Sequence Convergence

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If a sequence [tex]\{f_n\}[/tex] is convergent in [tex]\left(C[0,1],||\cdot||_{\infty}\right)[/tex] then it is also convergent in [tex]\left(C[0,1],||\cdot||_1\right)[/tex].

I think I understand why this is true. (In my own words) The relationship between the supremum norm and the usual norm (really any p-norm) is that the supremum norm is the greatest value in all p-norms. So, for all sequences, if the sequence is convergent in the supremum norm it's convergent in all norms on the same space. Is this true?

Also, for a sequence to converge it means
[tex]\exists \, f \,\ni \,\forall \,\epsilon>0 \,\exists\, N\ni\, \forall\, x\in C[0,1][/tex]
[tex]||f_n-f||_{\infty}<\epsilon \quad \forall n>N[/tex]

This is a given, but how could I use that to prove the implied part? This is for my own edification.

Also, I can think of a counter example to show the other direction is not true.
Such as, [tex]f_n(t)=t^n \quad \text{then} \quad ||f_n||_1\rightarrow 0[/tex]
but, [tex]||f_n||_{\infty}\rightarrow 1[/tex]
 

Answers and Replies

  • #2
mathman
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The first statement is not true in general, only for measure spaces where the total measure is finite.

Counterexample on the real line:

fn(x)=1 -n<x<n, =0 |x|>n+1, and connect up to be continuous in between.
sup |fn(x)|=1, while all Lp norms become infinite.
 
  • #3
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Right, thank you. That is a good point. I haven't yet thought about this proclamation in all spaces. My statement was about the behavior was concerning this normed metric space.
 
  • #4
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Since (C[0,1],[tex]||\cdot||_{\infty}[/tex]) is complete and [tex]\{f_n\}[/tex] is convergent,
we know every sequence in (C[0,1],[tex]||\cdot||_{\infty}[/tex]) is Cauchy convergent and converges uniformly [tex]\Rightarrow \quad ||f_n-f||_{\infty}\rightarrow \, 0[/tex].
Because of this we also know:
[tex]||f_n-f||_1=\int_0^1|f_n(t)-f(t)|dt\leq\int_0^1||f_n-f||_{\infty}dt=||f_n-f||_{\infty}[/tex]
Therefore,
[tex]||f_n-f||_1\rightarrow \, 0[/tex]

Is my reasoning correct?
 
  • #5
mathman
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Yes, You can use the same idea for all Lp norms.
 
  • #6
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Thanks, I appreciate your input.
 

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