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

• MHB
• Math Amateur
In summary: It seems that you're saying that if $\epsilon_a(th)$ is small, then $\left\|\frac{1}{t}\epsilon_a(th)\right\| \ll \left\|h\|$. Is that what you're saying?
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MHB
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 $$\displaystyle a + th \in U$$ lead to demonstrating that $$\displaystyle \text{Df}(a)$$ is uniquely determined ...

and

how do we get

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

Peter said:
I do not understand the strategy (or overall idea) of this proof ... how does considering $$\displaystyle a + th \in U$$ lead to demonstrating that $$\displaystyle \text{Df}(a)$$ 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 $h \in \mathbb{R}^n$.

Peter said:
and

how do we get

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

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

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

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).

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

See above.

Krylov said:
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 $$\displaystyle a + h \in U$$ why don't we begin the proof with a statement like the following:

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

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

Can you help further ... ... ?

Peter

Krylov said:
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

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

nor

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

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

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

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

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

Peter said:
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

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

nor

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

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 $\frac{\|\epsilon_a(\hat{h})\|}{\|\hat{h}\|} \to 0$)

Krylov said:
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

## What is the definition of differentiability for mappings from R^n to R^p?

The definition of differentiability for mappings from R^n to R^p is that a function f: R^n -> R^p is differentiable at a point x if there exists a linear transformation L: R^n -> R^p such that the following limit exists: lim0 (f(x+h) - f(x) - L(h)) / ||h|| = 0. This means that the function has a unique tangent plane at that point.

## What is the significance of the D&K Lemma 2.2.3 in differentiability of mappings?

The D&K Lemma 2.2.3 states that if a mapping f: R^n -> R^p is differentiable at a point x, then all directional derivatives of f at x exist and are given by the directional derivative formula: D_vf(x) = L(v), where L is the linear transformation from the definition of differentiability. This lemma is important because it establishes a relationship between differentiability and directional derivatives, which are crucial in understanding the behavior of a function at a specific point.

## Can a mapping be differentiable at a point but not continuous?

Yes, it is possible for a mapping to be differentiable at a point but not continuous. This can occur when the limit in the definition of differentiability exists, but the function is not continuous at that point. In this case, the function may have a jump or a discontinuity at that point, but still satisfy the criteria for differentiability.

## What is the difference between partial differentiability and differentiability?

The main difference between partial differentiability and differentiability is that partial differentiability refers to the ability of a function to have partial derivatives with respect to each variable, while differentiability refers to the existence of all directional derivatives at a point. In other words, a function can be differentiable without being partially differentiable, but it cannot be partially differentiable without being differentiable.

## How is differentiability related to the continuity of a mapping?

Differentiability is closely related to the continuity of a mapping. If a function is differentiable at a point, it must also be continuous at that point. However, a function can be continuous at a point without being differentiable at that point. In general, differentiability is a stronger condition than continuity, as it requires the existence of all directional derivatives.

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