Norm of a Linear Transformation .... Junnheng Proposition 9.2.3 .... ....

[Math Amateur]

I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on \(\displaystyle \mathbb{R}^n\)"

I need some help with the proof of Proposition 9.2.3 ...

Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows:
https://www.physicsforums.com/attachments/7889
View attachment 7890
In the above proof we read the following:

" ... ... If \(\displaystyle \mathbf{x} \neq \mathbf{0}\) then \(\displaystyle \| \mathbf{x} \|^{-1} \mathbf{x}\) has a norm \(\displaystyle 1\), hence


\(\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1\) ... ... "
Now I know that \(\displaystyle T( c \mathbf{x} ) = c T( \mathbf{x} ) \)... BUT ...... how do we know that this works "under the norm sign" ...... that is, how do we know ...\(\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \)
... and further ... how do we know that ...
\(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1 \)Help will be appreciated ...

Peter

 

[Opalg]

I know that \(\displaystyle T( c \mathbf{x} ) = c T( \mathbf{x} ) \)

... BUT ...

... how do we know that this works "under the norm sign" ...

... that is, how do we know ...

\(\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \)

If you make the substitution $c = \| \mathbf{x} \|^{-1}$ in the equation $\| c \mathbf{y}\| = c\| \mathbf{y}\|$ (where $c$ is a positive constant), then it becomes $\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when $\mathbf{y} = T\mathbf{x}$, to get $\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$.

... and further ... how do we know that ...

\(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1 \)

That is a mistake. It should read \(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\| \).

 

[Math Amateur]

If you make the substitution $c = \| \mathbf{x} \|^{-1}$ in the equation $\| c \mathbf{y}\| = c\| \mathbf{y}\|$ (where $c$ is a positive constant), then it becomes $\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when $\mathbf{y} = T\mathbf{x}$, to get $\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$.That is a mistake. It should read \(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\| \).

Thanks Opalg ...

Appreciate your help ...

Peter

 

[Math Amateur]

If you make the substitution $c = \| \mathbf{x} \|^{-1}$ in the equation $\| c \mathbf{y}\| = c\| \mathbf{y}\|$ (where $c$ is a positive constant), then it becomes $\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when $\mathbf{y} = T\mathbf{x}$, to get $\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$.That is a mistake. It should read \(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\| \).

Hi Opalg ...

Just realized that I don't fully understand why \(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\| \) ...

Can you please help further and demonstrate why this is the case ...

Peter

 

[Opalg]

Hi Opalg ...

Just realized that I don't fully understand why \(\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\| \) ...

Can you please help further and demonstrate why this is the case ...

Peter

The definition $\|T\| = \sup\{\|T \mathbf{y} \| : \| \mathbf{y} \| = 1\}$ says that if $\| \mathbf{y} \| = 1$ then $\|T \mathbf{y} \| \leqslant \|T\|$. But $ \| \mathbf{x} \|^{-1} \mathbf{x} $ has norm $1$, so you can substitute that vector for $ \mathbf{y}$, to get $\bigl\|T( \| \mathbf{x} \|^{-1} \mathbf{x} )\bigr\| \leqslant \|T\|$.

 

[Math Amateur]

The definition $\|T\| = \sup\{\|T \mathbf{y} \| : \| \mathbf{y} \| = 1\}$ says that if $\| \mathbf{y} \| = 1$ then $\|T \mathbf{y} \| \leqslant \|T\|$. But $ \| \mathbf{x} \|^{-1} \mathbf{x} $ has norm $1$, so you can substitute that vector for $ \mathbf{y}$, to get $\bigl\|T( \| \mathbf{x} \|^{-1} \mathbf{x} )\bigr\| \leqslant \|T\|$.

Hi Opalg ... thanks again for the help ...

Peter