Let $X$ be a Banach space. Let $\phi: [0,\, 1] \to X $ be a sequentially weakly continuous function, that is, $$\forall \,(t_n) \subset [0, 1],\,\,\, t_n \to t \Rightarrow \phi(t_n) \rightharpoonup \phi(t).$$
Consider the function $ f : [0, 1] \to \mathbb{R} $ defined by $f(t) = \|φ(t)\| $ for every $t \in [0,\, 1]$. Prove that
$f$ is bounded,
there exists the minimum of $f$ on $[0,\, 1]$.
My solution.
Since $\|x\|\geq 0$, the function $f$ is bounded from below and the infimum $\, I:=\inf_{[0,1]} f \,\,$ is finite. Now we apply the direct method in calculus of variations.
Let $t_n$ be a minimizing sequence, that is $f(t_n)\to I$. Since $[0, \,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $\overline{t}\in [0,\,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular $$f(\overline t) \leq \liminf_{k\to\infty}f(t_{n_k})=\lim_{n\to\infty}f(t_{n})=I,$$ but by definition of infimum $I\leq f(\overline t)$ and we get $$f(\overline t)=I.$$
Questions:
It is necessary "$X$ is a Banach space" ? I think the answer is no.
It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).
Thanks in advance.
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n \to t$ such that $f(t_n) > n,$ but since $\varphi(t_n) \to \varphi(t)$ it turns out $\varphi(t_n)$ is bounded (which means $\|\varphi(t_n)\|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.