Hw help: real analysis proof

[coffeesneeze]

Hello beautiful smartypantses,

I'm having some trouble (running around in circles, pulling hair out) trying to figure out how to prove something on my current real analysis problem set.

Assume f is continuous, differentiable on R. Given f(0) is greater or equal to 1 and f'(x)>f(x) for all x in R. Show that f(x)>exp(x) for all x>0.

It seemed natural to try this by contradiction. Assume there exists a>0 for which f'(a) is less than or equal to exp(a). Define g(x)=f(x)-exp(x). We have g(0) is greater than or equal to 0 and g(a) is less than or equal to 0. So on (0,a) g is decreasing. By MVT there must be some b in (0,a) where g'(b)<0. And from here g(b)<g'(b)<0. So for some b in this interval f(b)<exp(b).

Where the crap is my contradiction though? The statement itself is VERY OBVIOUSLY TRUE, making it all the more frustrating. Does anyone have any suggestions? Looking for a direct proof has been even less productive. The previous part of this problem was to show that if f(x)=f'(x) for all x then f(x)=Cexp(x) for some constant C. I don't see where to apply this, if it's necessary.

Thanks, forum community. You are the coolest.

[letterX]

It seemed natural to try this by contradiction. Assume there exists a>0 for which f'(a) is less than or equal to exp(a). Define g(x)=f(x)-exp(x). We have g(0) is greater than or equal to 0 and g(a) is less than or equal to 0. So on (0,a) g is decreasing. By MVT there must be some b in (0,a) where g'(b)<0. And from here g(b)<g'(b)<0. So for some b in this interval f(b)<exp(b).

Where the crap is my contradiction though?

This is nearly the entire proof. All you have to do to make it work is choose 'a' a little more carefully...

[jestingrabbit]

That can you tell me about integrals of derivatives, and definite integrals of functions positive everywhere but one point?

Assuming you've got the fundamental theorem of calculus to use, that is.

[coffeesneeze]

It seemed natural to try this by contradiction. Assume there exists a>0 for which f'(a) is less than or equal to exp(a). Define g(x)=f(x)-exp(x). We have g(0) is greater than or equal to 0 and g(a) is less than or equal to 0. So on (0,a) g is decreasing. By MVT there must be some b in (0,a) where g'(b)<0. And from here g(b)<g'(b)<0. So for some b in this interval f(b)<exp(b).

Where the crap is my contradiction though?

This is nearly the entire proof. All you have to do to make it work is choose 'a' a little more carefully...

I think what you are suggesting is to take the least a for which f'(a) is less than or equal to exp(a). Can I simply assume that there is a least a with this property? It seems like I'd have to do some more work to show that there actually is a least a. Assuming the negative of the statement, it could be the case that f'(x) is less or equal to exp(x) for values of x arbitrarily close to x=0, in which case I can't just say I chose the least one. There could always be one smaller.
I've seen some of my classmates settle on this as a step in their proof, but I'm not convinced it works. Am I mistaken or maybe just misinterpreting your hint?

Jesting, I'm thinking over your hint now.
The integral of the derivative of f is equal to f up to an arbitrary constant. I'm not sure what you mean by

definite integrals of functions positive everywhere but one point?

What I'm assuming for contradiction gets me that there is at least one point, not just one. But a definite integral of an everywhere positive (except at one point) function would be... positive... /shrug.
The FTC is fair game. I will keep fussing with it, I know I've gotta be close. If anyone thinks of anything else that could help, please share.

Thanks to the both of you.

[letterX]

I think what you are suggesting is to take the least a for which f'(a) is less than or equal to exp(a). Can I simply assume that there is a least a with this property? It seems like I'd have to do some more work to show that there actually is a least a.

Well, what sorts of sets do or do not contain their own greatest lower bounds?

Also, under further reflection, I'm thinking my method only gets you as far as f(x) >= exp(x), and not a strict inequality. To get a strict inequality you have to do a bit more work.

