Let $X$ be a separable reflexive real Banach space and $\{\psi_n\}$ be a dense sequence in $$\{\psi\in X' : ||\psi||_{X'} \leq 1\}.$$ Consider in $X$ the scalar product defined by $$(x | y)_0 = \sum_{n=1}^\infty 2^{-n} \langle \psi_n,x \rangle\langle \psi_n,y \rangle.$$

Show every bounded sequence in $X$ admits a Cauchy subsequence with respect to the norm $||\cdot||_0 $ (norm induced by $(x| x)_0$) .


Since $X$ is reflexive, by Banach–Alaoglu theorem, every bounded sequence $\{x_j\}$ in $X$, there exists a weakly convergent subsequence $\{{x_j}_k\}$ to some $x$.

To show that ${{x_j}_k}$ is Cauchy under $||\cdot||_0$, it is sufficient to show that the sequence ${{x_j}_k}$ converges to $x$ under $||\cdot||_0$.

Observe that if ${x_j}_k \rightharpoonup x$, then $||{x_j}_k||_X \leq C$ and $$|\langle \psi_n,{{x_j}_k} - x \rangle| \leq ||\psi_n||_{X'}||{{x_j}_k} - x ||_X \leq 2C$$

Now let $\epsilon$ be given, $$||{x_j}_k - x||_0^2 = \sum_{n=1}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2$$ we split the sum into two parts at $N$ such that $$\sum_{n=N}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2\leq \sum_{n=N}^\infty 2^{-n} (2C)^2 \leq \epsilon/2.$$ Now for the first $N-1$ terms, choose $K$ such that for $k\geq K$ we have $$\sum_{n=1}^{N-1} 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2 \leq \epsilon/2,$$ the reason we could choose such $K$ is because $\langle \psi_n,{x_j}_k - x \rangle$ goes to zero as $k$ goes to $\infty$ for each of the $N-1$ terms.

Combine the two, we have for each $k\geq K$ $$||{x_j}_k - x||_0^2 = \sum_{n=1}^{N-1} 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2+ \sum_{n=N}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2\leq \epsilon/2 + \epsilon/2.$$


  1. Is my proof correct? Is there an easier way to do this? I know my proof is quite long.. Thank you for reading it!

  2. What is the significance of $\psi_n$ being dense? I did not use this fact in my proof.

  • $\begingroup$ A quick comment: In your solution, you are basically proving that the inner product $(\cdot|\cdot)_0$ induces the weak topology on bounded subsets of $X$. $\endgroup$ – Luiz Cordeiro Aug 1 '14 at 16:49
  • $\begingroup$ @LuizCordeiro And that is why I am a bit unsure about my proof, since I know that weak topology is not metrizable. Could you explain this please, thank you! $\endgroup$ – Xiao Aug 1 '14 at 21:36
  • 1
    $\begingroup$ When equipped with the weak topology, the whole space $X$ is not metrizable, but if we restrict the weak topology to a bounded subset of $X$, we obtain a metrizable topological space (as you proved). This says that there exist unbounded nets (not sequences!) in $X$ which converge weakly. See here for an example (the same argument works with "weak" in place of "weak*") $\endgroup$ – Luiz Cordeiro Aug 2 '14 at 7:53
  • $\begingroup$ @LuizCordeiro Thank you very much! $\endgroup$ – Xiao Aug 2 '14 at 14:59
  1. It looks fine.
  2. I think the denseness is used implicitly when you want to show that $\|\cdot\|_0$ is a norm.
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