Proof understanding: given unit vector $x$, find $y$, such that $y^* x = ||y|| = 1$ I was reading this proof, section III $\to$ II, where the situation is as follows:


*

*a vector $x\in \mathbb{C}^n$ is given, such that $||x|| = 1$, for some vector norm $||x||$.

*we are looking for a $y\in \mathbb{C}^n$, such that $y^*x = 1$ and $||y|| = 1$.


I can find $y$ if  $||\cdot|| = ||\cdot||_2$ or $||\cdot|| = ||\cdot||_\infty$, but I do not understand, how can $y$ exist in the case when $||\cdot|| = ||\cdot||_1$, since
$$|y^* x| = \left|\sum_{i} x_i \overline{y_i}\right|\leq
\sum_{i} |x_i \overline{y_i}|\leq\sum_i M_y |x_i| = M_y\text{,}$$
where $M_y = \max_i |y_i|$. Since $||y||_1 = 1$, we know that $M_y\leq1$. From the upper formula, it follows that we must have


*

*$M_y = 1$

*$|y_i| = M_y$, for all $i$,


otherwise $|y^* x| < 1$. Hence, $||y||_1 = \sum_i |y_i| = n$ and $y\neq 1$.
The proof says that For general norms, the existence of such a
vector follows from the theory of dual norms (a corollary to the Hahn-Banach theorem).
 A: This is a proof from Matrix Analysis, by Johnson and Horn. It has a great chapter on norms, if you're curious.

Theorem 5.5.9 (Duality theorem) [Let $V=\mathbb{R}^n$ or $\mathbb{C}^n$]
If $f$ is a norm, and $x_0\in V$ is given, then there is some $z\in V$ (not necessarily unique) such that $f^D(z)=1$ and $f(x_0)=z^*x_0$, that is, $|z^*x_0|\le f(x)$ for all $x\in V$ and $f(x_0)=z^*x_0$.

Proof.
For each given $x_0\in V$, [the equality of $f$ and $f^{DD}$] ensures that $f(x_0)=\max_{f^D(y)=1}\operatorname{Re}y^*x_0$, and compactness of the unit sphere of the norm $f^D$ ensures that there is some $z$ such that $f^D(z)=1$ and $\max_{f^D(y)=1}\operatorname{Re}y^*x_0=\operatorname{Re}z^*x_0$. If $z^*x_0$ were not real and nonnegative, there would be a real number $\theta$ such that $\operatorname{Re}(e^{-i\theta}z^*x_0)>0>\operatorname{Re}z^*x_0$ (of course, $f^D(e^{i\theta}z)=f^D(z)=1$), which would contradict maximality: $\operatorname{Re}z^*x_0\ge\operatorname{Re}y^*x_0$ for all $y$ in the unit sphere of $f^D$. $\qquad\qquad\Box$
