proof of criterion for convexity II
If was not convex, then there was a point such that for some in . Since is continuous, there would be a neighborhood , of such that for all in this neighborhood. (I.e., was “above” the line segment joining and .) Let .
Using the two points , we construct another line segment whose equation is given by ; we have for . In particular,
(One easily verifies .) This contradicts hypothesis.
Note that we have tacitly used the fact that for some and for some .
|Title||proof of criterion for convexity II|
|Date of creation||2013-03-22 18:25:25|
|Last modified on||2013-03-22 18:25:25|
|Last modified by||yesitis (13730)|