# $\phi_n \rightarrow \phi$ weakly-$*$, then $\|\phi\|\leq \limsup_n \|\phi_n\|$.

$\phi_n \rightarrow \phi$ weakly-$*$, then $\|\phi\|\leq \limsup_n \|\phi_n\|$.

My attempt:

$$\|\phi\| = \sup_{\|x\| = 1} |\phi(x)| \leq \sup_{\|x\| = 1} \lim_n |\phi_n(x)|$$ Using an epsilon-argument, there exists an $x_0$ with norm one such that $$\|\phi\| \leq \lim_n |\phi_n(x_0)| + \epsilon$$ then $$\|\phi\| \leq \limsup_n \sup_{\|x\| = 1} |\phi_n(x)| + \epsilon = \|\phi_n\| + \epsilon.$$

Question: could you please provide an explicit example such that $\phi_n \rightarrow \phi$ weakly-$*$ and $$\|\phi\| > \liminf \|\phi_n\|.$$ Thank you very much!

Edit: For the counter example, $\phi_n$ has to converge weakly-$*$ to something that is not the zero functional. We want $$\|\phi\| > \liminf \|\phi_n\|,$$ not $$\|\phi\| < \liminf \|\phi_n\|.$$

I know $\|\phi\|\leq \liminf \|\phi_n\|$ holds for $\phi_n \rightarrow \phi$ weakly. When the space is not reflexive, weak-$*$ convergence does not imply weak convergence, so I was hoping to find a counter example for this inequality.

Basically, I want to know why the inequality in the problem is a $\limsup$ but not $\liminf$ (which holds for weak convergence).

• Are you sure you want $\|\phi\|>\liminf \|\phi_n\|$? This is not true. Commented Sep 8, 2014 at 15:18
• This inequality is not true. With a little adapatation your proof, you can prove that $\|\phi\|\le \liminf \|\phi_n\|$. Commented Sep 8, 2014 at 15:22
• @Tomás yeah I was wondering about this as well. I know $\|\phi\|\leq \liminf \|\phi_n\|$ holds for $\phi_n \rightarrow \phi$ weakly. When the space is not reflexive, weak-$*$ convergence does not imply weak convergence, so I was hoping to find a counter example for this inequality.
– Xiao
Commented Sep 8, 2014 at 15:29
• math.stackexchange.com/questions/274764/… Commented Sep 8, 2014 at 15:33
– Xiao
Commented Sep 9, 2014 at 21:10

The correct inequality is $||\phi|| \le \lim \inf ||\phi_n||\$. $\$Let's prove it. Let $M = ||\phi||$. Take $\epsilon >0$ arbitrary. There exists $x$ of norm $1$ so that $|\phi(x)| > M -\epsilon$. Since $\phi_n(x)$ converges to $\phi(x)$ we also have $|\phi_n(x)| \to |\phi(x)|$. Therefore $|\phi_n(x)|> M-\epsilon$ for $n$ large enough. But $||\phi_n|| \ge |\phi_n(x)|$ since $||x||=1$. We conclude that $||\phi_n|| > M-\epsilon$ for $n$ large enough and therefore $\lim \inf ||\phi_n|| \ge M -\epsilon$. Since $\epsilon > 0$ was arbitrary we conclude $\lim \inf ||\phi_n|| \ge M$ that is $$\lim \inf ||\phi_n|| \ge ||\phi||$$ This inequality can be strict. Consider an infinite dimensional Hilbert space. A Hilbert space is self dual so we may consider vectors instead of functionals. Let $(\phi_n)$ an orthonormal system. Then $\phi_n \to 0$ weakly and $||\phi_n|| = 1$ for all $n$.

Also, $\phi_n + \phi_0 \to \phi_0$ weakly and $||\phi_n + \phi_0 ||= \sqrt{2}$, $||\phi_0|| =1$.

• How can $0>1$.? Commented Sep 8, 2014 at 15:18
• orangeskid presumably wanted to give a counterexample to the correct version of the inequality. Commented Sep 8, 2014 at 15:29
• just added a complete proof of the correct statement. Commented Sep 8, 2014 at 16:16

I guess that the OP meant:

$$\|\phi\| \leqslant \liminf_{n\to\infty}\|\phi_n\|,$$

which is always true. The inequality can be however strict.

Take $$(\phi_n)_{n=1}^\infty$$ to be the canonical basis of $$\ell_2 = \ell_2^*$$. It converges weak* to 0 but it consists of unit vectors. A similar situation takes place in dual spaces of all infinite dimensional Banach spaces; this is the famous Josefson–Nissenzweig theorem:

If $$X$$ is an infinite-dimensional Banach space, then there exists a sequence $$(\phi_n)_{n=1}^\infty$$ of unit vectors in $$X^*$$ which converges to 0 weak*.

For the references, see

B. Josefson, Weak sequential convergence in the dual of a Banach space does not imply norm convergence, Ark. Mat. 13 (1975), 79–89.

A. Nissenzweig, $$w^*$$ sequential convergence, Israel J. Math. 22 (1975), 266–272.