MHB Differentialbility & Continuity of Multivariable Vector-Valued Functions .... D&K Lemma 2.2.7 ....

[Math Amateur]

Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

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 an aspect of the proof of Lemma 2.2.7 (Hadamard...) ... ...

Duistermaat and Kolk's Lemma 2.2.7 and its proof read as follows:https://www.physicsforums.com/attachments/7829
View attachment 7830In the above proof we read the following:

" ... ... Or, in other words since $$(x - a)^t y = \langle x - a , y \rangle \in \mathbb{R}$$ for $$y \in \mathbb{R}^n$$,$$\phi_a(x) y = Df(a)y + \frac{ \langle x - a , y \rangle }{ \| x - a \|^2 } \epsilon_a ( x _ a ) \in \mathbb{R}^p \ \ \ \ \ (x \in U \setminus \{ a \} , \ y \in \mathbb{R}^n )$$.

Now indeed we have $$f(x) = f(a) + \phi_a(x) ( x - a )$$. ... ... "
My question is as followsow/why does $$\phi_a(x) y = Df(a)y + \frac{ \langle x - a , y \rangle }{ \| x - a \|^2 } \epsilon_a ( x _ a ) \in \mathbb{R}^p$$

... imply that ...

$$f(x) = f(a) + \phi_a(x) ( x - a )$$ ... ... ... ?
Help will be much appreciated ...

Peter

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NOTE:

The start of D&K's section on differentiable mappings may help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:View attachment 7831
View attachment 7832

The start of D&K's section on linear mappings may also help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:
View attachment 7833
View attachment 7834
View attachment 7835Hope the above helps readers understand the context and notation of the post ...

Peter

 

[S.G. Janssens]

Re: Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

My question is as followsow/why does $$\phi_a(x) y = Df(a)y + \frac{ \langle x - a , y \rangle }{ \| x - a \|^2 } \epsilon_a ( x _ a ) \in \mathbb{R}^p$$

... imply that ...

$$f(x) = f(a) + \phi_a(x) ( x - a )$$ ... ... ... ?

Clearly this is true for $x = a$, so we may assume $x \neq a$ and use the line preceding this statement in the proof.

Note that in that statement, $\phi_a(x)$ is in $L(\mathbb{R}^n,\mathbb{R}^p)$. Act with this on $y = x - a \in \mathbb{R}^n$ according to the right-hand side of the formula to obtain
\[
\phi_a(x)(x - a) = Df(a)(x - a) + \frac{\langle x - a, x - a\rangle}{\|x - a\|^2}\epsilon_a(x - a),
\]
and use that $\langle x - a, x - a\rangle = (x - a)^t(x - a) = \|x - a\|^2$. (The inner product of a vector with itself is the vector's norm squared.)

 

[Math Amateur]

Re: Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

Clearly this is true for $x = a$, so we may assume $x \neq a$ and use the line preceding this statement in the proof.

Note that in that statement, $\phi_a(x)$ is in $L(\mathbb{R}^n,\mathbb{R}^p)$. Act with this on $y = x - a \in \mathbb{R}^n$ according to the right-hand side of the formula to obtain
\[
\phi_a(x)(x - a) = Df(a)(x - a) + \frac{\langle x - a, x - a\rangle}{\|x - a\|^2}\epsilon_a(x - a),
\]
and use that $\langle x - a, x - a\rangle = (x - a)^t(x - a) = \|x - a\|^2$. (The inner product of a vector with itself is the vector's norm squared.)

Thanks Krylov,

... followed your advice and obtained the result ...

Thanks again ...

Peter

 

[Math Amateur]

Re: Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

Clearly this is true for $x = a$, so we may assume $x \neq a$ and use the line preceding this statement in the proof.

Note that in that statement, $\phi_a(x)$ is in $L(\mathbb{R}^n,\mathbb{R}^p)$. Act with this on $y = x - a \in \mathbb{R}^n$ according to the right-hand side of the formula to obtain
\[
\phi_a(x)(x - a) = Df(a)(x - a) + \frac{\langle x - a, x - a\rangle}{\|x - a\|^2}\epsilon_a(x - a),
\]
and use that $\langle x - a, x - a\rangle = (x - a)^t(x - a) = \|x - a\|^2$. (The inner product of a vector with itself is the vector's norm squared.)

Thanks again Krylov ...

BUT ... just a comment on D&K's proof ...

I have to say that D&K's proof of Hadamard's Lemma though valid, is not very intuitive, starting as it does with a fairly complicated definition of \phi_a(x) ... ...

Do you have any comments ... ?Peter

 

[S.G. Janssens]

Re: Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

Thanks again Krylov ...

BUT ... just a comment on D&K's proof ...

I have to say that D&K's proof of Hadamard's Lemma though valid, is not very intuitive, starting as it does with a fairly complicated definition of \phi_a(x) ... ...

Do you have any comments ... ?Peter

Hi Peter,

In the proof of "Hadamard" they refer to (2.9) which occurs in the proof of Prop. 2.2.1. I think that is indeed how the definition of $\phi_a$ in "Hadamard" gets motivated: Go back to Prop. 2.2.1 (which applies to the case $n = p = 1$) and notice how there $\phi_a$ is really obtained by "solving"
\[
f(x) = f(a) + \phi_a(x)(x - a) \qquad (*)
\]
for $\phi_a(x)$ to obtain the difference quotient (when $x \neq a$) or the derivative $f'(a)$ itself (when $x = a$). Then, returning to the general case of "Hadamard", the choice for $\phi_a$ in the proof is a quite natural generalization of the choice made in the proof of Prop. 2.2.1. (And, indeed, we know now that it does satisfy (*) because of the answer in post #2 to your original question in post #1.)

P.S. Sorry it took a few days.

 

[Math Amateur]

Re: Differentialbility & Continuity of Multivariable Vector-Valued Functions ... D&K Lemma 2.2.7 ...

Hi Peter,

In the proof of "Hadamard" they refer to (2.9) which occurs in the proof of Prop. 2.2.1. I think that is indeed how the definition of $\phi_a$ in "Hadamard" gets motivated: Go back to Prop. 2.2.1 (which applies to the case $n = p = 1$) and notice how there $\phi_a$ is really obtained by "solving"
\[
f(x) = f(a) + \phi_a(x)(x - a) \qquad (*)
\]
for $\phi_a(x)$ to obtain the difference quotient (when $x \neq a$) or the derivative $f'(a)$ itself (when $x = a$). Then, returning to the general case of "Hadamard", the choice for $\phi_a$ in the proof is a quite natural generalization of the choice made in the proof of Prop. 2.2.1. (And, indeed, we know now that it does satisfy (*) because of the answer in post #2 to your original question in post #1.)

P.S. Sorry it took a few days.

Thanks for the help, Krylov ... much appreciated!

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

Peter