As far as shapes go, one of the most useful things about uniform convergence is that if $f_n\to f$ uniformly on $[a,b]$ then ${\int }_a^b{f}_n\to {\int }_a^{b}f$. So, for the polygon example this would tell you that the limit of the areas of the polygons with $n$ sides as $n$ tends to infinity is the area of the circle.

The first of your messages is completely right. Also, what you had written about the disappearing triangle was what I thought was bothering you. Before I look at the second message, can I just refer you to Dan's comment about the limit of integrals. Consider the functions $f_1$, $f_2$, ... similar to the triangle functions I defined, but this time with the height of the triangle being $n$ (the ones I gave had height 1). i.e. make the gradient of the non-zero bit much steeper. Now think about these functions, their integrals and Dan's comment.

Regarding continuity, use,

$|f(x_m)-f(x)|\le |f(x_m)-f_n(x_m)|+|f_n(x_m)-f_n(x)|+|f_n(x)-f(x)|$.

Now given $\epsilon >0$ FIX $N$ large enough so that $\sup [|f_N(y)-f(y)|:y\in$ domain $]<\epsilon$. See if you can build from here with $n=N$ in the inequality above.

Regarding how you would come up with the inequality; the way I think of it is proving that $f$ is continuous via $f_N$. The uniform bit allows us to swap from $f$ to $f_N$ and we know that $f_N$ is continuous and this property then carries over to $f$.

Sorry, that's not too useful, it's a common method in analysis. The next thing to do (perhaps) is to prove Dan's lemma on the limit of integrals of a uniformly convergent sequence of functions.

i.e. show that if $f_1$, $f_2$, ... are a sequence of continuous functions on [0,1] that converge uniformly to $f$ then also

${\int }_0^1{f}_n\to {\int }_0^{1}f$

Bear in mind that $|{\int }_0^{1}g|\le \sup [|g(x)|:x\in [0,1]]$

Here's how that property can be useful. Consider the continuous functions $f_1$, $f_2$, ... on [0,1] as usual. This time we'll add them all together to form a series $\sum _{i=0}^{\infty }{f}_i(x)$

Suppose this sum exists. That is, if you fix any particular $x$ in [0,1] then the sum converges. In other words, the sequence of functions

$F_n(x)=\sum _{i=0}^n{f}_i(x)$

converges pointwise on [0,1].

If the $F_n$s converge uniformly we say that the series converges uniformly. It is then true that

${\int }_0^1\sum _{i=0}^{\infty }{f}_i=\sum _{i=0}^{\infty }{\int }_0^1{f}_i$.

Can you show this from what you have just derived. (This explains what I meant earlier about limit swapping properties associated with uniform convergence.)

Ian

Another good proof. Only slight point I'd make is that line 3, we don't need to assume pointwise convergence on [0,1], this follows since the other side of the equation converges. I'll write out the proof in stages, my main point is that it follows straight from Dan's lemma. The reason I do this is merely as confirmation. Idea is, as you have spotted, to use uniform convergence to revert to limit swapping for only the finite sum (which we know we can do). Incidentally, it's not universally known as Dan's lemma.

(1) The hypotheses are that $F_n(x)=\sum _{i=1}^n{f}_i$ is uniformly convergent to $F(x)$, where $f_i$ are continuous on [0,1]. Note that we then write $F(x)=\sum _{i=1}^{\infty }{f}_i$.

(2) Dan's lemma gives ${\int }_0^{1}F(x)={\lim }_{n\to \infty }{\int }_0^1{F}_n(x)$.

(3) Now we just write out the conclusion of (2) with sigma signs.

${\int }_0^1\sum _{i=1}^{\infty }{f}_i$

$={\lim }_{n\to \infty }{\int }_0^1\sum _{i=1}^n{f}_i$

$={\lim }_{n\to \infty }\sum _{i=1}^n{\int }_0^1{f}_i$

$=\sum _{i=1}^{\infty }{\int }_0^1{f}_i$.

Regarding:

# More Potatoes! ....What? :) I'll look this up

# Mock exams. Nrich will not be held responsible... no, really I think you'll absolutely fly through university maths being able to understand this stuff now.

# My time. I work for Nrich, this is what I'm meant to do! Amongst other things. That's besides the point, it's interesting to see how you work your way through everything. I like analysis, this is my subject and I enjoy discussing it. Also, you contribute hugely to Nrich for which we are greatly appreciative.

# The coins problem. I hadn't forgotten. I rewrote my article and the co-author said it was not clear enough so I have to write it again! This I will do at Easter. Honestly, I didn't forget!!

# I think another thread has started where Yatir asks what uniform convergence is. The definition seems different, but reconciling the two is important. I'll explain for the usual sequence $f_1$, $f_2$, ... defined on a set of real numbers $S$.

Uniform and pointwise convergence are nicely compared by looking at the two expressions:

$|f_n(x)-f(x)|\to 0$

$\sup [|f_n(y)-f(y)|:y\in S]\to 0$.

This is how I think of uniform convergence, just add the sup in.

Another (essentially the same) way of thinking about it is as follows.

For pointwise convergence, you FIX an $x\in S$ then

Given $\epsilon >0$ there exists $N$ where $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$.

For uniform convergence, you work with all $x\in S$ at once. i.e.

Given $\epsilon >0$ there exists $N$ where $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$ FOR ALL $x$.

The '' $N$'' in the pointwise case is dependent on $\epsilon$ AND $x$, whereas the '' $N$'' in the uniform case is only dependent on $\epsilon$.

You with me?

In fact $f$ must be continuous on a dense set - that is, in every non-empty open interval $f$ has a point of continuity.

