property of a bounded linear functional on a Hilbert space let $f$ be a bounded linear functional on a Hilbert space $H$. i.e. $(f : H \to \mathbb{C})$
then there exists an $M$ in $\mathbb{R}$ such that $| f(x)| \leq M \| x\|_H \; \forall x \in H$
in a certain proof they use the following inequality and claim that it holds because $f$ is a bounded linear functional
$| f(x)| \leq \| f\|_{H^{*}} \| x\|_H$ $(\star)$
the norm of the dual space is defined as follows $\| f\|_{H^{*}} = sup_{\| x\|_H \leq 1}  |f(x)|$
I couldn't figure out how to get to $(\star)$ using these properties and defintions
any help would be appreciated. thanks !
 A: The inequality is trivial for $x=0$. Take $x\ne 0$. Then $\|x\|_H^{-1}x$ has norm one:
$$
|f(x)| = \|x\|_H |f(\|x\|_H^{-1}x)| \le \|x\|_H \sup_{\|y\|_H\le 1}|f(y)|= \|x\|_H \|f\|_{H^*}.
$$
A: By definition(!) $\| f\|_{H^{*}}$ is the smallest $M \ge 0$ such that $| f(x)| \leq M \| x\|_H \; \forall x \in H$.
A: I offer a sketch of the proof of a standard fact, which seems to address the OP's question in a more general setting. 
Let $f:X\to Y$ be a bounded operator, $X$ and $Y$ Banach spaces. If
$N_1=\sup_{\|x\|=1} {\|f(x)\|}$
$N_2= \inf\lbrace C > 0:\|f(x)\|\le C\|x\|\ \forall x\in X\rbrace$
$N_3=\sup\left \{ \frac{\|f(x)\|}{\|x\|}:x\neq 0 \right \}.$
Then, $N_1=N_2=N_3$.
$a).\ $ We have $\left \{ \|f(x)\|:\|x\|=1 \right \}\subseteq \left \{ \frac{\|f(x)\|}{\|x\|}:x\neq 0 \right \}\Rightarrow N_1\le N_3,$ on taking sups. On the other hand, since $\frac{\|f(x)\|}{\|x\|}=\left \| f\left ( \frac{x}{\|x\|} \right ) \right \|$ and $ \left\|\frac{x}{\|x\|}\right\|=1,$ we get  $\frac{\|f(x)\|}{\|x\|}\le \sup_{\|y\|=1} {\|f(y)\|}=N_1.$ Suping over the left hand side, we get $N_3\le N_1$ from which we conclude that $N_3= N_1.$
$b).\ $ Now, let $c<N_3$. By definition of $N_3,$ there is an $x_c$ such that $\frac{\|f(x_c)\|}{\|x_c\|}> c\Rightarrow \|f(x_c)\|>c\|x_c\|,$ from which we conclude that if $c<N_3$ then $c\notin \lbrace C > 0:\|f(x)\|\le C\|x\|\ \forall x\in X\rbrace.$ Thus, if $d\in \lbrace C > 0:\|f(x)\|\le C\|x\|\ \forall x\in X\rbrace$ then it must be the case that $d\ge N_3$ and so, infing over all such $d$, we get  $N_2\ge N_3.$ On the other hand, we have, for all $x\in X,\ \frac{\|f(x)\|}{\|x\|}\le N_3$ and so  $\|f(x)\|\le N_3\|x\|$ which means that $N_3\in \lbrace C > 0:\|f(x)\|\le C\|x\|\ \forall x\in X\rbrace,$ from which we infer, by definition of the infimum, that $N_3\ge N_2$ and so $N_3=N_2.$
Combining $a).$ and $b).$ we get $N_1=N_2$ which finishes the proof. 
