^2f(z) (y-x) \ge m (y-x)^T(y-x)$ by the assumption about the eigenvalues, and hence we recover the second strong convexity equation above.

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

$f\,\!$ convex if and only if $f''(x) \ge 0$ for all $x\,\!$

$f\,\!$ strictly convex if $f''(x) > 0 \,\!$ for all $x\,\!$ (note: this is necessary, but not sufficient)

$f\,\!$ strongly convex if and only if $f''(x) \ge m > 0$ for all $x\,\!$

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

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.

Examples

The function $f (x) = x 2$ has $f\,''(x)=2>0$ at all points, so ƒ is a convex function. It is also strongly convex (and hence strictly convex too), with strong convexity constant 2.
The function $f (x) = x 4$ has $f\,''(x)=12x^2\ge0$ , so ƒ is a convex function. It is strictly convex, even though the second derivative is not strictly positive at all points. It is not strongly convex.
The absolute value function $f (x) = | x |$ is convex, even though it does not have a derivative at the point x = 0. It is not strictly convex.
The function $f (x) = | x | p$ for 1 ≤ p is convex.
The exponential function $f (x) = e x$ is convex. It is also strictly convex, since $f''(x) = e x > 0$ , but it is not strongly convex since the second derivative can be arbitrarily close to zero. More generally, the function $g (x) = e f (x)$ is logarithmically convex if ƒ is a convex function.
The function ƒ with domain [0,1] defined by ƒ(0) = ƒ(1) = 1, ƒ(x) = 0 for 0 < x < 1 is convex; it is continuous on the open interval (0, 1), but not continuous at 0 and 1.
The function x³ has second derivative 6x; thus it is convex on the set where x ≥ 0 and concave on the set where x ≤ 0.
Every linear transformation taking values in $\mathbb{R}$ is convex but not strictly convex, since if f is linear, then $f (a + b) = f (a) + f (b).$ This statement also holds if we replace "convex" by "concave".
Every affine function taking values in $\mathbb{R}$ , i.e., each function of the form $f (x) = a T x + b$ , is simultaneously convex and concave.
Every norm is a convex function, by the triangle inequality.
If ƒ is convex, the perspective function $g (x, t) = t f (x / t)$ is convex for t > 0.
Examples of functions that are monotonically increasing but not convex include $f(x) = \sqrt x$ and g(x) = log(x).
Examples of functions that are convex but not monotonically increasing include $h (x) = x 2$ and $k (x) = - x$ .
The function ƒ(x) = 1/x², with f(0) = +∞, is convex on the interval (0, +∞) and convex on the interval (-∞,0), but not convex on the interval (-∞, +∞), because of the singularity at x = 0.

References

^ Sierpinski Theorem, Donoghue (1969), p. 12

Moon, Todd. "Tutorial: Convexity and Jensen's inequality". http://www.neng.usu.edu/classes/ece/7680/lecture2/node5.html. Retrieved 2008-09-04.
Rockafellar, R. T. (1970). Convex analysis. Princeton: Princeton University Press.
Luenberger, David (1984). Linear and Nonlinear Programming. Addison-Wesley.
Luenberger, David (1969). Optimization by Vector Space Methods. Wiley & Sons.
Bertsekas, Dimitri (2003). Convex Analysis and Optimization. Athena Scientific.
Thomson, Brian (1994). Symmetric Properties of Real Functions. CRC Press.
Donoghue, William F. (1969). Distributions and Fourier Transforms. Academic Press.

Hiriart-Urruty, Jean-Baptiste, and Lemaréchal, Claude. (2004). Fundamentals of Convex analysis. Berlin: Springer.
Krasnosel'skii M.A., Rutickii Ya.B. (1961). Convex Functions and Orlicz Spaces. Groningen: P.Noordhoff Ltd.
Borwein, Jonathan, and Lewis, Adrian. (2000). Convex Analysis and Nonlinear Optimization. Springer.

External links

Stephen Boyd and Lieven Vandenberghe, Convex Optimization (PDF)

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