Condition on definition of dual norm In a review I'm reading, the dual norm is defined as $$||z||_*=\max_{||w||\leq 1}\langle w,z\rangle.  $$
Though I'm having a hard time understanding why $||w||\leq 1$ isn't equivalent to $||w||=1$, as given a $w$ such that $\langle w,z\rangle\geq 0$, if $||w||<1$, then using $\tilde{w}=1/||w||*w$ will give a larger inner product.
 A: when $T$ is a linear operator $X \to Y$ between two normed vector spaces 
$$\max_{\|z\| \le 1} \|T z\| \qquad\qquad \text{and} \qquad\qquad\max_{\|z\| = 1} \| T z \|$$  are the same. 
this is because a norm asks for $\| a z\| = |a| \ \|z\| $ for any constant $a$ and $T$ is linear so  $\| T (a z)\| = |a| \ \|T z\| $.
here your linear operator is $$\begin{array}{ll}T : &X \to \mathbb{R} \ \ \ (\text{ or } \mathbb{C}) \\&z \to \langle w,z\rangle\end{array}$$ and its (operator) norm is $$\| T\| = \max_{\|z\|=1} | T z| = \max_{\|z\|=1} | \langle w,z\rangle | $$
here it is not the case but don't forget many linear operators are unbounded, sometimes  $\|T\| = \infty$ 
A: \begin{equation}
\begin{aligned}
\parallel u \parallel_{*} & = \enspace \max\limits_{\parallel x \parallel \leq 1} <u, x>\\
\leq & \enspace \max\limits_{\parallel x \parallel \leq 1} \frac{<u, x>}{\parallel x \parallel}\\
\leq & \enspace \max\limits_{\parallel x \parallel \neq 0} \frac{<u, x>}{\parallel x \parallel}\\
= & \enspace \max\limits_{\parallel y \parallel = 1} <u, y>\\
\leq & \enspace \max\limits_{\parallel y \parallel \leq 1} <u, y> = \enspace \parallel u \parallel_{*}
\end{aligned}
\end{equation}
Essentially, this means 
$$
\parallel u \parallel_{*} = \max\limits_{\parallel x \parallel \neq 0} \frac{<u, x>}{\parallel x \parallel}
$$
and it implies that norm of "other vector" is exactly 1. I am guessing there is some corner case (may be depending on space, norm...) for which the inequality is needed. But for the normal cases, we can just assume that we are dealing with unit vector to find dual norm of $u$.
A: Consider the case where $z$ has negative components, or a certain set of 
negative components. We may want to consider minimizing them by inputing
a smaller length $w$ vector rather than choosing the largest possible $w$
vector available, as an example, consider when $w = <.6,.2>, z = <-.6,-.2>.$
We see that
$$\langle w,z \rangle \approx -.52$$
while
$$\langle \frac{w}{\|w\|}, z \rangle \approx -.822.$$
In this case, we get a strictly greater solution when we do not normalize
$w$.