[jestingrabbit]

definite integrals of functions positive everywhere but one point?

What I'm assuming for contradiction gets me that there is at least one point, not just one. But a definite integral of an everywhere positive (except at one point) function would be... positive... /shrug.

Well, I wouldn't go for contradiction, and I would try to make f(a) - exp(a) into an integral from 0 to a... of something (assuming that f(0) = 1).

[skeptical scientist]

There's a really nice, really easy way to do this that nobody has thought of yet.

Spoiler:

Let g(x)=ln(f(x)).

To be completely rigorous, you should prove that f(x)>0 first, but this should be easy for you.

What can you tell me about integrals of derivatives?

They might not exist.

I would try to make f(a) - exp(a) into an integral from 0 to a... of something (assuming that f(0) = 1).

This seems problematic, given the above.

[Talith]

I was under the impression that the derivative of any differentiable function on the reals had a well defined Riemann integral on any interval.

[PM 2Ring]

There's a really nice, really easy way to do this that nobody has thought of yet.

Spoiler:

Let g(x)=ln(f(x)).

To be completely rigorous, you should prove that f(x)>0 first, but this should be easy for you.

Hey, I thought of that approach. I didn't post though, because I couldn't think of a simple way of proving f(x)>0, apart from using a Taylor's expansion, and that seems like overkill for this problem.

Spoiler:

I tried using (f(x + a) - f(x)) / a = f'(x + c) for some c in 0 <= c <= a, and f'(x + c) > f(x + c), but that seems to lead to an infinite regress.

But I suspect that the solution will seem obvious after I've had some sleep.

[skeptical scientist]

I was under the impression that the derivative of any differentiable function on the reals had a well defined Riemann integral on any interval.

Nope. Volterra's function is a counterexample.

Hey, I thought of that approach.

Well, nobody who posted thought of it.

I didn't post though, because I couldn't think of a simple way of proving f(x)>0, apart from using a Taylor's expansion, and that seems like overkill for this problem.

This is problematic, for the same reason the fundamental theorem of calculus is problematic: the existence of one derivative, which need not even be continuous, does not give you a Taylor expansion. At best you can use the n=1 version of Taylor's theorem (better known as the mean value theorem).

[jestingrabbit]

I was under the impression that the derivative of any differentiable function on the reals had a well defined Riemann integral on any interval.

Nope. Volterra's function is a counterexample.

Then every suggestion I've made in this thread is worse than useless. Sorry about that.

[coffeesneeze]

So it looks like I can't do something like this:
Define

g(x)=\ln{(f(x))}.

It's derivative is this:

g'(x)=\frac{f'(x)}{f(x)}>1.

So this

\int_0^x g'(t)\ dt>\int_0^x1\ dx = x.

And

\int_0^x g'(t)\ dt=g(x)-g(0)=\ln(f(x))-\ln((f(0))>x.

Which implies from assumptions about f(0)

\ln(f(x))>x.

And then of course

f(x)>\exp(x).

The problem being I can't assume g'(x) is integrable?

[jestingrabbit]

Yeah, I think that you'll need to use letterX's approach.

[coffeesneeze]

Is there anything from stopping me from starting with define g(x)=f'(x)/f(x), Note G(x)=ln(f(x)) is an antiderivative for g(x)?*

Then I could avoid invoking whatever that evil Volterra function is.

*posted before thinking, edited a mistake.

[jestingrabbit]

Is there anything from stopping me from starting with define g(x)=f'(x)/f(x), Note g(x) is an antiderivative for h(x)=ln(f(x))?

Then I could avoid invoking whatever that evil Volterra function is.

The Volterra function is also an antiderivative of its derivative, its just not unique. You need uniqueness, and you can't get that.

[coffeesneeze]

The Volterra function is also an antiderivative of its derivative, its just not unique. You need uniqueness, and you can't get that.

That makes sense, Thanks.

[skeptical scientist]

Define g(x)=\ln{(f(x))}.
It's derivative is this:

g'(x)=\frac{f'(x)}{f(x)}>1.

