Real Analysis - Mean Value Theorem Application

In summary: As derivatives of a function with respect to a given parameter, antiderivatives are important for finding limits and for calculating areas and volumes.
  • #1

O_o

32
3

Homework Statement


Let f: R -> R be a function such that [tex] \lim_{z\to 0^+} zf(z) \gt 0 [/tex] Prove that there is no function g(x) such that g'(x) = f(x) for all x in R.

Homework Equations


Supposed to use the mean value theorem. If f(x) is continuous on [a,b] and differentiable on (a,b) then [tex] \exists c \in (a,b), f'(c) = \frac{f(b) - f(a)}{b-a}[/tex]

The Attempt at a Solution


Assume there exists a function g(x) such that g'(x) = f(x) for all x in R. By the mean value theorem:
[tex] \exists c \in (0,x), f(c) = g'(c) = \frac{g(x) - g(0)}{x - 0}[/tex]
This means [tex] \lim_{x\to 0^+} \left(\frac{g(x) - g(0)}{x} - f(c) \right)= 0
\\
\lim_{x\to 0^+} \left(g(x) - g(0) - xf(c)\right) = 0
\\
\lim_{x\to 0^+} \left(g(x) - g(0)\right) = \lim_{x\to 0^+} xf(c)
[/tex]Since g(x) is differentiable everywhere (and hence continuous), the left side is equal to 0.
Now the right side is where I'm not completely comfortable. If it was the limit of x(fx) then it's exactly what I'm given in the question. I'm thinking since x is going to 0 from the right then c also has to be going to 0 and so the limit on the right is greater than 0, which is a contradiction.

Can anyone tell me if I'm using faulty reasoning? I'm not really looking for the correct answer if I'm wrong. I'd just like to know where the flaw in my logic is.

Thanks
 
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  • #2
o_O said:
Now the right side is where I'm not completely comfortable. If it was the limit of x(fx) then it's exactly what I'm given in the question. I'm thinking since x is going to 0 from the right then c also has to be going to 0

You are treating [itex] c [/itex] as a function of [itex] x [/itex], which is permissible provided you define the function. For a given [itex] x [/itex] there might be more than one value [itex] c [/itex] that satisfies [itex] g'(c) = \frac{ g(x) - g(0) }{x - 0 } [/itex]. Unless you specify which [itex] c [/itex] to choose you haven't defined [itex] c [/itex] as a function of [itex] x [/itex].
 
  • #3
Thanks that's helpful. So if I can define a function c(x), is the following true:

If [tex] \lim_{x\to 0^+} c(x) = 0 [/tex] then [tex] \lim_{x\to 0^+} x f(c(x)) = \lim_{x\to 0^+} x f(x)[/tex]

And if so, does c(x) have to have any properties like being continuous or injective?
 
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  • #4
o_O said:
Thanks that's helpful. So if I can define a function c(x), is the following true:

If [tex] \lim_{x\to 0^+} c(x) = 0 [/tex] then [tex] \lim_{x\to 0^+} x f(c(x)) = \lim_{x\to 0^+} x f(x)[/tex]

And if so, does c(x) have to have any properties like being continuous or injective?
No.
Consider f(x) = 1/x, c(x) = x2.
 
  • #5
o_O said:
is the following true:

If [tex] \lim_{x\to 0^+} c(x) = 0 [/tex] then [tex] \lim_{x\to 0^+} x f(c(x)) = \lim_{x\to 0^+} x f(x)[/tex]

Among other things, that depends on a theorem that says [itex] lim_{x \rightarrow a} w(q(x)) = w ( lim_{x \rightarrow a } q(x)) [/itex]

And if so, does c(x) have to have any properties like being continuous or injective?


For that theorem to apply, [itex] w(x) [/itex] must be continuous at [itex] x= a [/itex], but I don't think that [itex] q(x) [/itex] must be. Look it up, under "composite limit theorem" or "limit of the composition of functions".

In addition to that theorem you'll need a theorem about the limit of a product [itex] x f(x) [/itex] being the product of the limits of the individual factors.
 
