Undergrad Derivative of a Vector-Valued Function of a Real Variable ...

[Math Amateur]

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

$\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$$\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

 

[fresh_42]

Let $\mathcal{E}$ note the expression in the norm, resp. absolute value (because of laziness, not because it has a certain meaning). If you have all the $f_j$ differentiable, then $|\mathcal{E}(f_j)|^2 < \frac{1}{n}\varepsilon$ in a small neighborhood of $a$ which gives you the estimation for $||\mathcal{E}(f)||^2$ and vice versa.

Now the first question is, why is $|x_j|^2 \leq ||x||^2 \leq \sum_j |x_j|^2$ for any vector $x=(x_1,\ldots,x_n) \in \mathbb{R}^n$.
Can you prove this?

 

[Math Amateur]

Let $\mathcal{E}$ note the expression in the norm, resp. absolute value (because of laziness, not because it has a certain meaning). If you have all the $f_j$ differentiable, then $|\mathcal{E}(f_j)|^2 < \frac{1}{n}\varepsilon$ in a small neighborhood of $a$ which gives you the estimation for $||\mathcal{E}(f)||^2$ and vice versa.

Now the first question is, why is $|x_j|^2 \leq ||x||^2 \leq \sum_j |x_j|^2$ for any vector $x=(x_1,\ldots,x_n) \in \mathbb{R}^n$.
Can you prove this?

Thanks fresh_42 ...

To prove $|x_j|^2 \leq ||x||^2 \leq \sum_j |x_j|^2$ for any vector $x=(x_1,\ldots,x_n) \in \mathbb{R}^n$.

... well ... we have ...

$\| x \|^2 = ( x_1^2 + x_2^2 + \ ... \ ... \ + x_m^2 )$

... so clearly ...

$\mid x_j \mid^2 \le \| x \|^2$ ... ... ... ... ... (1)

Now ...

$\sum_j \mid x_j \mid^2 \ = \ \mid x_1 \mid^2 + \mid x_2 \mid^2 + \ ... \ ... \ + \mid x_m \mid^2 \ = \ x_1^2 + x_2^2 + \ ... \ ... \ + x_m^2$

hence

$\sum_j \mid x_j \mid^2 = \| x \|^2$ ... ... ... ... ... (2)and hence (2) implies

$\| x \|^2 \le \sum_j \mid x_j \mid^2$ ... ... ... ... ... (3)So ... (1) and (3) imply that $\mid x_j \mid^2 \le \| x \|^2 \le \sum_j \mid x_j \mid^2$ ...
Is that correct?

Peter

 

[fresh_42]

Yes, that's correct, and the other claim is even easier, as we only have to change the order of reasoning. In one direction with the first inequality and in the other with the second, depending on what is given. We only have to use the fact, that the dimension $n$ is a constant, so it doesn't really affect the $\varepsilon$.

 

[Math Amateur]

Yes, that's correct, and the other claim is even easier, as we only have to change the order of reasoning. In one direction with the first inequality and in the other with the second, depending on what is given. We only have to use the fact, that the dimension $n$ is a constant, so it doesn't really affect the $\varepsilon$.

Thanks for all your help on this matter fresh_42 ...

But I am still struggling to relate what you are saying to differentiation of functions from \mathbb{R} to \mathbb{R} ... that is when yu write:

" ... ... If you have all the $f_j$ differentiable, then $|\mathcal{E}(f_j)|^2 < \frac{1}{n}\varepsilon$ in a small neighborhood of $a$ which gives you the estimation for $||\mathcal{E}(f)||^2$ and vice versa. ... ..."

... how does this arise out of the definition of differentiation of functions from $\mathbb{R}$ to $\mathbb{R}$ ... ? This seems to me to be important since the only definition/discussion of differentiation in Junghenn before the above case of a the derivative of a vector-valued function of a real variable is the case of functions from $\mathbb{R}$ to $\mathbb{R}$ ... ...

Junghenn's introduction to the differentiation of functions from $\mathbb{R}$ to $\mathbb{R}$ reads as follows:

Peter

 

[fresh_42]

Differentiability of $g(x)$ at a point $x=a$ according to definition 9.1.1 means
$$
\lim_{h \to 0} \dfrac{g(a+h)-g(a)}{h} = g\,'(a)
$$
which means that for all $\varepsilon > 0$ we can find a $\delta(\varepsilon) > 0$ such that - as soon as $|h|< \delta(\varepsilon)$ - we have $||\dfrac{g(a+h)-g(a)}{h} - g\,'(a)|| < \varepsilon$. Now we can set likewise $g=f_j$ or $g=f$.
If the $f_j$ are differentiable, we find for every $\varepsilon > 0$ a $\delta_j(\varepsilon) > 0$ with $||\dfrac{f_j(a+h)-f_j(a)}{h} - f_j\,'(a)|| < \varepsilon$ and $||\dfrac{f(a+h)-f(a)}{h} - f\,'(a)||^2 \leq \sum_j |\dfrac{f_j(a+h)-f_j(a)}{h} - f_j\,'(a)|^2 < \sum_j \varepsilon^2 = n\cdot \varepsilon^2 = \varepsilon'$ for all $|h| < \delta := \min\{\,\delta_j(n \cdot \varepsilon^2) \,\}$. So for every $\varepsilon'$ there is a $\delta = \delta(\varepsilon' )$ with the required property. The other direction, if we start with the differentiability of $f$ is according.

