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 $||w||&lt;1$, then using $\tilde{w}=1/||w||*w$ will give a larger inner product.
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 $$\begin{array}{ll}T : &amp;X \to \mathbb{R} \ \ \ (\text{ or } \mathbb{C}) \\&amp;z \to \langle w,z\rangle\end{array}$$ 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  $\|T\| = \infty$
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 $w = &lt;.6,.2&gt;, z = &lt;-.6,-.2&gt;.$
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
$w$.