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08jmliu

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[½»Á÷] Let a function f : R => R be convex. Is f necessarily continuous£¿ ÒÑÓÐ3È˲ÎÓë

Let a function f : R => R be convex.  Is f necessarily continuous£¿
Why£¿ Give me some details or references.

Thanks!
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СÓêÃÈÃÈ(½ð±Ò+3): 3Q 2011-11-23 09:10:13
Yes. For u (f(y)-f(x0))/(y-x0)<=(f(x0)-f(u))/(x0-u)
and
(f(x0)-f(x))/(x0-x)<=(f(w)-f(x0))/(w-x0).
Fix u and w and let x,y tend to x0.
2Â¥2011-11-20 20:56:56
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xxxfield

Òø³æ (СÓÐÃûÆø)

Sorry, the second inequality should be
(f(x0)-f(x))/(x0-x)>=(f(w)-f(x0))/(w-x0).
3Â¥2011-11-20 20:58:16
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zhangzujin

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Сľ³æ(½ð±Ò+0.5):¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
ÒýÓûØÌû:
1Â¥: Originally posted by 08jmliu at 2011-11-20 15:39:12:
Let a function f : R => R be convex.  Is f necessarily continuous£¿
Why£¿ Give me some details or references.

Thanks!

Year. Indeed, the left and right derivative exists for each interior point...
4Â¥2011-11-20 21:21:09
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

jfili

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СÓêÃÈÃÈ(½ð±Ò+2): 3Q 2011-11-23 09:10:25
µÚÒ»²½£ºÉèF(x)=[f(x)-f(a)]/(x-a)ÔÚ£¨x²»µÈÓÚaʱÓж¨Ò壩£¬ÏÈÖ¤F(x)µ¥µ÷Ôö¼Ó£¬ÓÐÉÏϽ磨ÔÚº¬aµÄijÁÚÓòÉÏ£©
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5Â¥2011-11-22 20:45:53
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