 

[Math Amateur]

Differentiability of $g(x)$ at a point $x=a$ according to definition 9.1.1 means
$$
\lim_{h \to 0} \dfrac{g(a+h)-g(a)}{h} = g\,'(a)
$$
which means that for all $\varepsilon > 0$ we can find a $\delta(\varepsilon) > 0$ such that - as soon as $|h|< \delta(\varepsilon)$ - we have $||\dfrac{g(a+h)-g(a)}{h} - g\,'(a)|| < \varepsilon$. Now we can set likewise $g=f_j$ or $g=f$.
If the $f_j$ are differentiable, we find for every $\varepsilon > 0$ a $\delta_j(\varepsilon) > 0$ with $||\dfrac{f_j(a+h)-f_j(a)}{h} - f_j\,'(a)|| < \varepsilon$ and $||\dfrac{f(a+h)-f(a)}{h} - f\,'(a)||^2 \leq \sum_j |\dfrac{f_j(a+h)-f_j(a)}{h} - f_j\,'(a)|^2 < \sum_j \varepsilon^2 = n\cdot \varepsilon^2 = \varepsilon'$ for all $|h| < \delta := \min\{\,\delta_j(n \cdot \varepsilon^2) \,\}$. So for every $\varepsilon'$ there is a $\delta = \delta(\varepsilon' )$ with the required property. The other direction, if we start with the differentiability of $f$ is according.

Hi fresh_42,

Just now reflecting on yur above reply ...

... BUT ... just a minor clarification ...

In the above post, you are dealing with expressions like $\left\| \dfrac{f_j(a+h)-f_j(a)}{h} - f_j\,'(a) \right\|$ while Junghenn's inequalities deal with expressions like $\left\| \dfrac{f_j(a+h)-f_j(a)}{h} - x_j \right\|$ ... ...

Can you explain this apparent difference...

I find the difference quite perplexing ... how do they mean the same ...?

Peter

 

[fresh_42]

I find the difference quite perplexing ... how do they mean the same ...?

I have chosen the derivatives to concentrate on the argument. Junghenn's notation is a bit better. The point is, that at the start, we do not know whether all derivatives exist, so he has chosen $x_j,x$ as placeholders. The arguments formally go:
$f_j$ differentiable $\Rightarrow |\ldots - f_j'(a)| < \varepsilon_j \Rightarrow ||\ldots - (f_1'(a),\ldots ,f_n'(a)) || < \varepsilon$ with the according choices for the $\varepsilon_j,\varepsilon$ and deltas and so on. Now finally, by the uniqueness of the derivative, we get that $f\,'(a) = (f_1'(a),\ldots ,f_n'(a))$. Conversely, we have $||\ldots - f\,'(a)|| < \varepsilon \Rightarrow |\ldots - (f\,'(a))_j| < \varepsilon_j$ again with the according choices, and uniqueness again gives us $f_j'(a) = (f\,'(a))_j \,.$ By writing $x_j , x$ instead, he just saved all these details and could write the two inequalities in one line. Otherwise, you might have objected: "But we don't know the existence of the derivative yet! We want to prove it!" The dummy variables for the derivatives is a short notation, and mine with the derivatives, was rigorously wrong, as I assumed their existence from the start, in order to show how the inequalities work. But in detail it is a combination of the inequalities and the uniqueness argument.

 

[Math Amateur]

I have chosen the derivatives to concentrate on the argument. Junghenn's notation is a bit better. The point is, that at the start, we do not know whether all derivatives exist, so he has chosen $x_j,x$ as placeholders. The arguments formally go:
$f_j$ differentiable $\Rightarrow |\ldots - f_j'(a)| < \varepsilon_j \Rightarrow ||\ldots - (f_1'(a),\ldots ,f_n'(a)) || < \varepsilon$ with the according choices for the $\varepsilon_j,\varepsilon$ and deltas and so on. Now finally, by the uniqueness of the derivative, we get that $f\,'(a) = (f_1'(a),\ldots ,f_n'(a))$. Conversely, we have $||\ldots - f\,'(a)|| < \varepsilon \Rightarrow |\ldots - (f\,'(a))_j| < \varepsilon_j$ again with the according choices, and uniqueness again gives us $f_j'(a) = (f\,'(a))_j \,.$ By writing $x_j , x$ instead, he just saved all these details and could write the two inequalities in one line. Otherwise, you might have objected: "But we don't know the existence of the derivative yet! We want to prove it!" The dummy variables for the derivatives is a short notation, and mine with the derivatives, was rigorously wrong, as I assumed their existence from the start, in order to show how the inequalities work. But in detail it is a combination of the inequalities and the uniqueness argument.

Thanks fresh_42 ...

... just reflecting on what you have written ...

Thanks for all your help ... it is much appreciated...

Peter