MHB How does Lemma 2.2.3 Demonstrate the Uniqueness of the Derivative?

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 2: Differentiation ... ...

I need help with the proof of Lemma 2.2.3 ... ...

Duistermaat and Kolk's Lemma 2.2.3 and its proof read as follows:View attachment 7796
I do not understand the strategy (or overall idea) of this proof ... how does considering $$a + th \in U$$ lead to demonstrating that $$\text{Df}(a)$$ is uniquely determined ...

and

how do we get

$$\text{Df}(a) h = \frac{1}{t} ( f ( a + th ) - f(a) ) - \frac{1}{t} \epsilon_a (th)$$ ... ... ... ... ... (1)follow from (2.10) ... ... and then, how does (1) lead to ...$$\text{Df}(a) h = \lim_{ t \rightarrow 0 } ( f ( a + th ) - f(a) )$$and, indeed, how does the above show that $$\text{Df}(a)$$ is uniquely determined ... .. ?
Hope someone can help ... ...

Peter==========================================================================================The above post mentions (2.10) which is in the notes following Definition 2.2.2 ... so I am providing Definition 2.2.2 and the accompanying notes ... ... as follows ... ...
View attachment 7797
https://www.physicsforums.com/attachments/7798

 

[S.G. Janssens]

I do not understand the strategy (or overall idea) of this proof ... how does considering $$a + th \in U$$ lead to demonstrating that $$\text{Df}(a)$$ is uniquely determined ...

If we show that for any $h \in \mathbb{R}^n$ (i.e. any element in the domain of the linear mapping $Df(a)$) the expression $Df(a)h$ depends only on $f$ (and of course on the points $a$ and $h$), then $Df(a)$ is uniquely determined by $f$. For, any other derivative $L$ of $f$ at $a$ is then such that
\[
Lh = Df(a)h
\]
for all $h \in \mathbb{R}^n$.

and

how do we get

$$\text{Df}(a) h = \frac{1}{t} ( f ( a + th ) - f(a) ) - \frac{1}{t} \epsilon_a (th)$$ ... ... ... ... ... (1)follow from (2.10) ... ...

This is a matter of taking the first equality in (2.10) with $th$ in place of $h$.

and then, how does (1) lead to ...$$\text{Df}(a) h = \lim_{ t \rightarrow 0 } ( f ( a + th ) - f(a) )$$

Note that
\[
\left\|\frac{1}{t}\epsilon_a(th)\right\| = \|h\| \cdot \frac{\|\epsilon_a(th)\|}{\|t h\|},
\]
and now take the limit $t \to 0$ (keeping $h$ fixed!) and use the second equality in (2.10).

and, indeed, how does the above show that $$\text{Df}(a)$$ is uniquely determined ... .. ?

See above.

 

[Math Amateur]

If we show that for any $h \in \mathbb{R}^n$ (i.e. any element in the domain of the linear mapping $Df(a)$) the expression $Df(a)h$ depends only on $f$ (and of course on the points $a$ and $h$), then $Df(a)$ is uniquely determined by $f$. For, any other derivative $L$ of $f$ at $a$ is then such that
\[
Lh = Df(a)h
\]
for all $h \in \mathbb{R}^n$.
This is a matter of taking the first equality in (2.10) with $th$ in place of $h$.
Note that
\[
\left\|\frac{1}{t}\epsilon_a(th)\right\| = \|h\| \cdot \frac{\|\epsilon_a(th)\|}{\|t h\|},
\]
and now take the limit $t \to 0$ (keeping $h$ fixed!) and use the second equality in (2.10).
See above.

Thanks for the help, Krylov ...

Working through your post line by line and reflecting ...

... just a small clarification ... at the start of your reply, you write:

" ... ... If we show that for any $h \in \mathbb{R}^n$ (i.e. any element in the domain of the linear mapping $Df(a)$) the expression $Df(a)h$ depends only on $f$ (and of course on the points $a$ and $h$), then $Df(a)$ is uniquely determined by $f$. For, any other derivative $L$ of $f$ at $a$ is then such that
\[
Lh = Df(a)h
\]
for all $h \in \mathbb{R}^n$. ... ... "
If we wish to show that the Lemma holds for any $$a + h \in U$$ why don't we begin the proof with a statement like the following:

"Consider any $$h \in \mathbb{R}$$ such that $$a + h \in U$$ ... ... "

then surely what we prove will hold for any/every $$h$$ such that $$a + h \in U$$ ... so why do we need to bring $$t$$ into the proof ... ...

Can you help further ... ... ?

Peter

 

[Math Amateur]

If we show that for any $h \in \mathbb{R}^n$ (i.e. any element in the domain of the linear mapping $Df(a)$) the expression $Df(a)h$ depends only on $f$ (and of course on the points $a$ and $h$), then $Df(a)$ is uniquely determined by $f$. For, any other derivative $L$ of $f$ at $a$ is then such that
\[
Lh = Df(a)h
\]
for all $h \in \mathbb{R}^n$.
This is a matter of taking the first equality in (2.10) with $th$ in place of $h$.
Note that
\[
\left\|\frac{1}{t}\epsilon_a(th)\right\| = \|h\| \cdot \frac{\|\epsilon_a(th)\|}{\|t h\|},
\]
and now take the limit $t \to 0$ (keeping $h$ fixed!) and use the second equality in (2.10).
See above.

