Spring 1996 #4

More studying of sequences and series of functions. These boogers are beginning to be a little more manageable.

Problem Statement: Consider the sequence of functions \{f_n(x)=xe^{-nx}\} for x\geq 0. (a) Prove \{f_n\} converges. (b) Is the convergence uniform? Prove your answer.

(a) I claim that \{f_n\} converges pointwise to 0.

Proof:Let x\geq 0 and consider


=x\lim\limits_{n\rightarrow\infty}\dfrac{1}{e^{nx  }}

Since e^n\rightarrow\infty as n\rightarrow\infty it follows that e^{nx}\rightarrow\infty as n\rightarrow\infty since x\geq 0 and x is fixed. Thus we may conclude that \dfrac{1}{e^{nx}}\rightarrow 0 as n\rightarrow\infty for x>0. If x=0 then f_n(0)=0 and so \lim\limits_{n\rightarrow\infty}f_n(0)=0. Thus, for x\geq 0 it follows that \{f_n\} converges pointwise to f(x)=0.


(b) I claim that the convergence is uniform.

Proof: To show uniform convergence we wish to show that the uniform norm ||f_n-f||_{x\geq 0}=sup\{|f_n(x)-f(x)|:x\geq 0\} goes to 0. Note that since we showed in (a) that the sequence converges to 0 we really only want to show that ||f_n(x)||_{x\geq 0} goes to 0.

Let us consider f'_n(x) for x\geq 0 and a fixed n\in\mathbb{N}.


So there is only one zero of f'_n(x) and it occurs when x=\dfrac{1}{n}. Checking points on either side of x=\dfrac{1}{n} tells us that f_n(x) obtains a maximum when x=\dfrac{1}{n}. Futhermore we can see that this maximum is f_n(\dfrac{1}{n})=\dfrac{1}{ne}. Since each f_n(x)\geq 0 for every n\in\mathbb{N} and x\geq 0 we have shown that each of the f_n are bounded. Moreover, we have found that ||f_n(x)||_{x\geq 0}=sup\{|f_n(x)|:x\geq 0\} goes to 0 as n\rightarrow \infty since \dfrac{1}{ne}\rightarrow 0 as n\rightarrow\infty.

A sequence of bounded functions converges uniformly on a domain A if and only if the uniform norm on A goes to zero, thus we have shown that the sequence \{f_n\} converges to f=0 uniformly for x\geq 0


Reflection: This was a great problem! I feel like I learned so much in working through this question. It always a little frustrating when they ask you to decide if something converges uniformly or not. At least in a test situation it is. Why can’t they just tell me what to prove?! 🙂 The way I went about doing this problem was a kind of round-about way, but in doing so it led me to better understand this idea of uniform norm. When I learned uniform convergence I was taught to take the derivative of the difference f_n-f and find where it’s maximum was. Then to show that the difference went to zero as n\rightarrow\infty. We never called it a uniform norm, we just did it. I always felt a little shaky about it though. Today I went about solving this problem in the same way and then something clicked. The light went on and suddenly the idea of a uniform norm and what I had been taught meshed together to make this super idea. What I had been doing all along was showing that the uniform norm goes to zero! I just never knew it. So now, after having discussed and chewed the fat of this problem for quite a while I am feeling much better about uniform convergence.

Only 25 (eh, it’s 10pm, let’s call it 24) days until the exam. I think I can, I think I can, I think I can…

This entry was posted in Analysis, Sequence of Functions, Uniform Convergence. Bookmark the permalink.

One Response to Spring 1996 #4

  1. Jason says:

    Erin – I really like the proof. Great job! My only suggestion would be to change the line “Checking points on either side… obtains a maximum” to say local maximum rather than just maximum. The subsequent line argues that the local maximum itself is a global maximum (with respect to your domain), and so serves as a proper upper bound of your sequence. Protect those 5 points!!!

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