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  1. Prove that a closed subspace of a Banach space is also a Banach space.
  2. Show that the linear space of all polynomials in one variable is not a Banach space in any norm.
t.b.
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RANI JOSE
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    Welcome to math.SE! What have you tried so far? For example, for the first problem, you have to show that a closed subspace is complete with respect to the norm inherited from the norm on the initial Banach. But in fact it doesn't have anything to do with vector space (it's true that a closed part of a complete metric space is complete). For the second exercise, you can use Baire's categories theorem. – Davide Giraudo Apr 10 '12 at 18:51
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    Is it just me or does one of these problems seem substantially more difficult than the other? – David Mitra Apr 10 '12 at 19:23
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    @DavidMitra: it's not just you :) – t.b. Apr 10 '12 at 19:26

1 Answers1

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Hints:

  1. Prove that a closed subspace of a complete metric space is complete.

  2. 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.

t.b.
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    The argument in the second exercise is easily adapted to the statement that the vector space dimension of a Banach space is either finite or uncountable. With different methods one can establish that the dimension of an infinite-dimensional Banach space is at least the cardinality of the continuum. – t.b. Apr 10 '12 at 19:28
  • Yay : ) I just found another example of application of the Baire category theorem : ) – Rudy the Reindeer May 15 '12 at 09:35
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    Why does the set of polynomials of degree $\leq n$ have empty interior? – Rudy the Reindeer Jul 02 '12 at 13:28
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    Because a subspace with non-empty interior is open and an open subspace of a topological vector space is the entire space, for example by connectedness (an open subspace is closed). – t.b. Jul 02 '12 at 13:35
  • I'm not sure but I think you don't answer my question. My question (more generally) was: why does a closed subspace have to have empty interior? I know that finite dimensional spaces are complete and by Heine Borel we therefore have that polynomials of degre $\leq n$ is closed. So I have closedness. – Rudy the Reindeer Jul 02 '12 at 13:56
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    I think I do answer your question: I argue that a subspace with non-empty interior has to be the entire space and the space polynomials of degree $\leq n$ (or more generally a proper subspace, closed or not) is not the entire space. – t.b. Jul 02 '12 at 13:59
  • Yes, sorry. Thanks. And I have no excuse since I did think before posting the comment. But perhaps not enough. – Rudy the Reindeer Jul 02 '12 at 14:07
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    No worries :) Maybe you prefer this argument: if a subspace has non-empty interior it contains some ball $B_\varepsilon(x)$ with $x$ in the subspace. Translating by $-x$ we see that $B_\varepsilon(0)$ is contained in the subspace. By homogeneity of the norm and closedness of the the subspace under scalar multiplication the subspace has to be the entire space. – t.b. Jul 02 '12 at 14:11
  • Thanks. Actually, I have no preference : ) Pick $x \in X$. Then $\frac{\varepsilon}{2 |x|} x \in B_\varepsilon (0)$ is in the subspace and hence so is $x$, I assume. – Rudy the Reindeer Jul 02 '12 at 14:42
  • Yes, in retrospect I do have a preference: for the second argument. It explains why any subspace with non-empty interior has to be open. What I still don't understand is how non-empty interior implies it's open. But I'm quite satisfied as is. – Rudy the Reindeer Jul 29 '12 at 19:12
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    @MattN. Let $Y$ be the subspace of $X$ and let $y \in Y$ be an interior point. Then there is an open set $U$ of $X$ such that $y \in U \subset Y$. For every $y' \in Y$ we have that $y' - y + U$ is an open neighborhood of $y'$ in $X$ and contained in $Y$, so $Y$ is open. – t.b. Jul 29 '12 at 19:18
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    Then I'll upvote this one instead : ) – Rudy the Reindeer Jul 29 '12 at 19:18
  • Yeah, the other had pretty bad notation... – t.b. Jul 29 '12 at 19:19
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    better?${}{}{}{}$ – t.b. Jul 29 '12 at 19:21
  • Ok : ) ${}{}{}$ – Rudy the Reindeer Jul 29 '12 at 19:21
  • x${}{}{}{}{}{}{}$ – t.b. Jul 29 '12 at 19:22
  • x ${}{}{}{}{}$ blushes : ) – Rudy the Reindeer Jul 29 '12 at 19:23
  • Oops, but I should be doing commutative algebra really : ) Need to learn about filtrations (and instead here I am, thinking about vector spaces). – Rudy the Reindeer Jul 29 '12 at 19:23
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    Then go and filtrate :) Have fun – t.b. Jul 29 '12 at 19:24
  • Aye! closes tab (for now) – Rudy the Reindeer Jul 29 '12 at 19:25