Hi Krylov ... thanks again for the help ...

Just another clarification ...

You write:

" ... ... Note that
\[
\left\|\frac{1}{t}\epsilon_a(th)\right\| = \|h\| \cdot \frac{\|\epsilon_a(th)\|}{\|t h\|},
\]
and now take the limit $t \to 0$ (keeping $h$ fixed!) and use the second equality in (2.10)."The above expression is expressed in norms ... but neither

$$\text{Df}(a) h = \frac{1}{t} ( f ( a + th ) - f(a) ) - \frac{1}{t} \epsilon_a (th)$$ ... ... ... ... ... (1)

nor

$$\text{Df}(a) h = \lim_{ t \rightarrow 0 } ( f ( a + th ) - f(a) )$$

have norm signs in them ...Can you explain what is going on with the taking of norms in your explanation ...Hope that my question is clear ...

Peter

 

[S.G. Janssens]

If we wish to show that the Lemma holds for any $$a + h \in U$$ why don't we begin the proof with a statement like the following:

"Consider any $$h \in \mathbb{R}$$ such that $$a + h \in U$$ ... ... "

then surely what we prove will hold for any/every $$h$$ such that $$a + h \in U$$ ... so why do we need to bring $$t$$ into the proof ... ...

The purpose of the proof is to derive an expression for $Df(a)h$ where $h$ is arbitrary but fixed. This expression should depend only on $f$, $a$ and $h$. Namely, we will then have established that the derivative of $f$ at $a$ (if it exists) acting on an arbitrary element of its domain is uniquely determined.

However, the proof works by taking a limit. So in order to keep the direction $h$ arbitrary but fixed while at the same time being able to take the limit, we introduce the parameter $t$. Then we show that
\[
\frac{1}{t}{\left(f(a + th) - f(a)\right)} \to Df(a)(h),
\]
as $t \to 0$. Without $t$ in the game, we would have to somehow consider the limit $h \to 0$, which would spoil the purpose of keeping $h$ fixed.

You write:

" ... ... Note that
\[
\left\|\frac{1}{t}\epsilon_a(th)\right\| = \|h\| \cdot \frac{\|\epsilon_a(th)\|}{\|t h\|},
\]
and now take the limit $t \to 0$ (keeping $h$ fixed!) and use the second equality in (2.10)."The above expression is expressed in norms ... but neither

$$\text{Df}(a) h = \frac{1}{t} ( f ( a + th ) - f(a) ) - \frac{1}{t} \epsilon_a (th)$$ ... ... ... ... ... (1)

nor

$$\text{Df}(a) h = \lim_{ t \rightarrow 0 } ( f ( a + th ) - f(a) )$$

have norm signs in them ...Can you explain what is going on with the taking of norms in your explanation ...

Sure. I wanted to argue that in the limit $t \to 0$, the quantity $\frac{1}{t}\epsilon_a(th)$ that appears in the displayed equation in the proof actually goes to zero, because that is what the proof uses in its penultimate line.

Now, the above quantity goes to zero precisely when $\left\|\frac{1}{t}\epsilon_a(th)\right\| \to 0$. The latter expression however is easier to deal with, because we can divide by $\|h\|$ but not by $h$ itself. (In turn, the purpose of this division-and-multiplication manipulation is to recover the second equality in (2.10) in your text, but with $\hat{h} := th$ in place of $h$. Note that for fixed $h \in \mathbb{R}^n$ we have that $\hat{h} \to 0$ if $t \to 0$, so it would follow that $\frac{\|\epsilon_a(\hat{h})\|}{\|\hat{h}\|} \to 0$)

 

[Math Amateur]

The purpose of the proof is to derive an expression for $Df(a)h$ where $h$ is arbitrary but fixed. This expression should depend only on $f$, $a$ and $h$. Namely, we will then have established that the derivative of $f$ at $a$ (if it exists) acting on an arbitrary element of its domain is uniquely determined.

However, the proof works by taking a limit. So in order to keep the direction $h$ arbitrary but fixed while at the same time being able to take the limit, we introduce the parameter $t$. Then we show that
\[
\frac{1}{t}{\left(f(a + th) - f(a)\right)} \to Df(a)(h),
\]
as $t \to 0$. Without $t$ in the game, we would have to somehow consider the limit $h \to 0$, which would spoil the purpose of keeping $h$ fixed.
Sure. I wanted to argue that in the limit $t \to 0$, the quantity $\frac{1}{t}\epsilon_a(th)$ that appears in the displayed equation in the proof actually goes to zero, because that is what the proof uses in its penultimate line.

Now, the above quantity goes to zero precisely when $\left\|\frac{1}{t}\epsilon_a(th)\right\| \to 0$. The latter expression however is easier to deal with, because we can divide by $\|h\|$ but not by $h$ itself. (In turn, the purpose of this division-and-multiplication manipulation is to recover the second equality in (2.10) in your text, but with $\hat{h} := th$ in place of $h$. Note that for fixed $h \in \mathbb{R}^n$ we have that $\hat{h} \to 0$ if $t \to 0$, so it would follow that $\frac{\|\epsilon_a(\hat{h})\|}{\|\hat{h}\|} \to 0$)

Thanks so much for the help, Krylov ...

Just now reflecting on what you have written ...

Thanks again,

Peter