Let $X$ be a normed vector space and $g : \mathbb{R}^+\to \mathbb{R}$ a monotonic increase function that is lower semi continuous. Then, I want to show that $h := g(\|\cdot\|)$ is weak lower semi continuous.
I tried the following approach: take $\{x_n\}$ weakly converging to $x$. Then, given $\epsilon > 0$ there is $f\in X'$, $\|f\| = 1$ such that $$f(x-x_n) + \epsilon > \|x-x_n\|.$$
Since $$\|x\| \leq \|x-x_n\| + \|x_n\|,$$ then, once $g$ is monotone and increasing, $$g(\|x\|) \leq g(\|x-x_n\| + \|x_n\|) < g(f(x-x_n) + \epsilon + \|x_n\|).$$
Note that in principle $f$ depends on $n$.
But thinking carefully one sees that since $|f(x-x_n)| \leq 1$, then it has a convergent subsequence (Bolzano--Weierstrass). Which implies that $\lim\inf f(x-x_n) = 0$ indeed.
Therefore, note that there is $n_0$ such that for all $n > n_0$, $$f(x-x_n) < \epsilon.$$ Since, $$\lim\inf g(f(x-x_{\epsilon}) + \epsilon + \|x_n\|) = \lim_{N\to\infty}\inf_{n\geq N} g(f(x-x_{\epsilon}) + \epsilon + \|x_n\|),$$ for $N > n_0$, $$\lim\inf g(f(x-x_{\epsilon}) + \epsilon + \|x_n\|) \leq \lim\inf g(2\epsilon + \|x_n\|).$$
Now one needs to handle with $\epsilon \to 0^+$. But since $g$ is lower semicontinuous, then $$g(0) \leq \lim\inf_{\epsilon\to 0^+}g(\epsilon) \Rightarrow$$ given any $\delta > 0$ there is $\epsilon_0 > 0$ such that for any $\epsilon <\epsilon_0,$ $$g(0) < g(\epsilon) + \delta.$$
Therefore, there is no $\delta > 0$ such that $g(\|x_n\|) \geq \lim_{\epsilon\to 0^+}g(2\epsilon + \|x_n\|) + \delta,$ for some $n>0$.
Hence, $$g(x) \leq \lim_{\epsilon\to 0^+}\lim\inf (2\epsilon + \|x_n\|) \leq \lim\inf g(\|x_n\|).$$
Is this correct?