5
$\begingroup$
  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.
  • 4
    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.2012-04-10
  • 7
    Is it just me or does one of these problems seem substantially more difficult than the other?2012-04-10
  • 2
    @DavidMitra: it's not just you :)2012-04-10

1 Answers 1

21

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.

  • 4
    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.2012-04-10
  • 0
    Yay : ) I just found another example of application of the Baire category theorem : )2012-05-15
  • 1
    Why does the set of polynomials of degree $\leq n$ have empty interior?2012-07-02
  • 2
    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](http://math.stackexchange.com/a/137254/5363)).2012-07-02
  • 0
    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.2012-07-02
  • 2
    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.2012-07-02
  • 0
    Yes, sorry. Thanks. And I have no excuse since I *did* think before posting the comment. But perhaps not enough.2012-07-02
  • 2
    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.2012-07-02
  • 0
    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.2012-07-02
  • 0
    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.2012-07-29
  • 2
    @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.2012-07-29
  • 1
    Then I'll upvote this one instead : )2012-07-29
  • 0
    Yeah, the other had pretty bad notation...2012-07-29
  • 1
    better?${}{}{}{}$2012-07-29
  • 0
    Ok : ) ${}{}{}$2012-07-29
  • 0
    x${}{}{}{}{}{}{}$2012-07-29
  • 0
    x ${}{}{}{}{}$ *blushes* : )2012-07-29
  • 0
    Oops, but I should be doing commutative algebra really : ) Need to learn about filtrations (and instead here I am, thinking about vector spaces).2012-07-29
  • 1
    Then go and filtrate :) Have fun2012-07-29
  • 0
    Aye! *closes tab* (for now)2012-07-29