An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

# Homework 1

ediazloa

Hey Everyone! I figured I would start a discussion about the homework due Monday to hear some of your thoughts...
The homework is exercise 6.2.6:

Using the Cauchy Criterion for convergent sequences of real numbers (Theorem 2.6.4), supply a proof for Theorem 6.2.5. (First, define a candidate for $f(x)$, and then argue that $f_{n} \rightarrow f$ uniformly.)

So, Theorem 6.2.5 (what we need to actually prove) is:

A sequence of functions ($f_{n}$) defined on a set $A \subset \mathbb{R}$ converges uniformly on $A$ iff for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\vert f_{n}(x) - f_{m}(x) \vert < \epsilon$ for all $m,n \geq N$ and all $x \in A$.

Let's share some ideas!

Cromer

I'm finding it significantly harder to prove using the Cauchy Criterion for numerical sequences rather than just a definition-based proof. For example, to prove uniform convergence implies Cauchyness, it boils down to expanding

$$|f_n(x) - f_m(x)| = |f_n(x) - f(x) + f(x) - f_m(x)| \leq |f_n(x) - f(x)| + |f_m(x) - f(x)|.$$

Since this is true for all $x \in A$, and we assume for this direction of the proof that $(f_n) \rightarrow f$ uniformly, we can choose a single $N$ to make both of those terms $\epsilon$ small. This was straightforward. However, trying to use the Cauchy Criterion for numerical sequences seems trickier to me.

The way I have been trying to do this is to think of each value of $x$ defining a single numerical sequence $(f_n(x))$. Then, all of these must individually converge to the values $f(x)$, so they must individually all be Cauchy. However, breaking things up like that makes it less obvious how to show that one $N$ works for all $x$. It may not be all that hard, but I'm still unsure how to do it. Anyone have any ideas?

ediazloa

I agree, using a definition seems more straight forward. What's getting me is choosing a particular $f(x)$... Maybe when that is more clear, the Cauchy Criterion method will make more sense.

ediazloa

On second thought, could we not just say in the proof "uniform converges implies cauchyness so..."

Cromer

Working on the other half of the proof, I think I have an idea on what they mean by "choosing" an $f(x)$.

The other direction is to suppose $(f_n)$ is Cauchy in the functional sequence sense, and show $(f_n)$ converges uniformly. It is not totally clear what function should be the limit in this case, but since the functional sequence $(f_n)$ is Cauchy, it follows that for all $x$, each sequence $(f_n(x))$ must be Cauchy in the numerical sequence sense. Now, this is where I think we need the numerical Cauchy Criterion, as it tells us each of these sequences labelled by $x$ must converge to a number $L(x)$ (which also has to be labelled by $x$ as there's no reason the sequences need to converge to the same thing). This is what I believe our "candidate" for $f(x)$ is supposed to be. Now, all that's left to due is show $(f_n) \rightarrow L$ (where $L$ is a function) uniformly...

ediazloa

I am thinking the same! I am looking through my notes from analysis I and found the proof for the Cauchy Criterion.. I think it is going to be helpful. Defining $f(x)$ as a limit (i.e. $f(x)=lim_{n \rightarrow \infty} f_{n}(x)$), namely $f(x)=L$. Then using the triangle inequality from there...

Cromer

So did you (or anyone) figure out how to do the last part? I've been thinking about it but so far I haven't made any more progress.

nhodges

I believe we can use the fact that f_m is pointwise convergent to f(x) and f_n is uniformly cauchy to show that f_n is uniformly convergent for all x.

ediazloa

Yep, I'm still struggling with the last part...
@nhodges I'm not sure how we can use those two facts to show that $f_{n}$ is uniformly convergent. By the first part of our proof we see that $f_{n}$ is pointwise convergent by definition. How are you seeing that $f_{m}$ is uniformly cauchy? I might be missing something there...

ediazloa

In other news, I just got a notification from the school that classes beginning before 11am have been cancelled.

nhodges

Since $f_n$ is cauchy $[ f_n(x) - f_m(x) ]$can be made arbitrarily small for all $x$. Let$f_m$ converge to $f(x)$ pointwise for some $x \in A$. Since$f_m(x)$ approaches $f(x)$ as m approaches infinity for a given$x$ an m can be chosen so that$[f_m(x) - f(x)]$ can be made really really small. The idea I think is that pointwise convergence + cauchy implies uniformly continuous.

violincounter

Pretty sure you got it. I don't feel great about it though. $f(x)$ is entirely too mysterious to me.