So this

\int_0^x g'(t)\ dt>\int_0^x1\ dx = x.

...
The problem being I can't assume g'(x) is integrable?

Right. Depending on the level of rigor expected, your grader might let it go and mark it correct, or might give you a hard time because g' might not be integrable. However, you can prove that g(x)>x using the mean value theorem, and avoid mentioning integration entirely.

You still need to show that g(x) is well defined, i.e. that f(x)>0. (Really you only need this for x≥0, which is good, because it might be false for some negative x.) Try tweaking your original attempt at a proof by contradiction of f(x)>e^x that didn't quite work.

The Volterra function is also an antiderivative of its derivative, its just not unique. You need uniqueness, and you can't get that.

Erm, antiderivatives are unique, up to additive constants. It follows from the mean value theorem that if f and g are differentiable and f'=g' on [a,b], then f-g is constant on [a,b]. The problem is the fundamental theorem of calculus requires 1) f is differentiable, and 2) f' is integrable. Here you have 1), but don't necessarily have 2).*

*You can show that f is monotone and differentiable, which might be enough to guarantee that f' is integrable; I'm not sure. But I can't figure out how to do this, so it's probably at least much harder than just using MVT.

[Dark Avorian]

Wait. So, it's perhaps not as elegant as some of the other answers, nor does it use anywhere near as high power proof techniques. but here's how I did it using g(x) = f(x) - exp(x)

Spoiler:

Well first, I established that within some interval [0,a], all values of g(x)>=0 and further that there exists some point y on the interval such that g(y) > 0

Spoiler:

Obviously, if f(0)>1 then g(0)>0 and by elementary use of the definition of continuity there is suchan interval.if f(0)=1 we have that g(0)=0, but g'(0)>0, so we can use the definition of the derivative to establish both the existence of a positive interval (this is the same as where slope of secant lines are positive), and a positive value of g on that interval. Now we can prove that g is positive for all other x

Spoiler:

With the mean value theorem

Spoiler:

So now, if there is some x such that g(x) <= 0, then there is a least z such that g(z) = 0 and all x < z have g(x)>0. But, since there exists as shown a y such that g(y) > 0 then by the mean value theorem there is some x, y < x < z such that g'(x) < 0. but, for all such x, g(x)>0, and we know that g'(x)= f'(x)-exp(x)>f(x)-exp(x) = g(x)

I guess I assumed continuity of the derivative...so YMMV

[skeptical scientist]

I guess I assumed continuity of the derivative...so YMMV

Where? I only see continuity of f, so I think this works.

[Dark Avorian]

mean value theorem relies on continuity of derivative, no?
EDIT: wait, no, the equivalent of generalized Rolle's theorem produces the same result
EDIT: Oh right, ignore me, differentiability implies that no jump discontinuities appear in the derivative.

[skeptical scientist]

mean value theorem relies on continuity of derivative, no?

Nope. The hypothesis of the MVT is that f is continuous on [a,b] and differentiable on (a,b).

Oh right, ignore me, differentiability implies that no jump discontinuities appear in the derivative.

There could still be other kinds of discontinuities, however.

I'm not quite sure why I didn't come up with your proof. I was thinking along those lines for ~20 minutes and the best result I could get was f(x)≥e^x. Then I thought of the idea of working with ln f instead of f, and then the result just fell out.

I'm also not quite sure what you mean by "high power proof techniques". I don't think any of the ideas suggested require anything more advanced than the mean value theorem. It's true that I pulled out Volterra's function as an example to explain why you couldn't use the fundamental theorem of calculus, but the proof I was hinting at just used the mean value theorem.

[Dark Avorian]

Just my impostor syndrome acting up I guess. I always assume the forum knows way more than me and could crush me with high power proofs.

