MHB Differentiability of Multivariable Vector-Valued Functions .... ....

[Math Amateur]

In Theodore Shifrin's book: Multivariable Mathematics, he defines the derivative of a multivariable vector-valued function as follows:View attachment 8503
Lafontaine in his book: An Introduction to Differential Manifolds, defines the derivative of a multivariable vector-valued function slightly differently as follows:View attachment 8504
Although intuitively the definitions look similar, I need help to show (formally and rigorously) that Shifrin's definition implies Lafontaine's definition and vice versa ...

Can someone please (formally and rigorously) demonstrate that the two definitions are equivalent ...
Hope that someone can help ...

Peter

 

[Opalg]

Lafontaine explains that his usage of the notation $o(h)$ means that $\lim_{h\to0}\dfrac{\|o(h)\|_1}{\|h\|_2} = 0$. If you apply that to his Definition 1.1, he is saying that the condition for differentiability is \[\lim_{h\to0}\frac{\|f(a+h) - f(a) - L\cdot h\|}{\|h\|} = 0.\] But it is true in any normed space that the conditions $\lim_{x\to0}F(x) = 0$ and $\lim_{x\to0}\|F(x)\| = 0$ are equivalent. So Lafontaine's definition is equivalent to \[\lim_{h\to0}\frac{f(a+h) - f(a) - L\cdot h}{\|h\|} = 0.\] That is exactly the definition used by Shifrin, except for the difference in notation where Lafontaine calls the derivative $L$ whereas Shifrin uses the notation $Df(a)$.

 

[Math Amateur]

Lafontaine explains that his usage of the notation $o(h)$ means that $\lim_{h\to0}\dfrac{\|o(h)\|_1}{\|h\|_2} = 0$. If you apply that to his Definition 1.1, he is saying that the condition for differentiability is \[\lim_{h\to0}\frac{\|f(a+h) - f(a) - L\cdot h\|}{\|h\|} = 0.\] But it is true in any normed space that the conditions $\lim_{x\to0}F(x) = 0$ and $\lim_{x\to0}\|F(x)\| = 0$ are equivalent. So Lafontaine's definition is equivalent to \[\lim_{h\to0}\frac{f(a+h) - f(a) - L\cdot h}{\|h\|} = 0.\] That is exactly the definition used by Shifrin, except for the difference in notation where Lafontaine calls the derivative $L$ whereas Shifrin uses the notation $Df(a)$.

Hi Opalg,

Thanks for the help ... but just a minor point ...

I can see that if we treat $$o(h)$$ as an algebraic element or variable, then we can manipulate Lafontaine's equation as follows

$$f(a+h) = f(a) + L \cdot h+ o(h)$$$$\Longrightarrow o(h) = f(a+h) - f(a) - L \cdot h $$$$\Longrightarrow \frac{o(h) }{ \| h \| } = \frac{ f(a+h) - f(a) - L \cdot h }{ \| h \| }$$Therefore ... $$ \lim_{ h \to 0 } \frac{ \| o(h) \| }{ \| h \| } = 0 $$ ... ...... implies that ... $$\lim_{ h \to 0 } \frac{ \| f(a+h) - f(a) - L \cdot h \| }{ \| h \| } = 0$$
... and ... as we can see ... $$\lim_{ h \to 0 } \frac{ \| f(a+h) - f(a) - L \cdot h \| }{ \| h \| } = 0$$ ...

... is essentially Shifrin's definition ... ... BUT ...... how do we justify treating o(h) as an algebraic variable in the above manipulations ...
Hope someone can help ...

Peter

 

[Opalg]

... how do we justify treating o(h) as an algebraic variable in the above manipulations ...

Your comment illustrates why many people dislike the use of the $o(h)$ notation. It is sometimes hard to reconcile its use with the "formal and rigorous" approach to mathematics that you favour.

In essence, $X = o(h)$ means $\lim_{h\to0}X = 0$. That usage is then broadened so that (for example) $X = Y + o(h)$ is interpreted to mean $X-Y = o(h)$.

 

[Math Amateur]

Your comment illustrates why many people dislike the use of the $o(h)$ notation. It is sometimes hard to reconcile its use with the "formal and rigorous" approach to mathematics that you favour.

In essence, $X = o(h)$ means $\lim_{h\to0}X = 0$. That usage is then broadened so that (for example) $X = Y + o(h)$ is interpreted to mean $X-Y = o(h)$.

Thanks Opalg ...

Appreciate your help...

Peter