Let $(E, ||\cdot ||)$ a normed space over $\Bbb{C}$ and $f: E \rightarrow \Bbb{C}$ an unbounded linear operator. Prove that $\Bbb{C}= \{f(x) : ||x|| \leq 1 \}$
I am trying to prove this and I have seen this post where it has been done for $\Bbb{R}$. The problem is that I do not understund why $[0,f(x_n)] \subseteq f(X_1)$, so any possible explanation would be appreciated :)
Write $B=\{x\in E : \|x\| \leq 1 \}$. If you know that some $z\in\mathbb{C}$ is in $f(B)$, i.e. there exists $x\in B$ with $f(x)=z$, then you know that every $w\in\mathbb{C}$ with $|w|\leq|z|$ is also in $f(B)$, since there exists $\alpha\in\mathbb{C}$ with $|\alpha|\leq 1$ such that $$ w=\alpha z=\alpha f(x)=f(\alpha x) $$ and $\big\|\alpha x\big\|=|\alpha|\|x\|\leq 1$.
This is the part you didn't understand, maybe you can do the rest.
