Tidbit on Uniform Convergence

Something from first-year calculus: if a sequence of continuous functions f_n converges uniformly to limit function f (on interval I), then f is also continuous (over the same interval).

My Calculus 4* professor said that it's too hard to prove. I call bullshit. Here's the proof, courtesy of Calculus 2.

(As usual, fingers are crossed for the LaTeX to show up properly. I'm too spoiled by LyX to learn LaTeX seriously.)

First, a refresher. The sup-norm of a function f defined on a (real) interval I (mapping to the set of reals) is

 ||f ||_\infty = \displaystyle{\sup_{x\in I}}|f(x)|

Note that the supremum is precisely the maximum of the function if the maximum exists (on the interval); in particular, if interval is closed and the function is continuous, by the Extreme Value Theorem

||f||_\infty = \displaystyle{\max_{x\in I}}|f(x)|

Also, the norm depends on the interval.

I'm going to leave it as an exercise for the reader to show that this is indeed a norm.

Given a norm**, we can talk about sequences. That is, {f_n} converges uniformly to f on interval I if, for all &epsilon > 0, there exists a K > 0 such that, for all n >= K,

||f_n - f||_\infty < \varepsilon

Now we can prove the proposition.
Take p in I, &epsilon > 0. We seek a &delta > 0 such that |f(x) - f(p)| < &epsilon when |x - p| < &delta .

Since f_n converges to f uniformly on I, pick some n such that

||f_n - f||_\infty < \displaystyle{\frac{\varepsilon}{3}}

For any x in I, by the triangle inequality we get

|f(x)-f(p)| \le |f(x)-f_{n}(x)|+|f_{n}(x)-f_{n}(p)|+|f_{n}(p)-f(p)|\\
\le ||f_{n}-f||+|f_{n}(x)-f_{n}(p)|+||f_{n}-f||\\
< \displaystyle{\frac{\varepsilon}{3}+|f_{n}(x)-f_{n}(p)|+\frac{\varepsilon}{3}}

Since f_n is continuous for all n, there exists &delta > 0 such that

|f_{n}(x)-f_{n}(p)|<{\displaystyle \frac{\varepsilon}{3}}

when |x - p| < &delta. So

|f(x)-f(p)| < \frac{2\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon

when |x - p| < &delta.

As they say, QED.

[Edited to note]
My eqnarray won't work; extra dollar signs don't work too well either. Hence uglier formatting than usual. Oddly enough, neither \Vert or \| works but || gives the desired double-bars. Sigh.

*One of the reasons I really don't like this course is its lack of rigour. In particular, it has been mangling uniform convergence for the last few days. Seriously, convergence without epsilon-delta is but a toy - it loses alot of its power.
**To be pedantic, obviously a metric doesn't have to be a norm, but having a norm is certainly sufficient for us to define sequences and their convergence.