By the way, although I didn't say so explicitly, you can assume that $f_n$ and $f$ are defined on some non-empty open interval $I$ (possibly $R$). It is clear that proving $f$ has a point of continuity is really the same as proving $f$'s points of continuity are dense - if you could come up with an example in which $f$'s points of continuity were not dense then there would exist a subinterval $I<\mathrm{quote}/>$ of $I$ in which $f$ was nowhere continuous. This would then be a contradiction since $f_n\to f$ pointwise in $I<\mathrm{quote}/>$ and $f_n$ are continuous in $I<\mathrm{quote}/>$.

How did you get that $|f_N(x_M)-f(x_M)|<\mu /4$? This doesn't seem obvious to me. It is clear that if you fix $M$ first, then there exists $N$ which makes this inequality true - but the trouble is you've already chosen $N$.

One other point is that since $x$ is arbitrary in your proof then if it works it actually proves that $f$ is continuous everywhere - and we know this is not necessarily true.

What you actually need to do is find a way of constructing a point $x$ such that $f$ is continuous at $x$. Here is a crucial observation which helps you do this:

If $f$ is the pointwise limit of continuous functions then for every $\epsilon >0$, every non-empty open interval $I$ has a non-empty open subinterval $I<\mathrm{quote}/>$ such that for every $x$, $y$ in $I<\mathrm{quote}/>$ we have $|f(x)-f(y)|<\epsilon$.

In English: in every interval you can find a subinterval in which $f$ hardly varies at all. Any continuous function certainly has this property (can you see why?) - though there are discontinuous functions which also have this property e.g. any step function.

So there are two things left for us to do. 1) Prove the observation and 2) prove, using the observation, that $f$ has a point of continuity.

I suggest trying 2) first since it's quite a bit easier than 1). We need to show that $f$ has a point of continuity assuming that, for every $\epsilon >0$ and every interval $I$ (I'll not bother saying non-empty open from now on) there is a subinterval $I<\mathrm{quote}/>$ in which $f$ doesn't vary by more than $\epsilon$.

This sort of makes sense - everywhere you look you can find regions in which $f$ fluctuates by only a very small amount (as small as you like). So it's sort of intuitive that $f$ must be continuous at least one point (indeed on a dense set of points).

The problem is that you are actually going to have to _construct_ a point of continuity - you can't just take an arbitrary $x$ and hope you can prove $f$ is continuous at $x$ because it would then follow that $f$ is continuous everywhere, and this needn't be true.

I think you've got the right idea - but I don't think that quite works, as it stands. One problem is that just because there is a sequence $x_n\to x$ such that $f(x_n)\to f(x)$ that doesn't mean $f$ is continuous at $x$. (A counter-example is the indicator function of the rationals - this is discontinuous everywhere yet $f(1/n)=1$ which tends to $f(0)=1$ as $n\to \infty$.) You would need that to hold for _every_ sequence $x_n$ which tends to $x$ in order to conclude $f$ is continuous at $x$.

Take $I$ to be any old non-empty open interval. Set $\epsilon =1$. Find a non-empty open interval $I_1$ of $I$ in which $f$ varies by less than $\epsilon =1$. Now set $\epsilon =1/2$. Find a non-empty open interval $I_2$ of $I_1$ in which $f$ varies by less than $\epsilon =1/2$. Now set $\epsilon =1/3$. Find a non-empty open interval $I_3$ of $I_2$ in which $f$ varies by less than $\epsilon =1/3$. And so on.

This is essentially your zooming in idea - we have nested intervals $I_n$ in which $f$ varies by a smaller and smaller amount. We want to try and prove that there is a point $x$ in every $I_n$. Because then for every $n$ there exists a subinterval $I_n$ containing $x$ such that $|f(y)-f(z)|<1/n$ for all $y$, $z$ in $I_n$. In particular $|f(x)-f(z)|<1/n$ for all $z$ in $I_n$. It follows that $f$ is continuous at $x$.

So if such a point $x$ exists then we've proved 2).

Slight problem: as I've stated it there need not exist a point $x$ in every $I_n$. For example what if $I_n=(0,1/n)$? These are non-empty nested open intervals, but their intersection is empty.

The problem here is that $I_n$ share an endpoint - 0. If we can ensure that no two of $I_n$ share an endpoint then we should be OK.

Can you see why, in our construction of $I_1$, $I_2$, $I_3$, ... above, we can say: ''without loss of generality no two intervals $I_m$, $I_n$ share an endpoint''?

And once you've done that can you see why the intersection of $I_n$ must be non-empty?

I agree the proof of (2) is quite nice - I consider it to be the logical development of your zooming in idea. By the way, the fact we've proved (2) has other applications - for example it shows that every Riemann integrable function has a point of continuity (something usually proved by measure theory). Can you see why?

Anyway, any ideas on how to prove (1) (which would then complete the proof that the pointwise limit of continuous functions has a point of continuity)?

The best approach is probably contradiction. Assume that there exists a non-empty open interval $I$ and $\epsilon >0$ such that $f$ varies by $>\epsilon$ in every non-empty open subinterval of $I$, and try to prove that there exists $x$ such that $f_n(x)$ doesn't converge (which would then be a contradiction). In order to do this, try and observe the following:

There exists $\delta >0$ with the following property. Let $a$ be real. If $I<\mathrm{quote}/>$ is a non-empty open subinterval of $I$ and $n$ is a sufficiently large integer then there exists a non-empty open subinterval $I<\mathrm{quote}/><\mathrm{quote}/>$ of $I<\mathrm{quote}/>$ such that for all $x$ in $I<\mathrm{quote}/><\mathrm{quote}/>$ we have $|f_n(x)-a|>\delta$.

In other words (for any fixed $a$) anywhere you look (in $I$) for large enough $n$ you can find a region in which $f_n$ is bounded away from $a$ (by $\delta$).