Proof Using Mean Value Theorem

[toslowtogofast2a]

Homework Statement

Suppose f is diff. on R, 1<=f'(x)<=2 for x in R and f(0)=0. Prove x<=f(x)<=2x for all x > 0

Relevant Equations

MVT

See my image for my attempt at solving this problem. My approach varies significantly from the solution I have for this problem and I wanted to get feedback on if what I did is correct of where I went wrong. thanks.

 

[FactChecker]

Your proof looks ok to me. I wonder what the other proof is. If something is true, there can be multiple ways to prove it. Often, a particular proof is desired to teach the use of specific theorems or facts.

 

[pasmith]

Your argument is correct, but it is easier to note that for $x > 0$, $$1 \leq f'(x) \leq 2 \Rightarrow \int_0^x \,dt \leq \int_0^x f'(t)\,dt \leq \int_0^x 2\,dt$$ and hence $$x \leq f(x) - f(0) \leq 2x.$$

 

[toslowtogofast2a]

Your argument is correct, but it is easier to note that for $x > 0$, $$1 \leq f'(x) \leq 2 \Rightarrow \int_0^x \,dt \leq \int_0^x f'(t)\,dt \leq \int_0^x 2\,dt$$ and hence $$x \leq f(x) - f(0) \leq 2x.$$

Thanks for the reply. I agree that is easier and makes sense.

 

[toslowtogofast2a]

Your proof looks ok to me. I wonder what the other proof is. If something is true, there can be multiple ways to prove it. Often, a particular proof is desired to teach the use of specific theorems or facts.

Thanks for the reply. I was doing problem 29.14 out of elementary analysis by ross. When I finished the problem I googled the problem solution and got the link below. problem 29.14 is on the second page. thst is the solution i read before posting here.

https://www.math.stonybrook.edu/~olga/mat319-spr16/hw12sol.pdf

 

[FactChecker]

It's disturbing to me that a text in analysis uses integrals before formally defining them. But I guess it ain't no capital crime. ;-)

 

[pasmith]

It's disturbing to me that a text in analysis uses integrals before formally defining them. But I guess it ain't no capital crime. ;-)

Yes, this argument is clearly inadmissible if integrals have not been formally defined yet.

For the record, the book's argument is that $f(x) - x$ has positive derivative and is therefore increasing; since $f(0) - 0 = 0$ it follows that $f(x) - x \geq 0$ on $[0, \infty)$. Similarly $2x - f(x)$ is increasing and equal to zero at $x = 0$. That positive derivative implies increasing is proven in an earlier excersice by use of the mean value theorem.

 

[mathwonk]

I think your argument back in post #1 is the simplest most direct possible proof. I.e. you show that result follows from just substituting into the statement of the MVT. To me nothing could be easier than that.

 

[Infrared]

Your argument is correct, but it is easier to note that for $x > 0$, $$1 \leq f'(x) \leq 2 \Rightarrow \int_0^x \,dt \leq \int_0^x f'(t)\,dt \leq \int_0^x 2\,dt$$ and hence $$x \leq f(x) - f(0) \leq 2x.$$

I don't think this is correct without the additional assumption that $f'$ is integrable. In general $f$ being differentiable does not imply that $f'$ is integrable. See https://en.wikipedia.org/wiki/Volterra's_function

 

[WWGD]

I don't think this is correct without the additional assumption that $f'$ is integrable. In general $f$ being differentiable does not imply that $f'$ is integrable. See https://en.wikipedia.org/wiki/Volterra's_function

I think $f'$ is integrable precisely when $f$is Absolutely Continuous. You may too, have cases when $f'$ is integrable but not equal to $f$, such as with the Cantor function$C(x)$ https://en.m.wikipedia.org/wiki/Cantor_function which is a.e. $0$, thus $C'(x)=0$.

 

[pasmith]

I don't think this is correct without the additional assumption that $f'$ is integrable. In general $f$ being differentiable does not imply that $f'$ is integrable. See https://en.wikipedia.org/wiki/Volterra's_function

To apply the MVT, $f'$ must be at least continuous, and therefore integrable.

 

[Infrared]

To apply the MVT, $f'$ must be at least continuous, and therefore integrable.

No, MVT is valid for any differentiable function.

 

[renormalize]

No, MVT is valid for any differentiable function.

Wikipedia disagrees: it states that the function must also be continuous (https://en.wikipedia.org/wiki/Mean_value_theorem):

 

[WWGD]

Wikipedia disagrees: it states that the function must also be continuous (https://en.wikipedia.org/wiki/Mean_value_theorem):
View attachment 353413

Differentiability implies continuity.

 

[Infrared]

Wikipedia disagrees: it states that the function must also be continuous (https://en.wikipedia.org/wiki/Mean_value_theorem):
View attachment 353413

Any differentiable function is automatically continuous, so that's not an additional assumption. The question was whether $f'$ must also be continuous (it doesn't need to be).

The point of explicitly mentioning continuity in the wikipedia article is that the function doesn't need to be differentiable at the endpoints of the interval, as long as it's continuous there and differentiable on the interior, but being differentiable everywhere is still enough.

 

[FactChecker]

Any differentiable function is automatically continuous, so that's not an additional assumption. The question was whether $f'$ must also be continuous (it doesn't need to be).

The point of explicitly mentioning continuity in the wikipedia article is that the function doesn't need to be differentiable at the endpoints of the interval, as long as it's continuous there and differentiable on the interior, but being differentiable everywhere is still enough.

Good point. The wikipedia definition might have been less confusing about the requirements if it had said:
Differentiable in the open interval (a,b) and also continuous from the right at a and from the left at b.

 

[PeroK]

Wikipedia disagrees: it states that the function must also be continuous (https://en.wikipedia.org/wiki/Mean_value_theorem):
View attachment 353413

The important point is that a function could be differentiable on $(a,b)$ but not continuous at the end points. And for such a function the MVT may not apply.

In other words, differentiablity on $(a,b)$ implies continuity on $(a,b)$ but not necessarily continuity on $[a,b]$.

As @Infrared has pointed put, the MVT does not require that the derivative is integrable. So, the proof using the integral is not valid.

Moreover, this exercise is much earlier in the textbook than the development of the Riemann integral. That rules out the use of the integral in any case.

 

[PeroK]

I would have done this using the MVT directly, without using any previous results. Setting up a proof by contradiction:

Assume for some $0<c$ we have $f(c) < c$ or $f(c) > 2c$. Hence $\frac{f(c)}c<1$ or $>2$.

By the MVT, we can find $0 < d < c$ such that:
$$f'(d) = \frac{f(c)-f(0)}{c -0} = \frac{f(c)}c$$Hence $f'(d)<1$ or $>2$. QED