Cromer

To make sure I understand your argument, I'm going to repeat it and ask if I get it right:

Since $(f_n)$ is uniformly Cauchy (is this the term for 'Cauchy in a functional sequence sense'? Wikipedia says yes, but I don't recall if our book uses it), we can make $|f_n(x) - f_m(x)|<\epsilon$ for sufficiently large $n$ and $m$, and most importantly for all $x$. Using our arguments from above, we know $(f_m(x))\rightarrow f(x)$ in a pointwise fashion. Thus, we can take a limit of both sides of

$$|f_n(x) - f_m(x)|<\epsilon$$

to obtain

$$\lim_{m\rightarrow \infty} |f_n(x) - f_m(x)|< \lim_{m\rightarrow \infty} \epsilon.$$

We can be sure our limiting procedure is valid, since the inequality is true for all $m$ larger than some $N$, and we know $(f_m(x))$ converges. These are, as far as I can tell, the only things that could go wrong with the limit, so we are safe to take it and get

$$|f_n(x) - \lim_{m\rightarrow \infty} f_m(x)| = |f_n(x) - f(x)| < \epsilon.$$

But, this inequality held for $n \geq N$ (for some $N$) and for all $x$, so this is exactly the statement of uniform convergence.

Is this right, Nathan? The only thing that gives me pause is whether or not anything else can go wrong with the limit, but I can't think of anything, so I think it works.

ediazloa

@violincounter look at definition 6.2.1 for $f(x)$ !

violincounter

I just meant in the context of this problem it seemed strange.

ediazloa

Are you guys just sliding your homework under his door? @Cromer @violincounter

ediazloa

I think you got it. Taking the limit is the missing link! And it makes a lot of sense... @Cromer

Cromer

Sorry I didn't respond earlier! I did, but it seemed like it wouldn't be a big deal to get it in tomorrow since the snow kinda push things around today.

Cromer

Since our work above turned out to have one minor issue, it might be useful to post a corrected proof for reference. Here is mine with the correction:

First suppose that $(f_n) \rightarrow f$ uniformly. Let $\epsilon>0$. Then, pick $N$ so that $n \geq N$ implies $|f_n(x) - f(x)| < \epsilon/2$ for all $x$. Then, $n,m \geq N$ implies that

$$|f_n(x) - f_m(x)| = |f_n(x) - f(x) + f(x) - f_m(x)| \\ \leq |f_n(x) - f(x)| + |f_m(x) - f(x)|< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

for all $x$. So $(f_n)$ is uniformly Cauchy.

Now, suppose $(f_n)$ is uniformly Cauchy. Then, it is also true that for each $x$, $(f_n(x))$ is Cauchy. Thus, by the Cauchy Criterion for real sequences, $(f_n(x))$ converges for each $x$. Define

$$f: x \mapsto \lim_{n\rightarrow \infty} f_n(x).$$

Then, $(f_n) \rightarrow f$ pointwise. Let $\epsilon > 0$, and pick $N$ so that $n,m \geq N$ implies

$$|f_n(x) - f_m(x)| < \frac{\epsilon}{2} \quad \forall x \in A.$$

Then, when $n\geq N$,

$$\lim_{m\rightarrow \infty} |f_n(x) - f_m(x)| = |f_n(x) - \lim_{m\rightarrow \infty} f_m(x)| = |f_n(x) - f(x)|.$$

Since the inequality is valid for all $n \geq N$, it becomes a non-strict inequality in the limit, so

$$|f_n(x) - f(x)| \leq \lim_{m\rightarrow \infty} \frac{\epsilon}{2} = \frac{\epsilon}{2} < \epsilon$$

for all $x \in A$. Thus, $(f_n) \rightarrow f$ uniformly.

Ricky_Bobby

That looks pretty awesome to me. I like the way you wrote up the second part I struggled with that.