# 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?

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 $y$.

• The first inequality in the third line is not clear for me, could youclarify it please? – user426277 Nov 16 '17 at 1:31
• Since$$\left\{\bigl\|A(x)\bigr\|\,\middle|\,\|x\|=1\right\}\subset\left\{\bigl\|A(x)\bigr\|\,\middle|\,\|x\|\leqslant1\right\},$$we have$$\sup\left\{\bigl\|A(x)\bigr\|\,\middle|\,\|x\|=1\right\}\leqslant\sup\left\{\bigl\|A(x)\bigr\|\,\middle|\,\|x\|\leqslant1\right\}.$$ – José Carlos Santos Nov 16 '17 at 7:23
• Thank you so much for your so clear explanation. – user426277 Nov 16 '17 at 14:42

They are the same because $$||ax|| = |a| \cdot ||x|| .$$

• could you say more details please? – user426277 Nov 16 '17 at 1:52
• @Idonotknow The other correct answers here (including the one you accepted) use this idea to finish the proof. – Ethan Bolker Nov 16 '17 at 13:10

If $\|Tx\| \le M$ for all $\|x\|=1$, then, for $x\ne 0$, $$\|T \frac{1}{\|x\|}x\| \le M \implies \|Tx\| \le M\|x\| \le M.$$

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 $$\lVert A \rVert _2 = \sup\limits _{\lVert x \rVert \leq 1} \lVert A x \rVert = \sup _{\lVert x \rVert \leq 1} \lVert x \rVert \lVert A \frac{x}{\lVert x \rVert} \rVert \leq \sup\limits _{\lVert x \rVert \leq 1} \lVert x \rVert \sup\limits _{\lVert x \rVert \leq 1}\lVert A \frac{x}{\lVert x \rVert} \rVert = \sup\limits _{\lVert x \rVert = 1}\lVert A x \rVert = \lVert A \rVert _1.$$