# Real Analysis - Mean Value Theorem Application

1. Jan 13, 2015

### O_o

1. The problem statement, all variables and given/known data
Let f: R -> R be a function such that $$\lim_{z\to 0^+} zf(z) \gt 0$$ Prove that there is no function g(x) such that g'(x) = f(x) for all x in R.

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

3. 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:
$$\exists c \in (0,x), f(c) = g'(c) = \frac{g(x) - g(0)}{x - 0}$$
This means $$\lim_{x\to 0^+} \left(\frac{g(x) - g(0)}{x} - f(c) \right)= 0 \\ \lim_{x\to 0^+} \left(g(x) - g(0) - xf(c)\right) = 0 \\ \lim_{x\to 0^+} \left(g(x) - g(0)\right) = \lim_{x\to 0^+} xf(c)$$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

2. Jan 13, 2015

### Stephen Tashi

You are treating $c$ as a function of $x$, which is permissible provided you define the function. For a given $x$ there might be more than one value $c$ that satisfies $g'(c) = \frac{ g(x) - g(0) }{x - 0 }$. Unless you specify which $c$ to choose you haven't defined $c$ as a function of $x$.

3. Jan 13, 2015

### O_o

Thanks that's helpful. So if I can define a function c(x), is the following true:

If $$\lim_{x\to 0^+} c(x) = 0$$ then $$\lim_{x\to 0^+} x f(c(x)) = \lim_{x\to 0^+} x f(x)$$

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

Last edited: Jan 13, 2015
4. Jan 13, 2015

### haruspex

No.
Consider f(x) = 1/x, c(x) = x2.

5. Jan 13, 2015

### Stephen Tashi

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

For that theorem to apply, $w(x)$ must be continuous at $x= a$, but I don't think that $q(x)$ 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 $x f(x)$ being the product of the limits of the individual factors.

6. Jan 13, 2015

### Stephen Tashi

There is a theorem about the limit of a product being the product of the limits of the individual factors. If $f(x)$ were continuous at $x= 0$, what would that tell you about $lim_{z \rightarrow 0} z f(z)$

7. Jan 13, 2015

### O_o

If $$\lim_{x\to 0} f(x) = L$$ then $$\lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = L*0 = 0$$
So if f is continuous at x=0 then $$\lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = f(0)*0 = 0$$

So we can conclude that f isn't continuous at 0. And more generally $$\lim_{x\to 0} f(x)$$ does not exist.

8. Jan 14, 2015

### Stephen Tashi

Yes

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. Jan 14, 2015

### O_o

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. Jan 14, 2015

### Stephen Tashi

Antiderivatives are obviously differentiable functions. What are the significant properties of differentiable functions?  Or am I confusing matters? It's $g(x)$ that is differentiable, not $f(x)$.

11. Jan 14, 2015

### O_o

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. Jan 14, 2015

### Stephen Tashi

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 $lim_{x\rightarrow 0} f(x)$ does not exist.

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

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

If you apply the mean value theorem to the function $H(x) = x g(x)$ I think you can show that if $x \ne 0$ then $g(x) - g(c(x)) = c(x) g'(c(x)) = c(x) f(c(x))$. (Do that before taking any limits.) Then there is the question: Does $lim_{x \rightarrow 0} c(x) f(c(x)) = lim_{z \rightarrow 0} z f(z)$? In other words, can you do a continuous change of variable in function when taking a limit?

13. Jan 14, 2015

### O_o

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 :)