Uniform Convergence of g_n (x): Proof & Subset Analysis

[mscudder3]

1. Let g_n (x)=nx*exp(-nx). Is the convergence uniform on [0, ∞)? On what subsets of [0,∞) is the convergence uniform?


3. I am looking for a proof of how the convergence is uniform (possibly using Weierstrass' M Test?). I understand that the subset that determines uniform convergence is none other than the radius of convergence. My attempt is applying the ratio test: R=lim(a_n / a_(n+1))= lim(n/(n+1) * exp(x))=exp(x). This is confusing to me since the radius of convergence should not depend on x.

 

[micromass]

What do things like Weierstrass M-test and ratio-test have to do with this? These only apply to series, and you do not have a serie, but just a sequence.

As how you could start this question: try to find out if the sequence converges pointswise and work out it's pointswise limit...

 

[mscudder3]

That is actually the first part of the question. I omitted it due to already being solved. let x be some element in the domain, D. Thus, for any x, g_n=x(n/exp(nx)) where x is constant for all n.

There exists some N such that for any n>N, exp(nx)>n. Since exponential function increases much faster than n (polynomial of degree 1), lim(g_n)=0

 

[micromass]

OK, now you simply need to find the supremum of g_n(x) on [0,+\infty[ (this supremum will depend of n, but of course it will not depend on x). Then you must see whether the supremum converges to 0 as n becomes large.

 

[mscudder3]

(this supremum will depend of n, but of course it will not depend on x)

I thought that if I were to find the supremum, then I would use x? Would the sup not be dependent on x, and whether it was <1,=1, or >1?

Given that I do not, the sup(g_n(x)) seems difficult to analyze...

Edit: I assume u mean to use that ||g_n - g|| over D=sup{ g_n(x)-g(x) :x in D}. Since g(x) should be minimized (limit is 0), this equates to sup{g_n(x):x in D}. What next? because the sup is at n=1 so g_1 (x)=xexp(-x)...

 

[micromass]

I thought that if I were to find the supremum, then I would use x? Would the sup not be dependent on x, and whether it was <1,=1, or >1?

Given that I do not, the sup(g_n(x)) seems difficult to analyze...

The supremum (which is a maximum in our case) will just be the largest value of the function g_n. For example, the supremum of -x^2 is 0 and the supremum is attained in the point 0.
So, this obviously should not depend on x.

So, all you need to do is find the largest value of the function g_n. This is a basic calculus question...

 

[mscudder3]

So the derivative of g_1=exp(-x)-xexp(-x)=exp(-x)(1-x), so the max occurs at x=1. Thus the sup=1/e. So given that the sup occurs at x=1, it is obvious that g_n still converges to 0 (even at x=1), so this implies uniform convergence exists on [0,∞)? Where the convergence is only uniform for [0, 1]?

 

[micromass]

So given that the sup occurs at x=1

It won't occur at x=1 each time. You'll really have to calculate the sup for each g_n, not just g_1...

 

[mscudder3]

Shoot! Well the derivative =nexp(-nx)(1-nx). Set equal to 0 to obtain x=1/n. Thus, the max x is still at x=1 (for when n=1)...

 

[micromass]

Uuh, yes, the maximum will occur at x=1/n. So the supremum of the function is 1/e. This does not converge to 0, so the function does not converge uniformly...

 

[mscudder3]

I see! Thank you for being so patient with me lol. So as for the second part of the original question, is there any interval in which it converges uniformly? Is it where x is [0,1].

 

[Dick]

I see! Thank you for being so patient with me lol. So as for the second part of the original question, is there any interval in which it converges uniformly? Is it where x is [0,1].

Now why would you say [0,1]? If g_n has a max of 1/e at x=1/n why would you think that could converge uniformly to zero on [0,1]??

 

[mscudder3]

Now why would you say [0,1]? If g_n has a max of 1/e at x=1/n why would you think that could converge uniformly to zero on [0,1]??

Edit: i meant [0,1/e)

 

[Dick]

Edit: i meant [0,1/e)

That doesn't change the problem. There are an infinite number of values of n such that 1/n is less than 1/e. Did you sketch a graph of these functions? Do you think it might converge uniformly to zero on [1,infinity). If so, why?

 

[mscudder3]

My reasoning for my guess was that it would converge up until it reaches its maximum, which is generally the case for problems in my course (this is the first time I've encountered a function of this complexity, hence my confusion).

Since the maximum is at x=1/n, each x value beyond x=1/n would be less than the sup (by def.). Thus, the max(1/n : n in set natural numbers)=1, thus every value beyond 1, or [1,infinity) uniformly converges. The reason being each curve with x=>1 has g_n=nx/exp(nx), and since exponential function grows at a faster rate, g_n converges uniformly to 0?

 

[Dick]

My reasoning for my guess was that it would converge up until it reaches its maximum, which is generally the case for problems in my course (this is the first time I've encountered a function of this complexity, hence my confusion).

Since the maximum is at x=1/n, each x value beyond x=1/n would be less than the sup (by def.). Thus, the max(1/n : n in set natural numbers)=1, thus every value beyond 1, or [1,infinity) uniformly converges. The reason being each curve with x=>1 has g_n=nx/exp(nx), and since exponential function grows at a faster rate, g_n converges uniformly to 0?

That's not too clear. Read the definition of uniform convergence again. To converge uniformly on [1,infinity), it just requires that for sufficiently large n, that the function be close to zero everywhere in [1,infinity). For any n the max of g_n(x) ON [1,infinity) is g_n(1), right? To show this show g_n(x) is decreasing on [1,infinity) but bounded below by zero.

Now tell me if convergence is also uniform on [1/2,infinity), and if so, why? Hint: it is uniform. Pay attention to the 'sufficiently large n' part.

 

[micromass]

As a hint for the OP: download the free graphing program at http://www.padowan.dk/graph/
You can actually use the program to simulate convergence of sequences. So you can see whether a sequence converges uniformly or not.
This is no substitute for a real proof of course, but it could help you.

Now, for the second part. Try to show that your function converges uniformly on every interval [a,+\infty, with a not 0...

 

[mscudder3]

Well if it uniformly converges from 1/2 to infinity then using the min of 1/2 (bc all others from 1/2 to 1 will follow suit), g_(1/2)=(1/2)lim(n/exp(nx)) for large enough n, this term goes to 0. So does this conclude that it uniformly converges on (0,infinity)? Not including 0 since the sup{g_n-g: x in D} is greater than some epsilon>0.

 

[Dick]

Well if it uniformly converges from 1/2 to infinity then using the min of 1/2 (bc all others from 1/2 to 1 will follow suit), g_(1/2)=(1/2)lim(n/exp(nx)) for large enough n, this term goes to 0. So does this conclude that it uniformly converges on (0,infinity)? Not including 0 since the sup{g_n-g: x in D} is greater than some epsilon>0.

You already know limit of nx/exp(nx)=0 for any x. That doesn't tell you anything about uniform convergence. What is sup{g_1(x): x in D} for D=[1,infinity) and why? What is the sup for g_2(x), g_3(x), g_4(x)... And why? Now do the same problem for D=[1/2,infinity). Do D=[1/3,infinity) if you still aren't getting the picture. Finally, what about D=[0,infinity)? Do you see why that one is different from the others?