Linearity of the Differential .... Junghenn Theorem 9.2.1 ....

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SUMMARY

The discussion centers on the clarification of Theorem 9.2.1 from Hugo D. Junghenn's "A Course in Real Analysis," specifically its reliance on Theorem 9.1.10. The participants analyze the proof's concluding statement, which indicates that another application of Theorem 9.1.10 is necessary to establish that the differential of the linear combination of functions, represented as ##\alpha \mathbf{f} + \beta \mathbf{g}##, is valid. The conclusion reached is that the second application of Theorem 9.1.10 confirms the differentiability of the function, resolving the initial confusion regarding the proof's completeness.

PREREQUISITES
  • Understanding of real analysis concepts, particularly differentiation in Rn.
  • Familiarity with the definitions and implications of Theorems 9.1.10 and 9.2.1 from Junghenn's book.
  • Knowledge of differential calculus and linear combinations of functions.
  • Ability to interpret mathematical notation and limits in the context of real analysis.
NEXT STEPS
  • Review the implications of Theorem 9.1.10 in detail to understand its application in differentiability.
  • Study the proofs of related theorems in real analysis to gain a deeper understanding of differentiation.
  • Explore examples of linear combinations of differentiable functions to solidify comprehension of their properties.
  • Investigate the concept of limits in the context of differentiability and their role in proofs.
USEFUL FOR

Students of real analysis, mathematicians focusing on differentiation, and educators teaching advanced calculus concepts will benefit from this discussion.

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I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on Rn"

I need some help with an aspect of Theorem 9.2.1 ...

Theorem 9.2.1 reads as follows:
Junghenn - Theorem 9.2.1   ...  ... .png

Theorem 9.2.1 refers to and relies on Theorem 9.1.10 ... ... so I am providing the same ... as follows:

Junghenn - Theorem 9.1.10   ...  ... .png

At the end of the proof of Theorem 9.2.1 we read the following:

" ... ... Then

##( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} + \mathbf{h} ) = ( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} ) + ( \alpha \mathbf{ df_a } + \beta \mathbf{ dg_a } ) ( \mathbf{ h } ) + \| \mathbf{ h } \| ##and##\lim_{ \mathbf{ h } \rightarrow \mathbf{ 0 } } ( \alpha \eta + \beta \mu ) ( \mathbf{ h } ) = \mathbf{ 0 }##Another application of 9.1.10 completes the proof."
I don't understand the comment: "Another application of 9.1.10 completes the proof." ... ... why do we need another application of 9.1.10 ...

Doesn't the line" ##( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} + \mathbf{h} ) = ( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} ) + ( \alpha \mathbf{ df_a } + \beta \mathbf{ dg_a } ) ( \mathbf{ h } ) + \| \mathbf{ h } \| ##and##\lim_{ \mathbf{ h } \rightarrow \mathbf{ 0 } } ( \alpha \eta + \beta \mu ) ( \mathbf{ h } ) = \mathbf{ 0 }## ..."actually complete the proof?
Help will be appreciated ...

Peter*** EDIT ***

I think I see what the author meant ... he is arguing that the statement:" ##( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} + \mathbf{h} ) = ( \alpha \mathbf{f} + \beta \mathbf{g} ) ( \mathbf{a} ) + ( \alpha \mathbf{ df_a } + \beta \mathbf{ dg_a } ) ( \mathbf{ h } ) + \| \mathbf{ h } \| ##and##\lim_{ \mathbf{ h } \rightarrow \mathbf{ 0 } } ( \alpha \eta + \beta \mu ) ( \mathbf{ h } ) = \mathbf{ 0 }##BY 9.1.10 ! (THAT IS APPLYING 9.1.10 AGAIN)

proves that
##\alpha \mathbf{ df_a } + \beta \mathbf{ dg_a }## is the differential of ##\alpha \mathbf{f} + \beta \mathbf{g}## ...Hence we do apply 9.1.10 again ...

Is that right ...

(If it is correct ... then my apologies for the simple confusion ... )
 

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Yes that's correct. In the second application of 9.1.10 he is using the reverse direction of the iff in Theorem 9.1.10, arguing from the existence of the function ##(\alpha\eta+\beta\mu)## to the conclusion that ##(\alpha f+\beta g)## is differentiable.
 
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Thanks Andrew ...

That helps ...

Peter
 

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