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Difference between revisions of "Convex function"

(People! Not all functions are twice differentiable!!!)
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A [[function]] <math>f: I \mapsto \mathbb{R}</math> for some interval <math> \displaystyle I \subseteq \mathbb{R} </math> is ''convex'' (sometimes written ''concave up'') over <math> \displaystyle I </math> if and only if the set of all points <math> \displaystyle (x,y) </math> such that <math> \displaystyle y \ge f(x) </math> is [[convex set | convex]].  Equivalently, <math> \displaystyle f </math> is convex if for every <math> \lambda \in [0,1] </math> and every <math> x,y \in I</math>,
 
A [[function]] <math>f: I \mapsto \mathbb{R}</math> for some interval <math> \displaystyle I \subseteq \mathbb{R} </math> is ''convex'' (sometimes written ''concave up'') over <math> \displaystyle I </math> if and only if the set of all points <math> \displaystyle (x,y) </math> such that <math> \displaystyle y \ge f(x) </math> is [[convex set | convex]].  Equivalently, <math> \displaystyle f </math> is convex if for every <math> \lambda \in [0,1] </math> and every <math> x,y \in I</math>,
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<center><math> \displaystyle \lambda f(x) + (1-\lambda)f(y) \ge f\left( \lambda x + (1-\lambda) y \right) </math>.</center> Usually, when we do not specify <math> \displaystyle I </math>, we mean <math> I = \mathbb{R} </math>.
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<center><math> \displaystyle \lambda f(x) + (1-\lambda)f(y) \ge f\left( \lambda x + (1-\lambda) y \right) </math>.</center>
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We say that <math> \displaystyle f </math> is '''strictly convex''' if equality occurs only when <math> \displaystyle x=y </math> or <math> \lambda \in \{ 0,1 \} </math>.
  
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We say that <math> \displaystyle f </math> is '''concave''' (or, occasionally, that it is ''concave down'') if <math> \displaystyle -f </math> is convex.
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Usually, when we do not specify <math> \displaystyle I </math>, we mean <math> I = \mathbb{R} </math>.
  
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If <math> \displaystyle f </math> is differentiable, then it is convex if and only if <math> \displaystyle f' </math> is non-decreasing.  Similarly, if <math> \displaystyle f </math> is twice differentiable, we say it is convex over an interval <math> \displaystyle I </math> if and only if <math> f(x) \ge 0 </math> for all <math> x \in I </math>.
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We say that <math> \displaystyle f </math> is (strictly) '''concave''' (or, occasionally, that it is ''concave down'') if <math> \displaystyle -f </math> is (strictly) convex.
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If <math> \displaystyle f </math> is differentiable on an interval <math> \displaystyle I </math>, then it is convex on <math> \displaystyle I </math> if and only if <math> \displaystyle f' </math> is non-decreasing on <math> \displaystyle I </math>.  Similarly, if <math> \displaystyle f </math> is twice differentiable over an interval <math> \displaystyle I </math>, we say it is convex over <math> \displaystyle I </math> if and only if <math> f''(x) \ge 0 </math> for all <math> x \in I </math>.
  
 
Note that in our previous paragraph, our requirements that <math> \displaystyle f </math> is differentiable and twice differentiable are crucial.  For a simple example, consider the function
 
Note that in our previous paragraph, our requirements that <math> \displaystyle f </math> is differentiable and twice differentiable are crucial.  For a simple example, consider the function
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{{stub}}
 
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== Resources ==
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* [[Jensen's Inequality]]
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* [[Karamata's Inequality]]

Revision as of 09:17, 17 June 2007

A function $f: I \mapsto \mathbb{R}$ for some interval $\displaystyle I \subseteq \mathbb{R}$ is convex (sometimes written concave up) over $\displaystyle I$ if and only if the set of all points $\displaystyle (x,y)$ such that $\displaystyle y \ge f(x)$ is convex. Equivalently, $\displaystyle f$ is convex if for every $\lambda \in [0,1]$ and every $x,y \in I$,

$\displaystyle \lambda f(x) + (1-\lambda)f(y) \ge f\left( \lambda x + (1-\lambda) y \right)$.

We say that $\displaystyle f$ is strictly convex if equality occurs only when $\displaystyle x=y$ or $\lambda \in \{ 0,1 \}$.

Usually, when we do not specify $\displaystyle I$, we mean $I = \mathbb{R}$.

We say that $\displaystyle f$ is (strictly) concave (or, occasionally, that it is concave down) if $\displaystyle -f$ is (strictly) convex.

If $\displaystyle f$ is differentiable on an interval $\displaystyle I$, then it is convex on $\displaystyle I$ if and only if $\displaystyle f'$ is non-decreasing on $\displaystyle I$. Similarly, if $\displaystyle f$ is twice differentiable over an interval $\displaystyle I$, we say it is convex over $\displaystyle I$ if and only if $f''(x) \ge 0$ for all $x \in I$.

Note that in our previous paragraph, our requirements that $\displaystyle f$ is differentiable and twice differentiable are crucial. For a simple example, consider the function

$f(x) = \lfloor x \rfloor (x - \lfloor x \rfloor ) + {\lfloor x \rfloor \choose 2}$,

defined over the non-negative reals. It is piecewise differentiable, but at infinitely many points (for all natural numbers $\displaystyle x$, to be exact) it is not differentiable. Nevertheless, it is convex. More significantly, consider the function

$f(x) = \left( |x| - 1 \right)^2$

over the interval $\displaystyle [-2, 2]$. It is continuous, and twice differentiable at every point except $\displaystyle{} (0, 1)$. Furthermore, its second derivative is greater than 0, wherever it is defined. But its graph is shaped like a curvy W, and it is not convex over $\displaystyle [-2,2]$, although it is convex over $\displaystyle [-2,0]$ and over $\displaystyle [0,2]$.

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Resources

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