[lalop]

For my own practice

Spoiler:

Let x > 0 be arbitrary. Then by the FTC,

f(x) = f(0) + \int_0^x f'(t) \,dt > f(0) + \int_0^x f(t) \,dt

,

exp(x) = 1 + \int_0^x exp(t) \,dt

As in Dark Avorian's post, define g(x) = f(x) - exp(x). g(0) > 0 and thus, by continuity, there exists an open interval around it where g is strictly positive.

Now, assume for contradiction the set of positive reals where g(x) ≤ 0 is nonempty, and let m be its infimum. By continuity, g(m) ≥ 0 (otherwise, for example, take an increasing sequence x_n -> m, then f(x_n) -> f(m), and since all f(x_n) are positive, f(m) must be nonnegative). We show, in fact, that g(m) > 0; this would contradict continuity of g since, by definintion of infimum, there would then exist a sequence x_n such that x_n -> m but g(x_n) are all negative, hence could not converge to g(m) > 0.

Using the equations above,

g(m) = f(m) - exp(m) > \left( f(0) - 1 \right) + \int_0^m f(t) - exp(t) \,dt = g(0) + \int_0^m g(t) \,dt

Since g(0) > 0 and the integral is nonnegative (since g(t) > 0 in the entire interval (0,m)), the right hand side is, in fact, greater than zero. Thus g(m) > 0 and we have our contradiction.

[coffeesneeze]

Thanks everyone.
The MVT proof is pretty slick, I have to say. ln(f(x))/x>1 is pretty immediate.

I fully intend to bust out the Volterra function if any of my cohorts show up with integrals in their proofs tomorrow.

[lalop]

According to its wiki page, the Henstock-Kurzweil integral of an everywhere-existing derivative always exists (thus presumably allowing integration as a proof method without additional hypotheses). I was wondering if anyone was familiar enough to confirm or deny.

[skeptical scientist]

According to its wiki page, the Henstock-Kurzweil integral of an everywhere-existing derivative always exists (thus presumably allowing integration as a proof method without additional hypotheses). I was wondering if anyone was familiar enough to confirm or deny.

I tend to believe Wikipedia on math articles, as when I know the math well enough to judge (which is a lot of the time), it's almost always right. (Although I have found and corrected a few mistakes.) According to the Wiki, every derivative is H-K integrable and satisfies the FToC, and it's easy to verify that the H-K integral is monotone (as it is defined as a limit of Riemann sums). So I suppose you could use the Henstock-Kurzweil integral to do this exercise using the fundamental theorem of calculus, which I suppose should count as a proof "using high-powered proof techniques".

I much prefer the MVT proof. Here's my version of it, by the way, since it looks like the OP has solved the exercise, and someone else already posted a solution anyways:

Lemma: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>0 for x≥0.

Proof: Suppose f(x)≤0 for some x≥0, and let b=inf{x : x≥0 and f(x)≤0}. Then b≥0 and, by continuity, f(b)=0, so b>0. By choice of b, f(x)>0 on [0,b). We know f is continuous on [0,b] and differentiable on (0,b), so by the mean value theorem, there is some c in (0,b) such that f'(c)=(f(b)-f(0))/(b-0)≤0. But then f(c)>0≥f'(x), contradicting f'(x)>f(x) for all x in R.

Theorem: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>e^x for x>0.

Proof: By the lemma, the function g(x)=ln f(x) is well-defined for x≥0, and g'(x)=f'(x)/f(x)>1 for x≥0. By the mean value theorem, for all x>0 there is some c in (0,x) (depending on x) such that g(x)=g(0)+x*g'(c). Since g(0)≥0 and g'(c)>1, we have g(x)>x, whence f(x)=e^g(x)>e^x.

[PM 2Ring]

I think I get it, SS. But where did that a come from in there is some c in (a,b). I guess it's just a typo, and that the interval should be (0, b).

[skeptical scientist]

I think I get it, SS. But where did that a come from in there is some c in (a,b). I guess it's just a typo, and that the interval should be (0, b).

Yeah.

[PM 2Ring]

Sorry for the necro. I thought of ths last week, and then promptly forgot about it.

