the answer is for closed interval I want for open interval
can you help me please again ?
A function is continuous on an open interval if it is continuous on every closed subinterval. So the proof given for closed intervals implies that the result is also true for open intervals.
Originally Posted by Krizalid
you should know that the open interval is contained in the closed one; the closed interval just takes the endpoints of the interval, so the proof given above is a stronger result.
But a convex function defined on an open interval may not have an extension to the closure of the interval (it could go off to infinity there). So I don't think that you can deduce the result for open intervals by taking the closure of the interval.