Is it true $\left \| T(x) \right \| \le \left \| T \right \| \cdot \left \| x \right \| $ Let be $E$ a real Banach space and $E^*$ the real dual space. $\forall T \in E^*$
Is it true $\left \| T(x) \right \| \le \left \| T \right \| \cdot \left \|  x \right \| $? Why?
Let be $\{ T_n \}_n$ a Cauchy's sequence.
Is it true $\left \| T_n(x)-T_m(x) \right \| \le \left \| T_n-T_m \right \| \cdot \left \|  x \right \| $? Why?
Thanks to all.
NOTE: In my book (Brezis' one) the norm is define as:
$$ \left \| T \right \|= \sup_{\left \| x \right \| \le 1} \left \| T(x) \right \| $$
 A: Yes. For $x \neq 0$:
$$ \| T(x) \| = \left \| T \left(\frac{x \|x\|}{\|x\|} \right) \right \| = \| x \| \left \|T \left(\frac{x}{\|x\|} \right) \right \| \le \|T\| \|x\|$$
And it also obviously holds for $x=0$
A: In general, we can write this:
$$\|T\| = \sup_{z \neq 0} \frac{\|T(z)\|}{\|z\|}.$$
Suppose that $z^*$ is the value of $z$ for which you have the sup. Obvioulsy:
$$\|T\| = \frac{\|T(z^*)\|}{\|z^*\|}.$$
Consider a generic $x$. Then:
$$\frac{\|T(x)\|}{\|x\|} \leq \frac{\|T(z^*)\|}{\|z^*\|} = \|T\| \Rightarrow \|T(x)\|\leq\|T\| \|x\|.$$

Notice that using different definition as:
$$\|T\| = \sup_{\|z\| \leq 1} \frac{\|T(z)\|}{\|z\|},$$
doesn't change the situation. Indeed, suppose that $\|z^*\| = M > 1$. Then consider a vector $y^* = \frac{1}{M} z^*$, so that $\|y^*\| = 1$. In this case:
$$\frac{\|T(y^*)\|}{\|y^*\|} = \frac{\|T\left(\frac{1}{M}z^*\right)\|}{\|\frac{1}{M}z^*\|} = \frac{\frac{1}{M}\|T\left(z^*\right)\|}{\frac{1}{M}\|z^*\|} = \frac{\|T(z^*)\|}{\|z^*\|}.$$
This means that also $y^*$ is a point at which the sup is attained.
In general, the definition of sup is:
$$\|T\| = \sup_{\|z\| =1} \frac{\|T(z)\|}{\|z\|}.$$
Your book is giving an equivalent formulation. That's all.

Let's talk about the sequence. Since all $T_n$ are linear transformations, then $T_n - T_m$ is a linear transformation too (i.e. $T_n - T_m \in E^*$).
Using the previous results, you have that:
$$\|T_n(x) - T_m(x)\| \leq \|T_n - T_m\| \|x\|.$$