Lemma: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>0 for x≥0.

Proof: Suppose f(x)≤0 for some x≥0, and let b=inf{x : x≥0 and f(x)≤0}. Then b≥0 and, by continuity, f(b)=0, so b>0. By choice of b, f(x)>0 on [0,b). We know f is continuous on [0,b] and differentiable on (0,b), so by the mean value theorem, there is some c in (0,b) such that f'(c)=(f(b)-f(0))/(b-0)≤0. But then f(c)>0≥f'(x), contradicting f'(x)>f(x) for all x in R.

I do like your proof of that lemma, skeptical scientist, but I prefer to use direct proofs when practical. I don't think of myself as a Constructivist, but I do have some sympathy for their disdain of indirect proofs, although I'm happy to admit that I was duly impressed when I first understood the standard indirect proof of the irrationality of sqrt(2).

Anyway, is this sufficiently rigourous?

Let x₀ = 0
f(x₀) ≥1

If f(x_i) > 0 then f'(x_i) > 0, since f'(x) > f(x), so f(x) is increasing at x = x_i
Thus for all sufficiently small h > 0, f(x) > f(x_i), for all x, x_i < x ≤ x_i + h
Let x_i+1 = x_i + h,
So f(x) > f(x_i), where x_i < x ≤ x_i+1
Now f(x_i+1) > 0 and we can repeat the above argument indefinitely.
So by induction f(x) > 0 for all x ≥ 0

[letterX]

What if your h's happen to sum to some finite value? Then you only get that f(x) > 0 for all x up to whatever the h's happen to sum up to.

[PM 2Ring]

What if your h's happen to sum to some finite value? Then you only get that f(x) > 0 for all x up to whatever the h's happen to sum up to.

Ah. Good point. OTOH, f(x) hasn't decreased at all over that interval. And since f'(x) > f(x) >1, it ought to be possible to show that the function grows sufficiently rapidly to avoid that problem. But I can't immediately see a simple way to fix my proof up (unless I use skeptical scientist's indirect argument).

* goes back to the drawing board *

[skeptical scientist]

I think the mean and intermediate value theorems are constructively false, so I doubt you'll find a purely constructive proof.

[PM 2Ring]

I think the mean and intermediate value theorems are constructively false, so I doubt you'll find a purely constructive proof.

Ah, well. It's a little bit annoying that seemingly straight-forward and intuitive stuff like this can't be proved constructively.

[letterX]

I think the mean and intermediate value theorems are constructively false, so I doubt you'll find a purely constructive proof.

Ah, well. It's a little bit annoying that seemingly straight-forward and intuitive stuff like this can't be proved constructively.

Well, constructively the intermediate value theorem is neither straight-forward nor intuitive. The constructive version of the IVT says that given any continuous function on [0, 1] with f(0) < 0 < f(1) then you can actually find me an x with f(x) = 0, where actually find means give a routine for computing x to within any epsilon.

And there's no particular reason that should be true. It isn't hard to come up with examples of functions f which have f(x) = 0 only for some nasty uncomputable x.

[coffeesneeze]

For anyone interested, my professor gave us his proof. And he also commented that he was surprised that he saw so many "ingenious" things on this exercise because it was supposed to be one of the simpler ones.
Anyways...

Assume f is diff on R, f(0) >/= 1, and f'(x)>f(x) for all x in R.
Then f(x)>exp(x) for all x>0.

proof:
Set g(x)=exp(-x)f(x).
Then g'(x)=exp(-x)[f'(x)-f(x)]>0. We are using the fact that exp is always positive and f'>f.
So g(x) is always increasing, and g(0)=exp(0)f(0) >/= 1.
We have for all x>0, g(x)=exp(-x)f(x) >/= 1.
And thus f(x) >/= exp(x).

The obvious advantage to this proof over the other ones posted is you don't have to show f(x)>0 for all x>0. Not that there's anything particularly terrible about doing so, but it's nice that you don't have to.
I thought you guys deserved this nice little proof. later!