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?
Best Answer
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.