Differentialbility & Continuity of Multivariable Functions

[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 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:

In 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==========================================================================================

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:

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:

 

[andrewkirk]

My question is as follows:

How/why does $\phi_a(x) y = Df(a)y + \frac{ \langle x - a , y \rangle }{ \| x - a \|^2 } \epsilon_a ( x _ a ) \in \mathbb{R}^p\quad (1)$

... imply that ...

$f(x) = f(a) + \phi_a(x )( x - a )\quad (2)$ ... ... ... ?

Substitute $(x-a)$ for $y$ in (1) and simplify, noting that $\langle v,v\rangle = \|v\|^2$. Then make $Df(a)(x-a)$ the subject of that equation on the LHS. Insert the RHS of that equation into 2.10, replacing $h$ by $(x-a)$, do a bit of simplification and voilà!

 

[Math Amateur]

Substitute $(x-a)$ for $y$ in (1) and simplify, noting that $\langle v,v\rangle = \|v\|^2$. Then make $Df(a)(x-a)$ the subject of that equation on the LHS. Insert the RHS of that equation into 2.10, replacing $h$ by $(x-a)$, do a bit of simplification and voilà!

Thanks Andrew ...

Followed your advice and things fell into place as you indicated!

Thanks again,

Peter

 

[Math Amateur]

Substitute $(x-a)$ for $y$ in (1) and simplify, noting that $\langle v,v\rangle = \|v\|^2$. Then make $Df(a)(x-a)$ the subject of that equation on the LHS. Insert the RHS of that equation into 2.10, replacing $h$ by $(x-a)$, do a bit of simplification and voilà!

Thanks again Andrew ...

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

 

[fresh_42]

Thanks again Andrew ...

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

It's not that much lacking intuition. At $x=a$ we already have the differential as operator valued function: $a \mapsto D_a$. So what's left is a linear function for the entire neighborhood $U$ of $x=a$. This must be done continuously, so we have to approach $D_a$. $(x-a)(x-a)^t$ supplies the required linear function (a matrix), $||x-a||^2$ in the denominator norms it, so that it cannot "run away", and the $\varepsilon_a$ from the definition and precondition of differentialbility guarantees, that the additional term vanishes the closer we get to $x=a$. Thus the lack of intuition is more a lack of practice and experience, which the author probably doesn't lack of.

 

[Math Amateur]

It's not that much lacking intuition. At $x=a$ we already have the differential as operator valued function: $a \mapsto D_a$. So what's left is a linear function for the entire neighborhood $U$ of $x=a$. This must be done continuously, so we have to approach $D_a$. $(x-a)(x-a)^t$ supplies the required linear function (a matrix), $||x-a||^2$ in the denominator norms it, so that it cannot "run away", and the $\varepsilon_a$ from the definition and precondition of differentialbility guarantees, that the additional term vanishes the closer we get to $x=a$. Thus the lack of intuition is more a lack of practice and experience, which the author probably doesn't lack of.

Thanks fresh_42 ...

Now reflecting on what you have written ...

Peter