Convex function

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In mathematics, a real-valued function f defined on an interval (or on any convex subset C of some vector space) is called convex, 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.
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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. A function is also said to be strictly convex if

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

for any t in (0,1).

The opposite of a convex function is a concave function.

Contents

[edit] Properties of convex functions

A convex function f 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 f may fail to be continuous at the endpoints of C.

A continuous function on an interval C is convex if and only if

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

for all x and y in C.

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: f(y) ≥ f(x) + f'(x) (yx) for all x and y in the interval.

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, as shown by f(x) = x4.

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.

If two functions f and g are convex, then so is any weighted combination a f + b g with non-negative coefficients a and b. Likewise, if f and g are convex, then the function max{f,g} is convex.

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 level sets {x | f(x) < a} and {x | f(x) ≤ a} with aR are convex sets. However, a function whose level sets are convex sets may fail to be a convex function; such a function called a quasiconvex function.

Convex functions respect Jensen's inequality.

[edit] Examples

  • The second derivative of x2 is 2; it follows that x2 is a convex function of x.
  • The absolute value function |x| is convex, even though it does not have a derivative at x = 0.
  • The function f with domain [0,1] defined by f(0)=f(1)=1, f(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 for x ≥ 0 and concave for x ≤ 0.
  • Every linear transformation is convex but not strictly convex, since if f is linear, then f(a + b) = f(a) + f(b). This implies that the identity map (i.e., f(x) = x) is convex but not strictly convex. The fact holds if we replace "convex" by "concave".
  • An affine function is simultaneously convex and concave.

[edit] Useful theorems

[edit] Derivatives of convex functions

The function f(x) is convex if and only if its derivative f '(x) (if it exists) is monotonically increasing.

Proof: If f(x) is convex then choose two points x and y on the domain of f such that x < y and let k be

k = \frac {x+y}{2}

Using the mean value theorem:

f(x) − f(k) = f'(ξ1)(xk)
f(y) − f(k) = f'(ξ2)(yk)

We want to prove that f '(ξ1) ≤ f '(ξ2) or equivalently that

\frac { f(x) - f(y) }{ x - k } \le \frac { f(y)-f(k) }{ y - k }

Substituting k in the denominator

\frac { f(x) - f(k) }{ x - \frac {1}{2}x - \frac {1}{2}y } \le \frac { f(y)-f(k) }{ y - \frac {1}{2}x - \frac {1}{2}y }

Let's rewrite the inequality

f(x) - f(k)  \ge f(k)-f(y)
f(k) \le \frac {f(x) + f(y)}{2}

We have to show this last statement, but as we know that f is convex, we notice it's the convex function definition:

f \left (\frac {1}{2}x + \frac {1}{2}y \right ) \le \frac {1}{2}f(x) + \frac {1}{2}f(y)

as we wanted to prove.

Converse theorem: If f '(x) is monotonically increasing then choose two point x and y on the domain of f and let k be

k = tx + (1 − t)y

where t \in [0, 1] Using the mean value theorem we get:

f(x) − f(k) = f'(ξ1)(xk)
f(y) − f(k) = f'(ξ2)(yk)

we know that f '(ξ1) ≤ f '(ξ2), then

\frac {f(x)-f(k)}{x-k} \le \frac {f(y)-f(k)}{y-k}

Now, eliminate k

\frac {f(x)-f(t x + (1-t) y)}{(1-t) (x - y)} \le \frac {f(y)-f(t x + (1-t) y)}{t (y-x)}
f(t x + (1-t) y) - f (x) \le \frac {1-t}{t}  \Big (  f(y) - f(t x + (1-t) y)  \Big  )

and finally we get

f(t x + (1-t) y) \le t f(x) + (1-t) f(y)

which is the definition of convex function.

[edit] See also