Continuous and smooth on a compact set implies differentiability at a point

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SUMMARY

The discussion centers on proving that a function continuous on the closed interval [a,b] and smooth on the open interval (a,b) has at least one point x in (a,b) where the derivative f'(x) exists. The definition of smoothness used is that f is smooth at x if the limit \lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } equals zero. The user is attempting to demonstrate that the one-sided difference quotients converge as h approaches zero, but is encountering difficulties in establishing their existence.

PREREQUISITES
  • Understanding of limits and continuity in calculus
  • Familiarity with the concept of differentiability
  • Knowledge of the definition of smooth functions
  • Experience with one-sided difference quotients
NEXT STEPS
  • Study the properties of continuous functions on compact sets
  • Learn about the Mean Value Theorem and its implications for differentiability
  • Investigate the relationship between smoothness and differentiability in real analysis
  • Explore examples of functions that are continuous but not differentiable
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Mathematicians, calculus students, and anyone interested in real analysis, particularly those studying the properties of continuous and smooth functions.

resolvent1
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I'm trying to prove that if a function is continuous on [a,b] and smooth on (a,b) then there's a point x in (a,b) where f'(x) exists. The definition of smoothness is: f is smooth at x iff [itex]\lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0[/itex].

I'm starting with the simpler case where f(a) = f(b). I know that f has a max x and a min y on [a,b], and that if both occur at a or b, then f is a constant function. So I'm assuming wlog that the max x occurs somewhere in (a,b).

So at x,
[itex]- \frac{ (f(x) - f(x+h)) } {h} \leq 0[/itex] and
[itex]\frac{f(x-h) - f(x)}{h} \leq 0[/itex].

This is where I'm getting stuck. I'd like to show that the limits of the one sided difference quotients as [itex]h \rightarrow 0-[/itex] exist and then apply smoothness to show that these must be equal, hence f is differentiable at x. However, I haven't been able to show that they exist - taking the infima of both sides doesn't seem to yield anything, so I'm guessing I might be missign something else. Thanks in advance.
 
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resolvent1 said:
I'm trying to prove that if a function is continuous on [a,b] and smooth on (a,b) then there's a point x in (a,b) where f'(x) exists. The definition of smoothness is: f is smooth at x iff [itex]\lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0[/itex].

I'm starting with the simpler case where f(a) = f(b). I know that f has a max x and a min y on [a,b], and that if both occur at a or b, then f is a constant function. So I'm assuming wlog that the max x occurs somewhere in (a,b).

So at x,
[itex]- \frac{ (f(x) - f(x+h)) } {h} \leq 0[/itex] and
[itex]\frac{f(x-h) - f(x)}{h} \leq 0[/itex].

This is where I'm getting stuck. I'd like to show that the limits of the one sided difference quotients as [itex]h \rightarrow 0-[/itex] exist and then apply smoothness
What, exactly, is your definition of "smooth"?

to show that these must be equal, hence f is differentiable at x. However, I haven't been able to show that they exist - taking the infima of both sides doesn't seem to yield anything, so I'm guessing I might be missign something else. Thanks in advance.
 
The definition I'm using for smooth is:

A function [itex]f : \mathbb{R} \rightarrow \mathbb{R}[/itex] is said to be smooth at a point [itex]x \in \mathbb{R}[/itex] iff it is defined in a neighborhood of x and

[itex]\lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0[/itex].
 

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