Show that $|| \omega || \leq M$ if and only if $| \omega(x)| \leq M || x||$ for all $x \in X$

Question

Let $X$ be a Banach space. We defined its dual as $$ X^{*} = \left\{ \omega \mid \omega: X \rightarrow \mathbb{C} \ \text{is a continuous linear map} \right\}. $$ For a $\omega \in X^{*}$, we defined its norm as $$ || \omega || = \sup_{|| x || \leq 1} | \omega(x)|. $$

Problem: Prove that $|| \omega || \leq M$ if and only if $|\omega(x)| \leq M ||x|| $ for all $x \in X$.

Attempt: $\Leftarrow$: This direction I understand I think. I let $x \in X$. Then we have $|\omega(x)| \leq M || x||$ . In particular, this holds for all those $x \in X$ with $|| x || \leq 1$. Then $|\omega(x)| \leq M$, and since this holds for all $x$, we also have $|| \omega || \leq M$.

$\Rightarrow$: This I do not understand. I assume that $|| \omega || \leq M$. I let $x \in X$ be arbitrary. I wish to show that $| \omega(x) | \leq M || x ||$.

If $|| x || \leq 1$, then I have $$ | \omega(x) | \leq || \omega || \leq M $$ and from this I think I can conclude that $|\omega(x)| \leq M || x||$. But what if $|| x|| > 1$??

Answer 1

If $\|x\|>1$ let $y=\frac x {\|x\|}$. Then $\|y\| =1$ so $|\omega(y)| \leq M$. But $\omega$ is linear so $\omega(y)=\frac {\omega (x)} {\|x\|}$. Can you finish?.

Answer 2

Hint: Note that, for any non-zero $x \in X$, $$ \frac{|\omega(x)|}{\Vert x \Vert} = \left|\omega \left( \frac{x}{\Vert x \Vert} \right) \right|, $$ and $x / \Vert x\Vert$ has norm $1$.