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I am working on a problem right where I have a sequence $(x_n)$ which converges weakly to some $x$ in a Hilbert space $(H, \left<\cdot, \cdot\right>)$. Denote by $||\cdot||$ the norm induced by the inner-product. Since an induced norm is (sequentially) weakly lower-semicontinuous, we have that $||x|| \le \liminf_{n \to \infty} ||x_n||$.

Now consider the Banach space $(H, ||\cdot||_0)$ and suppose that $||\cdot||$ and $||\cdot||_0$ are bi-Lipchitz equivalent. That is, $\exists \alpha, \beta \in \mathbb{R}$ such that $\alpha ||v|| \le ||v||_0 \le \beta ||v||$ for all $v \in H.$ Can I similarly conclude that $||x||_0 \le \liminf_{n \to \infty} ||x_n||_0$?

It's my understanding that, since the norms are equivalent, $(H, ||\cdot||)$ and $(H, ||\cdot||_0)$ must have the same continuous dual space, and hence the same weak topology. Therefore, $x_n \rightharpoonup x$ with respect to the weak topology of $(H, ||\cdot||_0)$. I would also imagine that the reflexivity of $(H, \left<\cdot, \cdot\right>)$ is carried over to $(H, ||\cdot||_0)$ for similar reasons, after which I should have that $||\cdot||_0$ is (sequentially) weakly lower-semicontinuous as well. Does this hold?

infinitylord
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Hm, perhaps, you have thought too much?
For all normed vector space $E$, if $x_n$ converges weakly to $x$ ($x,x_1,x_2,... \in E$ ), we have $|x| \le \liminf |x_n|$

Now, if you have already the weak convergence of $x_n$ to $x$ in the new norm, then clearly, you have that norm inequality.

Paresseux Nguyen
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  • Ah, yes, I confused that result with the one about bounded sequences having a weakly convergent subsequence in a reflexive space. For the sake of closure, I am correct in assuming that weak convergence in $(H, \left<\cdot, \cdot\right>)$ implies the weak convergence in $(H, ||\cdot||_0)$ (given the norm equivalence), yes? – infinitylord Dec 14 '20 at 20:08
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    Yes. Argument can be done as follows
    1) Because $| |$ and $| |$ are equivalent, every continuous on $(H, | |)$ is a linear functional on $(H, | |_0)$ .Vice versa
    2) For every linear function $f$ in $(H, | |_0)^*$, $f$ is also an element in $(H, | |)$ . Thus we have the convergence $\langle f,x_n \rangle \rightarrow \langle f,x \rangle$.
    3) Hence, $(x_n)$ converges weakly to $x$ under the new space. Hence the conclusion.     – Paresseux Nguyen Dec 14 '20 at 20:13
  • Nonetheless, I agree this point is very confusing. – Paresseux Nguyen Dec 14 '20 at 20:13