The difference between 2 definitions of the norm of an operator. In Kreyszig the definition of the norm of an operator is given in the following picture:

While in Israel Gohberg it is written as:

Why in the first picture in the definition it is written that $||x|| = 1,$ while in the second picture in the definition it is written that $||x|| \leq 1,$ is this two things the same? if so why? could anyone explain this for me please? 
 A: Yes, it's the same thing.
Of course,$$\sup_{\|x\|=1}\bigl\|A(x)\bigr\|\leqslant\sup_{\|x\|\leqslant1}\bigl\|A(x)\bigr\|.$$Now, suppose that$$\sup_{\|x\|=1}\bigl\|A(x)\bigr\|<\sup_{\|x\|\leqslant1}\bigl\|A(x)\bigr\|.$$That means that there is a vector $y$ such that $\|y\|<1$ and that $\bigl\|A(y)\bigr\|>\bigl\|A(x)\bigr\|$ whenever $\|x\|=1$. But $\left\|\frac y{\|y\|}\right\|=1$ (note that $\bigl\|A(y)\bigr\|>0\implies y\neq0\implies\|y\|\neq0$) and$$\left\|A\left(\frac y{\|y\|}\right)\right\|=\frac{\bigl\|A(y)\bigr\|}{\|y\|}>\bigl\|A(y)\bigr\|.$$This contradicts the choice of $y$.
A: They are the same because
$$
||ax|| = |a| \cdot ||x|| .
$$
A: If $\|Tx\| \le M$ for all $\|x\|=1$, then, for $x\ne 0$,
$$
                 \|T \frac{1}{\|x\|}x\| \le M
    \implies \|Tx\| \le M\|x\| \le M.
$$
A: Let $\lVert A \rVert _1 = \sup\limits_{\lVert x \rVert = 1} \lVert Ax \rVert$ and $\lVert A \rVert _2 = \sup\limits_{\lVert x \rVert \leq 1} \lVert Ax \rVert$. We will show $\lVert A \rVert _ 1 = \lVert A \rVert _2$.
Obviously $\lVert A \rVert _1 \leq \lVert A \rVert _ 2$ (because there are more $x$ admissible for the second supremum). On the other hand for $x$ with $\lVert x \rVert \leq 1$ and $x \neq 0$ we have
$$ \lVert Ax \rVert = \lVert x \rVert \lVert A \frac{x}{\lVert x \rVert} \rVert$$
and $\frac{x}{\lVert x \rVert}$ has norm $1$. Thus
$$ \lVert A \rVert _2 = \sup\limits _{\lVert x \rVert \leq 1} \lVert A x \rVert = \sup _{\lVert x \rVert \leq 1} \lVert x \rVert \lVert A \frac{x}{\lVert x \rVert} \rVert \leq \sup\limits _{\lVert x \rVert \leq 1} \lVert x \rVert \sup\limits _{\lVert x \rVert \leq 1}\lVert A \frac{x}{\lVert x \rVert} \rVert =  \sup\limits _{\lVert x \rVert = 1}\lVert A x \rVert = \lVert A \rVert _1.$$
