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Interior of L^2 in L^1 is empty

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The set A = { f, such that f is an element of L^2 on [0,1] } can be considered as a subset of the metric space L^1 on [0,1]. Prove that A has empty interior as a subset of L^1 on [0,1].

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Solution Summary

We prove that the interior of the space L^2 as a subset of L^1 is empty in the topology induced by the L^1-norm.

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The interior of the set A is the set of all points x in A that are contained in A together with an open ball of radius epsilon for some epsilon>0.
To show that the interior of L^2([0,1]) in L^1([0,1]) is empty, we need to show that for every function f1 from L^2([0,1]) ...

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