Uniform Convergence Problem: Does fn converge uniformly to f?

futurebird
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I'm trying to better understand convergence so I made upa problem for myself based on an example from class. I want to know if I'm answering my own questions correctly.

Define a sequence of functions fn(x) = 1 if x is in {r1, r2, ... , rn} and 0 otherwise. Where r1, r2, ... , rn are the first n rational numbers in some enumeration of all rational numbers. fn converges pointwise to the dirichlet function. But, can we say anything else about how fn --> f?

Uniform Convergence
Given e > 0 is there an N such that when n > N |fn - f | < e for all x? No. Just let e=.5 we cal alays find an x value where |fn - f | = 1. That is, a rational number that has not yet been listed by the time we reach n.

Convergence in Measure
Yes. The measure of the set where f and fn are not the same is *always* 0.

Almost Uniform Convergence
Yes. If we let A, the set of measure less the any e where uniform convergence fails be Q, the rationals I think we have almost uniform convergence. since m{Q}=0 < e for all e > 0.

Convergence in LP
Yes, the Lp norm of the fn and f is always 0 anyway.
 
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Yup. You've got it.
 
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...
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