Uniform Convergence of Continuous Functions: A Proof?

simba31415
Messages
12
Reaction score
0

Homework Statement


As in the question -

Suppose that f_n:[0,1] -> Reals is a sequence of continuous functions tending pointwise to 0. Must there be an interval on which f_n -> 0 uniformly?

I have considered using the Weierstrass approximation theorem here, which states that we can find, for any continuous function [0,1] -> Reals, a uniform approximation by polynomials.

Because of this, it seems to me - though I could be wrong - that these f_n -> 0 uniformly if this series of polynomials (each p_n approximating an f_n to a sufficient degree of accuracy) tends to 0 uniformly - in which case it suffices to prove the result for any series of polynomials.

Even if this deduction -is- correct, which I'm not 100% confident about, I can't seem to follow through and show that there exists such an interval for a polynomial sequence. On the other hand, perhaps there is a counterexample and I'm going about this completely the wrong way! Could anyone lend a hand please?
 
Physics news on Phys.org


I think this property is false... But my counterexample is a little complicated (or wrong). I'll think of an easier one...
 


micromass said:
I think this property is false... But my counterexample is a little complicated (or wrong). I'll think of an easier one...

That sounds promising! What were you thinking of? :)
 


See www.math.ubc.ca/~feldman/m321/dini.pdf at example (c).

So basically, my counterexample consists of a triangle moving to the left.
Of course this has an interval at which the convergence is uniform. So the trick is to add more (smaller) triangles, such that the convergence on any interval is disturbed.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top