- Prove that a closed subspace of a Banach space is also a Banach space.
- Show that the linear space of all polynomials in one variable is not a Banach space in any norm.
Banach space in functional analysis
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$\begingroup$
functional-analysis
banach-spaces
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2@DavidMitra: it's not just you :) – 2012-04-10
1 Answers
21
Hints:
Prove that a closed subspace of a complete metric space is complete.
The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire category theorem.
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0Aye! *closes tab* (for now) – 2012-07-29