# Uniform convergence problem

1. May 6, 2010

### losin

let f is a continuous, real-valued function on [a,b]

then, for any e, there exist a polygonal function p such that

sup|f(x)-p(x)|<e

using uniform convergence, this might be shown... but i cannot figure it out...

2. May 6, 2010

### LCKurtz

Use the fact that a continuous function on [a,b] is uniformly continuous.

3. May 6, 2010

### losin

how does it have to d with polygonal function?

4. May 7, 2010

### cheeee

This is a basic rough sketch of the argument ...there are a few gaps that need to be filled in by yourself to be rigorous...

Since [a,b] is a closed and bounded interval, note that f(x) is uniformly continous on this interval.

So given any epsilon > 0 , then there must exist a delta such that
| f(x) - f(y) | < (epsilon/2) whenever |x-y| < delta. This is simply the definition of uniform convergence.

Now choose an appropriate partition for the interval [a,b] such that |x(n+1) - x(n)| < delta for all n.

Now by the triangle inequality |p(x)-f(x)|<= |p(x)-p(x(n))|+|p(x(n))-f(x(n))|+|f(x(n))-f(x)| =
|p(x)-p(x(n))|+|f(x(n))-f(x)|<= epsilon, since |x-x(n)|<delta.