Is the space $L^4([0,1])$ endowed with $3-$norm a Banach space?
I prepare for an exam in functional analysis (undergraduate) and I think that compactness of $[0,1]$ might allow this statement to be true.
$3-$norm on $L^4$ space
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integration
functional-analysis
measure-theory
banach-spaces
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0With $3$-norm do you mean $L^3$ norm? – 2017-02-05
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0yes 3-Norm is L³ norm – 2017-02-05
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0@ Open Ball: I thought The theorem of Fisher-Riesz ( "Normkonvergenzsatz") states L^p is a Banachspace with respect to L^p-norm ?? – 2017-02-05
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0Indeed it does. I had something else in mind. – 2017-02-05
1 Answers
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It is not true. Every $L^3$ function can be approximated in the $L^3$-norm by step-functions. Those step-functions are all in $L^4$. If $L^4$ was a Banach space with respect to the $L^3$-norm this would imply $L^3\subset L^4$, but this is not true, as you should know.