Differentiating Vector-Valued Function: Junghenn Prop 9.1.2 - Peter seeks help

[Math Amateur]

Derivative of a Vector-Valued Function of a Real Variable - Junghenn Propn 9.1.2 ...

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

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

I need some help with the proof of Proposition 9.1.2 ...

Proposition 9.1.2 and the preceding relevant Definition 9.1.1 read as follows:
https://www.physicsforums.com/attachments/7861
View attachment 7862In the above text from Junghenn we read the following:

" ... ... The assertions follow directly from the inequalities

\(\displaystyle \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \le \left\| \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right\|^2 \)

\(\displaystyle \le \sum_{ i = 1 }^m \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \) ...

... ... "
Can someone please show why the above inequalities hold true ... and further how they lead to the proof of Proposition 9.1.2 ... ...Help will be much appreciated ...

Peter

 

[Math Amateur]

Re: Derivative of a Vector-Valued Function of a Real Variable - Junghenn Propn 9.1.2 ...

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

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

I need some help with the proof of Proposition 9.1.2 ...

Proposition 9.1.2 and the preceding relevant Definition 9.1.1 read as follows:

In the above text from Junghenn we read the following:

" ... ... The assertions follow directly from the inequalities

\(\displaystyle \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \le \left\| \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right\|^2 \)

\(\displaystyle \le \sum_{ i = 1 }^m \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \) ...

... ... "
Can someone please show why the above inequalities hold true ... and further how they lead to the proof of Proposition 9.1.2 ... ...Help will be much appreciated ...

Peter

After some reflection I think that the following is a proof that the above inequalities hold true ...Put \(\displaystyle x_j = \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j\) and \(\displaystyle x = \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \)then we have to show \(\displaystyle |x_j|^2 \leq ||x||^2 \leq \sum_j |x_j|^2\) for any vector \(\displaystyle x=(x_1,\ldots,x_n) \in \mathbb{R}^n\).... well ... we have ...\(\displaystyle \| x \|^2 = ( x_1^2 + x_2^2 + \ ... \ ... \ + x_m^2 ) \)... so clearly ...\(\displaystyle \mid x_j \mid^2 \le \| x \|^2\) ... ... ... ... ... (1)Now ...

\(\displaystyle
\sum_j \mid x_j \mid^2 \ = \ \mid x_1 \mid^2 + \mid x_2 \mid^2 + \ ... \ ... \ + \mid x_m \mid^2 \ = \ x_1^2 + x_2^2 + \ ... \ ... \ + x_m^2 \)hence \(\displaystyle \sum_j \mid x_j \mid^2 = \| x \|^2 \) ... ... ... ... ... (2)

and hence (2) implies\(\displaystyle \| x \|^2 \le \sum_j \mid x_j \mid^2\) ... ... ... ... ... (3)

So ... (1) and (3) imply that \(\displaystyle \mid x_j \mid^2 \le \| x \|^2 \le \sum_j \mid x_j \mid^2 \) ... ...
Is that correct?
But even if the above is correct I still have a problem ... and that is how the above inequalities actually prove Proposition 9.1.2 ...

... indeed ... further to the proof of the proposition ... I am somewhat perplexed at expression like \(\displaystyle \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2\) ... particularly the \(\displaystyle x_j\) ... what is \(\displaystyle x_j\) and how do we interpret \(\displaystyle \left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \) in terms of a derivative ...?Note that we must prove this proposition from the definition of differentiation of functions from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) and the derivative of a vector valued function of a real variable ... ... since the only definition previous to the definition above of the derivative of a vector valued function of a real variable ... is the definition of differentiation of functions from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) ... ...Hope someone can help ...Peter

 

[math771]

Hello Peter,

I would be happy to help you with the proof of Proposition 9.1.2. Let's start by breaking down the inequalities mentioned in the text.

The first inequality,

\left\vert \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\vert^2 \le \left\| \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right\|^2

can be rewritten as

\left( \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right)^2 \le \left( \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right)^2

This is simply the Cauchy-Schwarz inequality for vectors, which states that for any two vectors u and v,

|u \cdot v| \leq \|u\| \|v\|

Applying this to our case, we have

\left| \left( \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right) \cdot \left( \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right) \right| \leq \left\| \frac{f_j ( a + h ) - f_j (a)}{ h } - x_j \right\| \left\| \frac{ f( a + h ) - f(a) }{ h } - ( x_1, \ ... \ ... \ , x_m ) \right\|

Simplifying this, we get

\left| \frac{1}{h} \sum_{i=1}^m (f_i (a+h) - f_i(a)) (x_i - f_i(a)) \right| \leq \left\| \frac{f(a+h) - f(a)}{h} \right\| \left\| x - f(a) \right\|

Now, using the definition of the norm,