Examples and counterexamples

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What would be an example (or c/ex)of a closed and bounded but not compact subset C of a complete metric space and why?

What would be an example of a sequence of functions which converges in L^2([0,1]), but which does not converge pointwise almost everywhere?
 

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morphism
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1. Can you find an example complete metric space that contains a bounded but not totally bounded closed set? Why will this help?

2. Think about this: how does convergence in the L^2 norm work?
 
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*I can only think about the metric space ([0,1],d*) where d* is the discrete metric. [0,1] is bounded but not totally bounded.
I know [0,1] is closed but I'm not sure if ([0,1],d*) is complete. (I know a space is complete if every Cauchy sequence in there converges).
I guess since a metric space is compact iff it is complete and totally bounded., we get that [0,1] is closed and bounded, but since it is not totally bounded, it cannot be compact.

*I think (fn) converges to f in L^2([0,1]) if lim as n tends to infinity of integral of |fn-f|^2 dm is 0.
My problem is that I cannot think of a not really complicated example.
 
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morphism
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*I can only think about the metric space ([0,1],d*) where d* is the discrete metric. [0,1] is bounded but not totally bounded.
I know [0,1] is closed but I'm not sure if ([0,1],d*) is complete. (I know a space is complete if every Cauchy sequence in there converges).
I guess since a metric space is compact iff it is complete and totally bounded., we get that [0,1] is closed and bounded, but since it is not totally bounded, it cannot be compact.
Yup.

*I think (fn) converges to f in L^2([0,1]) if lim as n tends to infinity of integral of |fn-f|^2 dm is 0.
My problem is that I cannot think of a not really complicated example.
We can come up with a simple example where f_n -> 0 in L^2[0,1] but f_n doesn't converge pointwise anywhere. Here's how: let f_1 = characteristic function of [0,1/2], f_2 = characteristic function of [1/2,1], f_3 = characteristic function of [0,1/3], f_4 = characteristic function of [1/3,2/3], f_5 = characteristic function of [2/3,1], and so on. I'll let you check that this does the job.
 
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  • #5
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Thanks morphism it really helps.
 

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