Differentiation on R^n ....need/ use of norms ....

[Math Amateur]

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

I am currently focused on Chapter 9: "Differentiation on [FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Math]n[/FONT]"

I need some help with an aspect of Theorem 9.1.10 ...

Theorem 9.1.10 reads as follows:
View attachment 7883The proof of Theorem 9.1.10 relies on the definition of the derivative of a vector-valued function of several variables ... that is, Definition 9.1.6 ... so I am providing the same ... as follows:
View attachment 7884
In Junghenn's proof of Theorem 9.1.10 above, we read the following:

" ... ... and

\(\displaystyle \eta (h) = \frac{ f(a + h ) - f(a) - df_a (h) }{ \| h \| }\) if \(\displaystyle h \neq 0\)

... ... "Now there are no norm signs around this expression (with the exception of around \(\displaystyle h\) in the denominator ...) ... and indeed no norm signs around the expression \(\displaystyle \lim_{ h \rightarrow 0 } \eta(h) = 0\) ... nor indeed are there any norm signs in the limit shown in Definition 9.1.6 above (with the exception of around \(\displaystyle h\) in the denominator ...) ...

... BUT ...

... ... this lack of norm signs seems in contrast to the last few lines of the proof of Theorem 9.1.10 as follows ... where we read ...

" ... ... Conversely if (9.6) holds for some \(\displaystyle \eta\) and \(\displaystyle T\), then \(\displaystyle \lim_{ h \rightarrow 0 } \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| } = \lim_{ h \rightarrow 0 } \| \eta(h) \| = 0
\)

... ... "Here, in contrast to the case above, there are norm signs around the numerator and indeed around \(\displaystyle \eta(h)\) ... ...
Can someone please explain why norm signs are used in the numerator and around \(\displaystyle \eta(h)\) in one case ... yet not the other ...
Help will be appreciated ...

Peter

 

[Opalg]

In Junghenn's proof of Theorem 9.1.10 above, we read the following:

" ... ... and

\(\displaystyle \eta (h) = \frac{ f(a + h ) - f(a) - df_a (h) }{ \| h \| }\) if \(\displaystyle h \neq 0\)

... ... "Now there are no norm signs around this expression (with the exception of around \(\displaystyle h\) in the denominator ...) ... and indeed no norm signs around the expression \(\displaystyle \lim_{ h \rightarrow 0 } \eta(h) = 0\) ... nor indeed are there any norm signs in the limit shown in Definition 9.1.6 above (with the exception of around \(\displaystyle h\) in the denominator ...) ...

... BUT ...

... ... this lack of norm signs seems in contrast to the last few lines of the proof of Theorem 9.1.10 as follows ... where we read ...

" ... ... Conversely if (9.6) holds for some \(\displaystyle \eta\) and \(\displaystyle T\), then \(\displaystyle \lim_{ h \rightarrow 0 } \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| } = \lim_{ h \rightarrow 0 } \| \eta(h) \| = 0
\)

... ... "Here, in contrast to the case above, there are norm signs around the numerator and indeed around \(\displaystyle \eta(h)\) ... ...
Can someone please explain why norm signs are used in the numerator and around \(\displaystyle \eta(h)\) in one case ... yet not the other ...
Help will be appreciated ...

Peter

\(\displaystyle f(a + h ) - f(a) - df_a (h)\) is a vector in $\Bbb{R}^m$, and $h$ is a vector in $\Bbb{R}^n$. A vector can be multiplied (or divided) by a scalar, but not by another vector. So in the quotient \(\displaystyle \eta(h) = \frac{ f( a + h ) - f(a) - Th }{ \| h \| }\) it is essential to have norm signs in the denominator. There are no norm signs in the numerator, and so $\eta(h)$ is a vector in $\Bbb{R}^m$.

When it comes to taking the limit as $h\to0$, it is always the case that a vector goes to zero if and only if its norm goes to zero. Therefore the conditions $\eta(h) \to0$ and $\|\eta(h)\| \to0$ are equivalent.

Finally, it follows from one of the axioms for a norm that $$\|\eta(h)\| = \left\|\frac{ f( a + h ) - f(a) - Th }{ \| h \| } \right\| = \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| }.$$

 

[Math Amateur]

\(\displaystyle f(a + h ) - f(a) - df_a (h)\) is a vector in $\Bbb{R}^m$, and $h$ is a vector in $\Bbb{R}^n$. A vector can be multiplied (or divided) by a scalar, but not by another vector. So in the quotient \(\displaystyle \eta(h) = \frac{ f( a + h ) - f(a) - Th }{ \| h \| }\) it is essential to have norm signs in the denominator. There are no norm signs in the numerator, and so $\eta(h)$ is a vector in $\Bbb{R}^m$.

When it comes to taking the limit as $h\to0$, it is always the case that a vector goes to zero if and only if its norm goes to zero. Therefore the conditions $\eta(h) \to0$ and $\|\eta(h)\| \to0$ are equivalent.

Finally, it follows from one of the axioms for a norm that $$\|\eta(h)\| = \left\|\frac{ f( a + h ) - f(a) - Th }{ \| h \| } \right\| = \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| }.$$

Thanks Opalg ... your post was very helpful ...

It was particularly helpful to me to be reminded that ... ... " ... ... When it comes to taking the limit as $h\to0$, it is always the case that a vector goes to zero if and only if its norm goes to zero. ... ... "

But ... given what you have said, I am still a bit perplexed as to why the author bothered to put norm signs around the numerator of ... ...\(\displaystyle \lim_{ h \rightarrow 0 } \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| } = \lim_{ h \rightarrow 0 } \| \eta(h) \| = 0
\)

... given what you said, surely he need not have bothered .... can you comment ...Thank you again for your help ...

Peter

 

[Opalg]

I am still a bit perplexed as to why the author bothered to put norm signs around the numerator of

\(\displaystyle \lim_{ h \rightarrow 0 } \frac{ \| f( a + h ) - f(a) - Th \| }{ \| h \| } = \lim_{ h \rightarrow 0 } \| \eta(h) \| = 0 \)

... given what you said, surely he need not have bothered .

I can't see any need for the norm signs. It makes no difference whether they are there or not.

 

[Math Amateur]

I can't see any need for the norm signs. It makes no difference whether they are there or not.

Thanks Opalg ...

I understand ... but that is a very important point to me ...

THanks again for clarifying the issue ...

Peter