CONVEXITY AT A POINT
Undergraduate Research Project
by
Siwei Shen
Spring 2000
Relation between being convex at a point and having support at a point
Conclusion: Having a support at a point implies being convex at that point. More specifically, say function ¦ : U ® V has a support at a point X0 Î U, then, ¦ is convex at X0.
Proof: Let X1, X2 Î U, X0 = l X1 + (1- l )X2, where l Î [0, 1],
and let A(X) = ¦ (X0) + T(X - X0) be an affine function that supports ¦ at X0.
Then, A(X) £ ¦ (X), " X Î U. The equality only holds at point X0 as the affine function touches ¦ at that point by definition of support. That is:
¦ (X0) = A(X0) = A[l X1 + (1- l )X2];
Since A is affine, the right hand side is equal to
A(l X1) + A(1- l ) X2) = l A(X1) + (1- l ) A(X2);
I.e. ¦ (X0) = l A(X1) + (1- l )A(X2). Because A(X) is the support for ¦ on U, we have
¦ (X0) = l A(X1) + (1- l )A(X2) £ l ¦ (X1) + (1- l )¦ (X2), which establishes the convexity of the point X0.
Remark: Geometrically, convexity at a point requires that each chord linking two points on the graph of the function on two sides of the point lies above the point itself. Of course, this makes sense only in 2-dimension. Thus, for 2-dimension, we could approach the above proof in some (non-rigorous) geometric way. The easy case would be that the point at which the function has a support is the minimum point of the function, then the chord-over-point argument above automatically holds. Therefore, the question boils down to the case in which the point having support is not the minimum point. However, we play a trick by rotating the graph of the function until the support becomes parallel to x axis. Since every point on the graph lies above the support (straight line), any chord between two points on different sides of the point having support is guaranteed to be over the point. As we have not changed the shape of the function, it implies that in the original graph, the chord-over-point statement is true. This way, we know having a support assures convexity at a point. As said before, this so-called geometric approach does have a lot of restrictions (dimension, etc.) so that it should not be viewed as any formal proof or so.
Convexity at point vs. Convexity on set
Conclusion: Let U be a convex set, if a function ¦ is convex at each point on U, then ¦ is convex on U.
Proof: Let X1, X2 Î U, X0 = l X1 + (1- l )X2, where l Î [0, 1].
Since U is a convex set, it is guaranteed that X0 Î U. Then, as ¦ is convex at each point on U, we have: ¦ (X0) = ¦ [l X1 + (1- l )X2] £ l ¦ (X1) + (1- l )¦ (X2). This shows that ¦ is convex on U.
If a function is convex at a point that is not the minimum point, is it always possible to find a neighborhood of that point on which the function is convex?
Conclusion: No.
Proof: To prove the statement is not true, one counterexample is more than enough. One of such examples is shown below:
Consider function ¦ defined on [- 1, 1] and
¦ (X) = X^{2/3} X Î [- 1, 0)
¦ (X) = - X/3 X Î [0, 1].
We claim that ¦ is convex at X = 0 (not the minimum point in the domain), but there is no neighborhood of X = 0 such that ¦ is convex on.
To show this, we apply the geometric interpretation of being convex at a point, that is, any chord linking two points on the graph on different sides of ¦ (0) lies above the point X = 0. Arbitrarily choose a Î [- 1, 0) and b Î (0, 1] (¦ (a) & ¦ (b) are on different size of ¦ (0)), and look at the chord linking ¦ (a) and ¦ (b). The slope of the chord can be calculated as follows:
m = [a^{2/3} - (- b/3)] / (a – b).
a Î [-1, 0) Þ - a^{1/3} £ 1 £ 3 Þ 3a^{2/3} ³ - a Þ 3a^{2/3} + b ³ b - a Þ [3a^{2/3} + b] / (a - b) £ - 1
Þ [a^{2/3} - (- b/3)] / (a – b) £ - 1/3, i.e. m £ - 1/3.
This means that the chord is steeper than the right branch of graph of function ¦ . Thus, it must be above X = 0. Since a & b are arbitrary, it is safe to say that any chord of this kind satisfies this, which in turn implies the convexity at X = 0.
It is clear that the left branch of function ¦ is concave so that there does not exist any Î neighborhood of X = 0 on which ¦ can be convex.