Show that $\limsup_{n\rightarrow\infty} \|x_n\|^{\frac{1}{n}}=1$

Question

Let $(x_n)$ be weakly convergent, but not norm convergent, sequence from a Banach space. Show that $\limsup_{n\rightarrow\infty} \|x_n\|^{\frac{1}{n}}=1$.

Any help?

Answer 1

A weakly convergent sequence in a normed vector space is norm-bounded. So there is an $M \in [0,\infty)$ such that we have $\lVert x_n\rVert \leqslant M$ for all $n$. Then $\lVert x_n\rVert^{1/n} \leqslant M^{1/n} \to 1$, whence

$$\limsup_{n\to\infty} \lVert x_n\rVert^{1/n} \leqslant 1.$$

Now it remains to show the inequality in the other direction,

$$\limsup_{n\to\infty} \lVert x_n\rVert^{1/n} \geqslant 1$$

for a sequence that is not norm-convergent. For example by showing the contrapositive, if the $\limsup$ is smaller than $1$, then the sequence is norm-convergent.