Advanced calculus proof involving mean value theorem

[mrchris]

Homework Statement

If f is strictly decreasing and differentiable on R, then f '(x) ≤ 0 for all x.

Homework Equations

Mean value theorem

The Attempt at a Solution

if f is strictly decreasing, then for any a,b$\in$ℝ such that a<b, f(b)<f(a) or f(b)-f(a)<0. By the MVT, there exists a number c$\in$(a, b) such that
f '(c)= [f(b)-f(a)]/(b-a). Since we know b-a>0 and f(b)-f(a)<0, f '(c)<0. I'm kind of stuck here because intuitively I can say that since f is strictly decreasing, its derivative is always negative so since f '(c) is negative then all f '(x) must be negative, but that is just restating the statement I am supposed to prove.

 

[Fredrik]

Does the problem say that you have to use the mean value theorem? It seems easier to just use the definition of f'(x).

 

[mrchris]

No it does not say that, but I wasnt sure how to do it using only the definition of f '(x)

 

[Fredrik]

Let x be an arbitrary real number. Define g by
$$g(h)=\frac{f(x+h)-f(x)}{h},$$ for all h. What does the assumption about f tell you about g? Use the answer to prove that $\lim_{h\to 0} g(h)$ can't be positive. (You will need to use the ε-δ definition of limits of functions).

 

[mrchris]

I tried to do this as a proof without using epsilon delta: if f is differentiable over ℝ, we can take f '(x)= limit as h→0 of [f(x+h)-f(x)]/h. case 1: take h>0. by definition of strictly decreasing, we have f(x+h)-f(x)<0. since we know this and the fact that h>0, then we know f '(x) is always negative if h>0.
Case 2: h<0. then we have f(x+h)-f(x)>0. Then by the same logic from case 1, we know
f '(x) must also be negative if h<0.
Case 3: h=0. this will give us f '(x)= limit as h→0 of [f(x)-f(x)]/0 equals some number L because we are told f is differentiable. we know the limit as h→0 of h is 0, and we can use this to determine what f '(x) is. we can re-write the limit as h→0 of f(x)-f(x) as
limit as h→0 of [f(x)-f(x)]*[h/h], or {limit as h→0 of [f(x+h)-f(x)]/h}*{limit as h→0 of h}. since we know both limits exist, we know the product of their limits equals some number A. But since we have f '(x)*0, we know A=0 for case 3.

This should prove that f '(x)≤0 always.

 

[Bacle2]

Why not use MVT for f(c),any c, putting c in (c-h,c+h), for h>0 ?

 

[mrchris]

ok, so then we would have for any point c in ℝ, and c-h<c<c+h, we know f '(c)=
[f(c+h)-f(c-h)]/[(c+h)-(c-h)]. then by the definition of strictly decreasing, we know that
f(c+h)<f(c-h) or f(c+h)-f(c-h)<0. we also know (c-h)<(c+h) or (c+h)-(c-h)>0. Therefore, we can say that for any c in the open interval (c-h, c+h), f '(c)<0. Is this also enough to imply that f '(c)≤0? and what about a case where h<0, does this proof follow intuitively or do I need to write it out explicitly?

 

[mrchris]

I also realized that my previous post did not prove that f '(x)< 0 if h=0. When using that definition of the derivative, do I need to address the case of h=0?

 

[Bacle2]

But if h<0 , then (x+h)<x , so that f(x+h)>f(x).

 

[Fredrik]

I tried to do this as a proof without using epsilon delta: if f is differentiable over ℝ, we can take f '(x)= limit as h→0 of [f(x+h)-f(x)]/h. case 1: take h>0. by definition of strictly decreasing, we have f(x+h)-f(x)<0. since we know this and the fact that h>0, then we know f '(x) is always negative if h>0.
Case 2: h<0. then we have f(x+h)-f(x)>0. Then by the same logic from case 1, we know
f '(x) must also be negative if h<0.
Case 3: h=0. this will give us f '(x)= limit as h→0 of [f(x)-f(x)]/0 equals some number L because we are told f is differentiable. we know the limit as h→0 of h is 0, and we can use this to determine what f '(x) is. we can re-write the limit as h→0 of f(x)-f(x) as
limit as h→0 of [f(x)-f(x)]*[h/h], or {limit as h→0 of [f(x+h)-f(x)]/h}*{limit as h→0 of h}. since we know both limits exist, we know the product of their limits equals some number A. But since we have f '(x)*0, we know A=0 for case 3.

