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Agreement

All notions are up to null sets.

Limits are meant by simple functions.

Problem

Given a finite measure space $\mu(\Omega)<\infty$ and a Banach space $E$.

Consider bounded measurable functions $F:\Omega\to E: \|F\|_\infty<\infty$.

Precisely the pointwise limits have separable image: $$S_n\stackrel{0}{\to}F\iff \mathrm{im}F\text{ separable}$$ and precisely the uniform limits have precompact image: $$S_n\stackrel{\infty}{\to}F\iff \mathrm{im}F\text{ precompact}$$ (Is it right like this?)

Now, what is an example of a pointwise limit but not a uniform limit? $$S_n\stackrel{0}{\to}F,\,S'_n\stackrel{\infty}{\nrightarrow}F$$

Note that the special ingredients are: Finite Measure + Bounded & Measurable Function

freishahiri
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1 Answers1

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And again a variant of the famous example: $$F:(0,1]\to\mathcal{H}:F(\frac{1}{n+1}<t\leq\frac{1}{n}):=e_n$$ Clearly, it is pointwise limit but can't be uniform limit.

freishahiri
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  • Do you mean $L^2$? $\ell^2$ usually denotes the square summable sequence space. Also here the pointwise limit is not actually in $L^2$. – Ian Nov 30 '14 at 18:33
  • @Ian: I really meant $\ell^2$. Can you explain your doubts? Let me think about it, too. – freishahiri Nov 30 '14 at 19:09
  • @Ian: Ok, any infinite dimensional Hilbert space would do the job. The key fact is that orthonormal vectors have distance two. – freishahiri Nov 30 '14 at 19:21
  • I suppose I didn't understand what you meant by $\chi_{1/n}$. I assumed from context that you meant $\chi_{[0,1/n]}$. – Ian Nov 30 '14 at 19:26
  • And funnily, I thought so too. But I meant orthonormal elements. – freishahiri Nov 30 '14 at 19:33