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Given an infinitely dimensional normed space $X$, is the map $x\mapsto\|x\|$ weakly continuous?

I can prove for Hilbert space that $x\mapsto\|x\|$ is not weakly continuous. How about general Banach space?

Sunny. Y
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Take $X=\ell^p$ with the $\|\cdot\|_p$ norm for $p\in(1,\infty)$. Then, for $p\ne2$, $X$ is a Banach space that is not a Hilbert space. For $n\in\mathbb{N}$ set $e_n$ to be the sequence that is $0$ everywhere except the $n$th slot, where the value is $1$. Then $\|e_n\|=1$ for all $n$. However, $X^*\cong\ell^q$, where $q$ is the conjugate exponent, so every functional $\phi\in X^*$ is of the form $\phi(x)=\sum_kx_ky_k$ for all $x=(x_k)\in X$, where $(y_k)\in\ell^q$. Note that $\phi(e_n)=y_n\to0$, since $\sum_k|y_k|^q<\infty$ and thus $|y_k|^q\to0$ hence $y_k\to0$. This shows that $e_n\to0$ weakly.

Just dropped in
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Thanks so much for @David Mitra 's help. From this link we know the weak closure of unit sphere $S$ is the unit ball $B=\{x\in X\colon \|x\|\leqslant 1\}\not=S$. Note $\{1\}\subset\mathbf{R}$ is closed, however, the preimage of $\{1\}$ is $S$, which is not weakly closed in $X$, so this map $x\mapsto \|x\|$ is not weakly continuous.

Sunny. Y
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    However, one should be aware that there are spaces where $|\cdot|$ is sequentially weakly continuous, i.e., for any sequence $(x_n)$ with $x_n \rightharpoonup x$ we have $|x_n| \to |x|$. For example, in $\ell^1$ a sequence is weakly convergent if and only if it is strongly convergent. – gerw Mar 22 '22 at 07:28