# How to prove $f(a)\leq f(b)$?

• Nov 19th 2012, 05:30 PM
How to prove $f(a)\leq f(b)$?
Dear all, suppose $\displaystyle f$ is a continous function on [TEX[a,b][/TEX]. $\displaystyle E\subset[a,b]$ is countable. Assume $\displaystyle f$ is differentiable on $\displaystyle [a,b]\backslash E$, the the derivative $\displaystyle >=0$. Show that $\displaystyle f(a)\leq f(b).$

Note: We assume only that $\displaystyle f$ is differentiable on $\displaystyle [a,b]\backslash E.$

I do not know how to prove. Do you know? Help me, thank you...
• Nov 20th 2012, 01:23 PM
RBowman
Re: How to prove $f(a)\leq f(b)$?
Since E is a countable subset of (a,b), let E = {ei} with a < e1 < e2< ... < b.

Then we have that f is continuos on [a, e1], and differentiable on (a, e1). The mean value theorem applies and we find that f(a) $\displaystyle \leq$ f(e1). We can prove this by contradiction. Assuming f(a) > f(e1), we would have by the MVT a c1, s.t. f'(c1) = (f(e1​)-f(a))/(e1-a) < 0, a contradiction to f'(x) $\displaystyle \geq$ 0.

Similiarly, we can find that f(e1) $\displaystyle \leq$ f(e2). Continuing in this way we can argue that for any ei < ej, f(ei) $\displaystyle \leq$ f(ej).

I think this may be a step toward proving the desired, f(a) $\displaystyle \leq$ f(b), but I haven't yet worked it completely. Will finish and post the rest soon, unless you or someone else can finish it.
• Jan 8th 2013, 11:55 PM
Re: How to prove $f(a)\leq f(b)$?