This should prove that f '(x)≤0 always.

1 and 2 are what I had in mind for the first part. 3 is unnecessary, and doesn't really make sense, since g(0) is undefined. The result of 1 and 2 do in fact imply that f'(x) (which by definition is equal to $\lim_{h\to 0}g(h)$) is ≤0, but to argue that convincingly, you have to use the definition of $\lim_{h\to 0}g(h)$.

 

[Bacle2]

Sorry, I just realized my argument needs some work.

mrchris: you only consider the limit as h-->0 , not when h=0. If you do, you will end up with an expression 0/0 (which also does not match the definition of

derivative, and physically does not seem to measure anything --tho I would leave this last to the physicists.)

 

[mrchris]

i am not sure what is wrong with the proof using the MVT and c in (c-h, c+h). I also am not sure how to use the definition of the limit here. Is it not enough to say that since we are taking h>0, as h approaches 0, those values of h will all be positive? and also, if g(0) is undefined, doesn't this mean that f is not differentiable on R because then f '(x) is not defined?

 

[Bacle2]

i am not sure what is wrong with the proof using the MVT and c in (c-h, c+h). I also am not sure how to use the definition of the limit here. Is it not enough to say that since we are taking h>0, as h approaches 0, those values of h will all be positive? and also, if g(0) is undefined, doesn't this mean that f is not differentiable on R because then f '(x) is not defined?

Well, the problem is that it is not true that:

[f(c+h)-f(c-h)]/2h=f'(c).

We do know that there is some t in the interval (c-h,c+h) , with f'(t)=[f(c+h)-f(c-h)]/2h

but we cannot say that t=c, nor that , for any c, there are x1,x2 with:

f'(c)=[f(x2)-f(x1)]/(x2-x1)

 

[Bacle2]

This may be a way of patching up the problem, i.e., this is a way of finding x1,x2 as in my previous post:

We find the value of f'(c), and take the line y-f(c)=f'(c)(x-c) . We then translate this line

so that it hits the curve at any two points (since we translate, the slope remains

constant). The places where the translated curve hits the graph are x1,x2. It may

need some work (for one, it is not an actual proof), but I think it is essentially right.

 

[mrchris]

why couldn't you arbitrarily pick any two endpoints such that a<b in ℝ and then proceed with the definition of the MVT?

 

[mrchris]

If i try to use the proof I did earlier with the 3 cases, I can not have g(h) not defined at any points because we are told f is differentiable.

 

[Fredrik]

I also am not sure how to use the definition of the limit here. Is it not enough to say that since we are taking h>0, as h approaches 0, those values of h will all be positive?

It's not enough, because if you haven't used the definition of the limit, you can't claim to have proved anything about the limit. Consider the related but slightly simpler problem of proving that if every term of a convergent sequence is <0, then the limit is ≤0. It's the same thing here. The statement is pretty obvious*, but you still need to prove it.

*) Actually it only seems obvious because we tend to implicitly assume that the definition we're using is consistent with our intuition about limits. As a student, you understand that the way that the definition has been presented to you means that it's universally agreed to be the best definition of the term "limit", and that it wouldn't be if mathematicians hadn't already determined that it's consistent with our intuition. So to argue that "it's obvious" is the same thing as arguing that "this thing wouldn't even be called a limit if someone hadn't already proved this".

I'll do the proof for sequences. You should be able to figure out how to do the corresponding proof for functions.

Definition: x is said to be a limit of $\langle x_n\rangle_{n=1}^\infty$ if every open interval that contains x contains all but a finite number of terms of the sequence.

Theorem: If $\langle x_n\rangle_{n=1}^\infty$ is a convergent sequence in $\mathbb R$, and $x_n<0$ for all $n\in\mathbb Z^+$, then $\lim_n x_n\leq 0$.

Proof: Define $x=\lim_n x_n$. Suppose that x>0. Since $x_n\to x$, the open interval (x/2,3x/2) contains all but a finite number of terms. But none of the numbers in this interval are negative, so this contradicts the assumption that all the terms are <0.

and also, if g(0) is undefined, doesn't this mean that f is not differentiable on R because then f '(x) is not defined?

No, because f'(x) isn't defined as g(0), but as $\lim_{h\to 0}g(h)$. Note that a function doesn't have to be defined at 0 to have a limit at 0.