# Uniform convergence

1. Jan 28, 2006

### T-O7

Hello,

I have two questions to ask regarding uniform convergence for sequences of functions.

So I know that if a sequence of continuous functions f_n : [a,b] -> R converge uniformly to function f, then f is continuous.

Is this true if "continous" is replaced with "piecewise continuous"? (I am not assuming that the sequence functions are discontinuous at the same points)
i.e. if f_n are each discontinuous at finitely many points, is the uniform limit function f discontinuous at finitely many points as well?

Also, does anyone know for what kinds of metric spaces (if any) is "pointwise convergence" for sequences of functions equivalent to "uniform convergence"?

Thanx.

2. Jan 29, 2006

### cogito²

Using the L^p spaces have this property in a sense under certain restrictions (such has having an integrable function that is always greater than the sequence).

3. Jan 29, 2006

### benorin

e.g. Dominated Convergence Theorem

4. Jan 30, 2006

### cogito²

I'd like to add that something, but it has to do with functional analysis. If you were to consider "uniform convergence" the same thing as convergence in the norm, then there are cases where this cannot be true. In some cases, pointwise convergence cannot be the same as convergence in the norm (ie. there is no norm, that would have those qualities).

One example would be the set of all sequences (they don't need to be bounded) with either real or complex numbers. There is no norm that would have the following property: a sequence of sequences converges converges pointwise if and only if the sequence converges in the norm. The reason that is not possible can be proven pretty simply by a counterargument.

(Outline of Proof: Assume such a norm exists. For any bounded sequence of sequences, you can then choose a subsequence by induction in each variable--ie. the i-th place of the sequence--so that they converge because each component is bounded sequence of real or complex numbers. The subsequence chosen would then converge pointwise. But that means that it also converges in the norm. What that implies is that every bounded sequence has a convergent subsequence in this norm. That implies that the dimension of the set off all sequences is finite. This is obviously not true and therefore there cannot exist a norm with the mentioned properties.)

So basically it depends on what you mean by "convergence in the norm". Under certain restrictions it can be realized, but in many cases a general convergence from pointwise convergence cannot be realized (which is where topological vector spaces take over for normed vector spaces).

5. Jan 31, 2006

### benorin

Norm (or strong) convergence of, say, a sequence of vectors {xn} in a Hilbert space may be defined thus:

$$x_{n}\rightarrow x\mbox{ if, and only if }\|x_n - x\|\rightarrow 0\mbox{ as } n\rightarrow\infty$$