## 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

$\lim\limits_{n\rightarrow\infty}f_n(x)=\lim\limits_{n\rightarrow\infty}xe^{-nx}$

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

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 $f(x)=0$.

$\Box$

(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 $n\in\mathbb{N}$.

$f'_n(x)=e^{-nx}(-nx+1)$

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 $x\geq 0$

$\Box$

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…

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### 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!!!