Talk:Convex function

I'm not sure what the opinion at Wikipedia is on the matter of accessibility of the material. When I come here, however, I am not looking for mathematically concise definitions of what I look up, nor do I handle well the high level description that requires knowledge of many other technical terms.

A paragraph describing the significance of the defining equation would be nice.

For me at least, this page is exactly what I was looking for about convex functions.

Correct Definition

Should the definition read "for all x,y" instead of for any?

To anyone used to reading math (and maybe most who aren't), it's the same thing. But the change might improve clarity some small amount. Dchudz 17:08, 15 November 2006 (UTC)

But anyone writing math would never say 'for any', as strictly this requires the condition to be satisfied just for some, and not for all instances. In this article, I am missing the 'useful theorems' section for higher dimensions, as well as a generally better treatment for the high dimensions. But, I still appreciate your work!

Merge with concavity?

Honestly, think that's a bad idea. Although the two concepts are so completely intertwined as to be two sides of the same coin, people come to Wikipedia for answers, not answer-hunting (reference the post above). I suppose it's possible that the merge could be done in such a way as to not be too confusing, but take a look at supermodular to see what can happen when things get thrown together (read the title, and then the definition; but then I suppose I'm supposed to fix it, not just sit back and complain). All in all, it would probably be better just to interlink the two (concavity and convexity) religiously -- say a link to concavity in the main definition. This is a good page right now: concise and exactly what people are looking for. Don't see much need to change it. —The preceding unsigned comment was added by Semanticprecision (talk • contribs) .

OK, I did not do any merge, rather moved concavity (which is now disambig) to concave function. Things might still need merging in the future, but at least for now the meaning of concave set and concave function are separate and not in the same article (with concave function taking the lion share of the room). Let us see how it goes. Oleg Alexandrov (talk) 00:28, 30 December 2005 (UTC)

In February 2005, I merged convex and concave function in the German Wikipedia (de:konvexe und konkave Funktionen). IMHO it is very difficult to maintain the versions for convex/concave funtion in a consistent way, and it is easy to write the article in not too confusing way. To avoid problems like in supermodular, clearly the title should mention both (e.g. convex and concave functions, as in the German version). --NeoUrfahraner 09:15, 2 January 2006 (UTC)

Removing link to Additive Inverse

The link from "opposite" ("The opposite of a convex function is a concave function") to the additive inverses page seems a bit gratuitous. I'm removing it. --Dchudz 14:13, 25 July 2006 (UTC)

The example with f(x)=x⁴

I didn't want to modify the article, but there is something suspicious to me. It says:

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.

Shouldn't the Hessian reduce to 2nd derivative in one-dimensional case?
The example function f(x) = x⁴ is convex and its second derivative 12x² is nonnegative, so I don't see why the converse does not hold. Moreover in the previous sentence it says "if and only if", so the converse MUST hold. Did I miss something ? —Preceding unsigned comment added by 83.37.136.53 (talk) 09:55, 18 January 2008 (UTC)

As far as I see, the article says that if the second derivative is strictly positive, then the function is strictly convex, but if the function is strictly convex, its second derivative need not be strictly positive, with $x^{4}$ being a counterexample (this function is indeed strictly convex, but the second derivative is not strictly positive, it also takes the value 0). Oleg Alexandrov (talk) 15:36, 18 January 2008 (UTC)

semi-strictly convex

what is "semi-strictly convext" of a function? Jackzhp (talk) 02:09, 30 November 2008 (UTC)

Strongly convex

It would be great to add a section with the definition of a strongly convex function; or, make a page for "strongly convex" (since it appears that it doesn't yet exist) and then link to that page. 71.130.221.31 (talk) 02:46, 2 February 2009 (UTC)

Well, I'd like to know what a "strongly convex" function is? --Bdmy (talk) 12:27, 2 February 2009 (UTC)

If a function is twice continuously differentiable, then is is convex if and only if its second derivative is never negative (and it's strictly convex if in addition the second derivative is never zero). A strongly convex function's second derivative is bounded away from zero.

Following Boyd and Vandenberghe's book, we have:

A twice continuously differentiable function is "strongly convex" if

$\nabla ^{2}f(x)\geq mI$

for all $x$ in the domain. The inequality is with respect to the positive semidefinite cone.

This let's us extend one of the fundamental properties of a convex function (that a convex function is always above a tangent plane) to the following:

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

for all $x$ and $y$ in the domain. If the function were merely convex, the final term with the $m$ wouldn't be there, and if the function were merely strictly convex, the final term wouldn't be there either, but the inequality would become a strict inequality.

The constant $m$ depend on the function, and we need $m>0$ otherwise it's not useful. Strongly convex functions are generally nicer than just convex functions when solving an optimization problem. Lavaka (talk) 21:13, 22 February 2009 (UTC)

I added this to the main page, since it's not too long and I think it's useful (but probably doesn't deserve it's own page).

I'm sure about my comments regarding strong convexity, however I'm unsure about some of the strict convexity comments (as the x^4 discussion above shows. update: I'm fixing my strict convexity comments Lavaka (talk) 02:06, 10 September 2009 (UTC)

Equivalent definition

I was just told that an equivalent definition is the following: a function is convex if at any point we can find a suppporting line (a line through the point that lies entirely below the function), i.e. f is convex if for any a, there exists an m such that $f(x)\geq f(a)+m(x-a)$ for all x. (It's easy to see why if that condition is true the function is convex, but I can't immediately think of a proof for the other direction.) Wondering if someone knows a source.... I could only find Springer EOM, which indicates that maybe this definition is equivalent to midpoint convex. Shreevatsa (talk) 01:15, 18 February 2009 (UTC)

In the real case you just need to take the value m to be between the left- and right-derivatives of ƒ at a (assuming a is an interior point of the domain, otherwise you can't do it in general). In several dimensions, what you need is more or less the finite dimensional part of the Hahn-Banach theorem. --Bdmy (talk) 11:21, 18 February 2009 (UTC)

Is it not possible that f does not have left- and right-derivatives at a? Anyway, so what is the condition equivalent to? (There is a cute proof of Jensen's inequality assuming this condition: take a to be E[X], then take expectation.) Shreevatsa (talk) 00:12, 20 February 2009 (UTC)

A (finite) convex function on an interval is continuous in the interior of the interval, and has left- and right-derivatives at every interior point a. This can be found in most books on functions of one real variable, together with an illustration of the fact that the "slopes" are increasing: when a < b < c,

${\frac {f(b)-f(a)}{b-a}}\leq {\frac {f(c)-f(a)}{c-a}}\leq {\frac {f(c)-f(b)}{c-b}}.$

(imagine, or draw for yourself, the picture that goes with these inequalities) --Bdmy (talk) 12:34, 20 February 2009 (UTC)

This is somehow related but have anyone see this equivalent definition, something like: a function is convex if and only if for any arbitrary E

$\sup _{\partial E}f=\sup _{E}f$

where I don't remember what E should be :) a compact set?. The above is incorrect in the way it is written; I don't remember the exact formulation. I also remember the converse of Jensen inequality is true (under some conditions). (This was an exercise in Rudin's real and complex analysis.) So, it should give another equivalent definition. -- Taku (talk) 13:42, 18 February 2009 (UTC)

Convex Decreasing

It would be good to have a picture of a convex decreasing function as well. See Indifference curve. Trogsworth (talk) 19:12, 5 May 2009 (UTC)

Counterexamples

It'd be really wonderful if along with some examples we could supply some interesting counterexamples.