Is the Lebesgue-Integral Continuous with Respect to the L^\infty Norm?

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SUMMARY

The Lebesgue integral is not continuous with respect to the L norm. Specifically, in the space of functions defined as 𝓟 = {f ∈ L1 : ||f||L1=1} ∩ L}, a Cauchy sequence f1, f2, ... can converge in L to a limit that is not integrable with L1 norm equal to one. An example provided is fn = (1/n) χ[0,n], which demonstrates that while ||fn||L1=1 and ||fn - 0||L=1/n converge to zero, the limit function does not maintain the L1 norm of one.

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Pere Callahan
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Hi,

I was wondering if the Lebesgue-integral is continuos with respect to the L^\infty norm.

More precisely, assume there is a space of functions

<br /> \mathcal{P}=\{f\in L^1 :||f||_{L^1}=1\}\cap L^\infty<br />

endowed with the essential supremum norm ||\cdot||_{L^\infty}. If there is then a Cauchy sequence f_1,f_2,\dots in \mathcal{P}, can one conclude that the limit (which exists in L^\infty) is also integrable with L^1 norm one?

Thank you very much.
Pere
 
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How about something like f_n = \frac{1}{n} \chi_{[0,n]} in L^1(\mathbb{R})? Here we have \|f_n \|_{L^1} = 1 and \|f_n - 0\|_{L^\infty} = \frac{1}{n} \to 0, while \| 0 \|_{L^1} = 0.
 
Thanks morphism, that seems to answer my question.
 

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