  • #6
There is a theorem about the limit of a product being the product of the limits of the individual factors. If [itex] f(x) [/itex] were continuous at [itex] x= 0 [/itex], what would that tell you about [itex] lim_{z \rightarrow 0} z f(z) [/itex]
 
  • #7
Stephen Tashi said:
There is a theorem about the limit of a product being the product of the limits of the individual factors. If [itex] f(x) [/itex] were continuous at [itex] x= 0 [/itex], what would that tell you about [itex] lim_{z \rightarrow 0} z f(z) [/itex]
If [tex]\lim_{x\to 0} f(x) = L[/tex] then [tex]\lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = L*0 = 0[/tex]
So if f is continuous at x=0 then [tex]\lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = f(0)*0 = 0[/tex]

So we can conclude that f isn't continuous at 0. And more generally [tex]\lim_{x\to 0} f(x)[/tex] does not exist.
 
  • #8
o_O said:
So we can conclude that f isn't continuous at 0.
Yes

And more generally [tex]\lim_{x\to 0} f(x)[/tex] does not exist.

I'm not sure we can't conclude that. Check what the if-part of the theorem about the limit of a product says. If it only requires the limits of the individual factors exist (as finite limits) then you could conclude it.

What are the significant properties of antiderivatives as functions? Have your course materials covered that?
 
  • #9
We haven't touched antiderivatives yet. So far we've only done the definition of derivative and associated laws (sum, product, quotient, chain rule) and theorems related to the mean value theorem (Rolle's Theorem, local maximums have a derivative equal to 0, mean value theorem, if f'(x) = 0 everywhere then f is a constant, if f'(x) > 0 then f is strictly increasing/f'(x) < 0 implies f is strictly decreasing, intermediate value theorem for derivatives, inverse function theorem).
 
  • #10
Antiderivatives are obviously differentiable functions. What are the significant properties of differentiable functions? [Edit] Or am I confusing matters? It's [itex] g(x) [/itex] that is differentiable, not [itex] f(x) [/itex].
 
  • #11
The big property I remember is that they're continuous. But the derivative itself doesn't necessarily have to be continuous.

edit: Yeah it's g(x) that is differentiable. f(x) is the derivative
 
  • #12
Well, at least we identified some difficulties in you original line of attack.:w

I looked up the theorem on the limit of products. It only requires that the limits of the individual factors exist, not that the factors be continuous functions. So we can conclude [itex] lim_{x\rightarrow 0} f(x) [/itex] does not exist.

Some thoughts (which may or may not lead to progress):

It think that once you define c(x), you can prove [itex] lim_{x \rightarrow 0} c(x) = 0 [/itex]
You can define [itex] c(0) = 0 [/itex] so that would show [itex] c(x) [/itex] is continuous at [itex] x = 0 [/itex]

If you apply the mean value theorem to the function [itex] H(x) = x g(x) [/itex] I think you can show that if [itex] x \ne 0 [/itex] then [itex] g(x) - g(c(x)) = c(x) g'(c(x)) = c(x) f(c(x)) [/itex]. (Do that before taking any limits.) Then there is the question: Does [itex] lim_{x \rightarrow 0} c(x) f(c(x)) = lim_{z \rightarrow 0} z f(z) [/itex]? In other words, can you do a continuous change of variable in function when taking a limit?
 
  • #13
Thanks a lot, you've really helped me get my head around the issues with this problem. I'm going to work on the below and hopefully get an answer. Cheers :)
Stephen Tashi said:
Then there is the question: Does [itex] lim_{x \rightarrow 0} c(x) f(c(x)) = lim_{z \rightarrow 0} z f(z) [/itex]? In other words, can you do a continuous change of variable in function when taking a limit?
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that for a continuous and differentiable function on a closed interval, there exists at least one point where the slope of the tangent line is equal to the average rate of change of the function over that interval.

2. How is the Mean Value Theorem applied in real analysis?

The Mean Value Theorem is used in real analysis to prove important results about continuous and differentiable functions, such as the existence of extreme values and the behavior of the function near critical points. It is also used to prove other theorems, such as the Fundamental Theorem of Calculus.

3. Can the Mean Value Theorem be applied to all functions?

No, the Mean Value Theorem only applies to continuous and differentiable functions on a closed interval. If a function is not continuous or differentiable, the Mean Value Theorem cannot be applied.

4. What are some real-world applications of the Mean Value Theorem?

The Mean Value Theorem has many practical applications, such as in physics and engineering to analyze the motion of objects, in economics to model supply and demand, and in finance to calculate rates of change in stock prices.

5. How does the Mean Value Theorem relate to other theorems in calculus?

The Mean Value Theorem is closely related to other important theorems in calculus, such as the Intermediate Value Theorem and the Rolle's Theorem. These theorems all deal with the behavior of functions on closed intervals and are essential tools in proving more complex results in calculus.

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