# Positive derivative implies growing function using Bolzano-Weierstrass

#### schniefen

Problem Statement
Using the Bolzano-Weierstrass theorem (for every bounded sequence, there exists a convergent subsequence), prove that a positive derivative implies a growing function.
Relevant Equations
The Bolzano-Weierstrass theorem: Every bounded sequence $c_n$ has a convergent subsequence $c_{n_k}$.
I'm stuck on a proof involving the Bolzano-Weierstrass theorem. Consider the following statement:
$$f'(x)>0 \ \text{on} \ [a,b] \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2$$ i.e. a positive derivative over an interval implies that the function is growing over the interval. I've been instructed to proceed with a proof by contradiction, yet am slightly confused over how one would start such a proof. I would assume $f'(x)>0$ over a given interval but then also assume the logical negotiation of what that implies, correct? In other words, that the function is then decreasing over the interval and show that this leads to a contradiction. I can also somewhat see how one would go about proving it... $\exists \ x_3 \in[x_1,x_2]$ ...but don't see how the Bolzano Weierstrass theorem comes into play.

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#### Math_QED

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I don't see a quick way to prove the claim using Bolzanno Weierstrass theorem. Are you familiar with the mean value theorem? It provides a proof that takes 2 lines...

#### PeroK

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Problem Statement
Using the Bolzano-Weierstrass theorem (for every bounded sequence, there exists a convergent subsequence), prove that a positive derivative implies a growing function.
How do you define a "growing function"?

#### Math_QED

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How do you define a "growing function"?
It's in the quantor statement of the theorem.

#### schniefen

I don't see a quick way to prove the claim using Bolzanno Weierstrass theorem. Are you familiar with the mean value theorem? It provides a proof that takes 2 lines...
Yes, I am familiar with the mean value theorem and the proof using it. This time I've been asked to use the Bolzano Weierstrass theorem instead of the mean value theorem.

#### PeroK

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It's in the quantor statement of the theorem.
I'll leave it to you then, as I have no idea what that means.

#### schniefen

How do you define a "growing function"?
This is how I would define a growing function over a given interval $[a,b]$: $\forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2$

#### PeroK

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This is how I would define a growing function over a given interval $[a,b]$: $\forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2$
Okay, that's strictly monotonic increasing in my terminology!

• Math_QED and schniefen

#### Math_QED

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Okay, that's strictly monotonic increasing in my terminology!
In mine as well, but there does not seem to be consensus sadly. I use the term "non-decreasing" for what others call "increasing", lately.

#### PeroK

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Problem Statement
Using the Bolzano-Weierstrass theorem (for every bounded sequence, there exists a convergent subsequence), prove that a positive derivative implies a growing function.
Relevant Equations
The Bolzano-Weierstrass theorem: Every bounded sequence $c_n$ has a convergent subsequence $c_{n_k}$.

I'm stuck on a proof involving the Bolzano-Weierstrass theorem. Consider the following statement:
$$f'(x)>0 \ \text{on} \ [a,b] \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2$$ i.e. a positive derivative over an interval implies that the function is growing over the interval. I've been instructed to proceed with a proof by contradiction, yet am slightly confused over how one would start such a proof. I would assume $f'(x)>0$ over a given interval but then also assume the logical negotiation of what that implies, correct? In other words, that the function is then decreasing over the interval and show that this leads to a contradiction. I can also somewhat see how one would go about proving it... $\exists \ x_3 \in[x_1,x_2]$ ...but don't see how the Bolzano Weierstrass theorem comes into play.
You need to generate a sequence in some way. Then, use the BWT to show that the sequence converges to a point. Then, conclude something about that point.

You could start with your positive derivative on $[a, b]$. Then, assume there exist $x < y \in [a, b]$ with $f(x) > f(y)$.

Now, what next?

#### fresh_42

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Yes, I am familiar with the mean value theorem and the proof using it. This time I've been asked to use the Bolzano Weierstrass theorem instead of the mean value theorem.
The mean value theorem is a corollary from Rolle. Thus the question reduces to: how to prove Rolle with Bolzano-Weiertsraß? We already have a bounded interval, so all what is left to do is, to find a sequence of the form $\frac{f(a)-f(b)}{a-b}$ converging to zero.

#### PeroK

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You need to generate a sequence in some way. Then, use the BWT to show that the sequence converges to a point. Then, conclude something about that point.

You could start with your positive derivative on $[a, b]$. Then, assume there exist $x_1 < y_1 \in [a, b]$ with $f(x_1) > f(y_1)$.

Now, what next?
... what about looking at the point $\frac{x_1+y_1}{2}$?

#### schniefen

You need to generate a sequence in some way. Then, use the BWT to show that the sequence converges to a point. Then, conclude something about that point.

You could start with your positive derivative on $[a, b]$. Then, assume there exist $x < y \in [a, b]$ with $f(x) > f(y)$.

Now, what next?
This feels like the right outline of the proof, but I can't see what we're looking for in the proof. What are we looking for in the point that a possible subsequence would converge to?

#### PeroK

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This feels like the right outline of the proof, but I can't see what we're looking for in the proof. What are we looking for in the point that a possible subsequence would converge to?
I can see how to get a proof but it doesn't actually need the BWT per se. And I'm not sure how enlightening it would be.

Maybe there's a slick way using the BTW, but I don't see it.

#### StoneTemplePython

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Gold Member
I'm stuck on a proof involving the Bolzano-Weierstrass theorem. Consider the following statement:
$$f'(x)>0 \ \text{on} \ [a,b] \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2$$ i.e. a positive derivative over an interval implies that the function is growing over the interval. I've been instructed to proceed with a proof by contradiction, yet am slightly confused over how one would start such a proof.
Since people are stuck, I'll offer an approach that precedes Rolle's.

However, it's still a little abusive on my part, because what I'm doing here is only closely related to what OP asks for-- in essence the below is a compactness argument. (i.e. in this case it's really the extreme value theorem which is frequently proven in short order with Bolzano Weierstrass.)
- - - - -
It's enough to consider two cases: either

$\text{(i)} f(a) \lt f(x)$ for $x \in (a,b]$ i.e. the left boundary is mapped to the the strict minimum in this closed set
$\text{(ii)}$ there is (at least one) $x^* \in(a,b]$ where $f(x^*) \leq f(x)$ for all $x \in [a,b]$.

in the case of $\text{(i)}$ everything works great, so you can skip to the end.
But $\text{(ii)}$ isn't true so proceed by contradiction: Consider the sequence of left difference quotients at $f(x^*)$ we know they converge to $f'(x^*) = c\gt 0$, select $\epsilon = \frac{c}{2}$ and we find a contradiction if $f(x^*)$ is a minimum.

the fact the $f$ is differentiable implies continuity, which combined with compactness guarantees a minimum value of $f$ in this $[a,b]$, thus we must have $\text{(i)}$.

with this result in hand, consider re-running the argument on a particular subset, i.e. $[x_1, x_2]$

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