Showing relatively compact and finite dimension

compactnessfunctional-analysislp-spacesreal-analysissolution-verification

I have a question where I have to show:

  1. Prove that a normed linear space is finite-dimensional if and only if every bounded subset is relatively compact.
  2. Show that the spaces $C[0,1]$ and $L^{2}(0,2 \pi)$ are infinite dimensional.

Now for the first part I'm not entirely sure if I understand the concept of relative compactness well. To show a set is relatively compact is it enough to take a sequence in it and show that it has a convergent subsequence? Would that imply that the closure of the set is compact (which is what I understand is the definition of relative compactness or precompactness)? If this is true here is my proof.

  1. Take any bounded subset B of a finite dimensional space X. Let
    $x^{(n)}$ be a sequence in B. Then due to equivalence of norms on
    finite dimensional spaces we have an $M$ and $K$ such that:

    $K \geq \sup_n ||x^{(n)}|| \geq \sup_n M \max_{k=1,2..N}
    \{|\alpha_k^{(n)}|\} > |\alpha_k^{(m)}|\quad \forall k$

    Where $x^{(n)} = \sum_{i\leq N} e_i \alpha^{(n)}_i$ where
    $\{e_i\}_{i\leq N}$ is a normalized finite basis. Then
    $|\alpha_k^{(m)}|$ has a convergent subsequence by Bolzano Weierstrass
    and we can inductively find a subsequence of $x^{(n)}$ with convergent
    coordinates. Now since convergence in coordinates implies
    convergence for finite dim sequences we have found a convergent
    subsequence and I think we are done(?).

    For the other direction we can see that the closed unit ball is
    compact so the space must be finite dimensional.

  2. For C[0,1] I take a sequence in the unit ball $f_n(t) = t^n$,
    suppose it has a convergent subsequence in sup norm then the
    sequence must also converge pointwise, but then the sequence must
    converge to $1_{\{1\}}$ which is not in C[0,1] hence can't be in
    closure of the unit ball either.

    For Lp we take again a sequence in the closed unit ball given by
    $f_n(x) = 1_{[0, 1/n]} x n$ again if we have a convergent
    subsequence then that convergent subsequence has a subsequence that
    converges pointwise (to the same limit), but that limit clearly must
    be 0, but $||f_n(x)|| = \sqrt{1/2}$ hence contradiction again, so
    closure of unit ball is not compact. Thank you in advance for the
    help!

Best Answer

The definition you provide for relative compactness is the definition of compactness. We say a set is relatively compact if it's closure is compact. In terms of sequences, we have the following definition:

Definition: $A$ is $\textit{relatively compact}$ if for each sequence in $\overline{A}$, there is a convergent subsequence in $\overline{A}$.

I think there is a much simpler proof for 1. Suppose $V$ is finite dimensional, then there exists $T: V\rightarrow \mathbb{R}^n$ an isometric isomorphism for some $n$. Let $A\subset V$ be bounded and note that $T^{-1}\cdot T(\overline{A})=\overline{A}$. Note that $T(\overline{A})$ is closed and bounded in $\mathbb{R}^n$, so it's compact. Therefore, $T^{-1}\cdot T(\overline{A})$ is compact.

The other parts of your answer are correct.

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