Functional Analysis Question

Asked Jul 06 '14 at 00:55

Active Jul 06 '14 at 10:58

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Let $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_1)$ be Banach spaces. Does it imply that $\|\cdot\|_1-\|\cdot\|_2$ (equivalent)?
It is know that if $\|\cdot\|_1-\|\cdot\|_2$ and $(X,\|\cdot\|_1)$, then $(X,\|\cdot\|_1)$ is also Banach space.
I actually want to prove that $(C[0,1],\|\cdot\|_{L^1}$ is not Banach space.

How can we prove the last statement? Is there any counterexample or we need to prove directly by using Cauchy sequence?

edited Jul 06 '14 at 10:58

Martin Sleziak

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asked Jul 06 '14 at 00:55

Analysis

Write down an example of a discontinuous function $[0,1] \rightarrow {\mathbf R}$ that is an $L^1$-limit of continuous functions. – KCd Jul 06 '14 at 00:59
For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jul 06 '14 at 10:58
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BTW $-$ is rather unusual to denote equivalence. Perhaps $\sim$ $\sim$ would be better? – Martin Sleziak Jul 06 '14 at 10:59

Functional Analysis Question

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