Strong Convex Function
A Strong Convex Function is a convex function that ...
- AKA: Strongly Convex Function.
- See: Elliptic Operator, Taylor's Theorem, Extreme Value Theorem, Hessian Matrix, Positive-Definite Matrix, Eigenvalue, Nesterov Algorithm.
References
2015
- (Wikipedia, 2015) ⇒ http://en.wikipedia.org/wiki/convex_function#Strongly_convex_functions Retrieved:2015-6-8.
- 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 is called strongly convex with parameter m > 0 if the following inequality holds for all points x, y in its domain: : [math]\displaystyle{ (\nabla f(x) - \nabla f(y) )^T (x-y) \ge m \|x-y\|_2^2 }[/math] or, more generally, : [math]\displaystyle{ \langle \nabla f(x) - \nabla f(y), (x-y) \rangle \ge m \|x-y\|^2 }[/math] where [math]\displaystyle{ \|\cdot\| }[/math] is any norm. Some authors, such as refer to functions satisfying this inequality as elliptic functions.
An equivalent condition is the following: : [math]\displaystyle{ f(y) \ge f(x) + \nabla f(x)^T (y-x) + \frac{m}{2} \|y-x\|_2^2 }[/math] It is not necessary for a function to be differentiable in order to be strongly convex. A third definition for a strongly convex function, with parameter m, is that, for all x, y in the domain and [math]\displaystyle{ t\in [0,1] }[/math] , : [math]\displaystyle{ f(tx+(1-t)y) \le t f(x)+(1-t)f(y) - \frac{1}{2} m t(1-t) \|x-y\|_2^2 }[/math] 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 is twice continuously differentiable, then is strongly convex with parameter m if and only if [math]\displaystyle{ \nabla^2 f(x) \succeq m I }[/math] for all x in the domain, where I is the identity and [math]\displaystyle{ \nabla^2f }[/math] is the Hessian matrix, and the inequality [math]\displaystyle{ \succeq }[/math] means that [math]\displaystyle{ \nabla^2 f(x) - mI }[/math] is positive semi-definite. This is equivalent to requiring that the minimum eigenvalue of [math]\displaystyle{ \nabla^2 f(x) }[/math] be at least m for all x. If the domain is just the real line, then [math]\displaystyle{ \nabla^2 f(x) }[/math] is just the second derivative [math]\displaystyle{ f''(x) }[/math] , so the condition becomes [math]\displaystyle{ f''(x) \ge m }[/math] . If m = 0, then this means the Hessian is positive semidefinite (or if the domain is the real line, it means that [math]\displaystyle{ f''(x) \ge 0 }[/math] ), 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 [math]\displaystyle{ \nabla^2 f(x) }[/math] implies that it is strongly convex. Start by using Taylor's Theorem: : [math]\displaystyle{ f(y) = f(x) + \nabla f(x)^T (y-x) + \frac{1}{2} (y-x)^T \nabla^2f(z) (y-x) }[/math] for some (unknown) [math]\displaystyle{ z \in \{ t x + (1-t) y : t \in [0,1] \} }[/math] . Then : [math]\displaystyle{ (y-x)^T \nabla^2f(z) (y-x) \ge m (y-x)^T(y-x) }[/math] by the assumption about the eigenvalues, and hence we recover the second strong convexity equation above.
A function is strongly convex with parameter m if and only if the function [math]\displaystyle{ x\mapsto f(x) -\frac{m}{2}\|x\|^2 }[/math] is convex.
The distinction between convex, strictly convex, and strongly convex can be subtle at first glimpse. If is twice continuously differentiable and the domain is the real line, then we can characterize it as follows:
: convex if and only if [math]\displaystyle{ f''(x) \ge 0 }[/math] for all .
: strictly convex if [math]\displaystyle{ f''(x) \gt 0 }[/math] for all (note: this is sufficient, but not necessary).
: strongly convex if and only if [math]\displaystyle{ f''(x) \ge m \gt 0 }[/math] for all .
For example, consider a function that is strictly convex, and suppose there is a sequence of points [math]\displaystyle{ (x_n) }[/math] such that [math]\displaystyle{ f''(x_n) = \frac{1}{n} }[/math] . Even though [math]\displaystyle{ f''(x_n) \gt 0, }[/math] the function is not strongly convex because [math]\displaystyle{ f''(x) }[/math] will become arbitrarily small.
A twice continuously differentiable function on a compact domain [math]\displaystyle{ X }[/math] that satisfies [math]\displaystyle{ f''(x) \gt 0 }[/math] for all [math]\displaystyle{ x\in X }[/math] 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 on compact sets.
- 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.