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Techniques in functional analysis


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#1 SFNQuestions

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Posted 24 January 2017 - 01:22 AM

Hello,

As I understand, one typically derives properties of functions that satisfy certain functional equations by substituting x for various values like 0, 1, y, -x, x^2 and so on. But what I am wondering is why algebraically solving for the unknown function doesn't work, or at least what it means when you do. 

 


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#2 wtf

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Posted 24 January 2017 - 01:31 AM

Can you give an example of what you mean? That doesn't sound like functional analysis to me. Functional analysis is basically linear algebra and calculus on infinite dimensional vector spaces. For example the Hilbert space of quantum physics is studied in functional analysis.
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#3 SFNQuestions

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Posted 24 January 2017 - 01:42 AM

Can you give an example of what you mean? That doesn't sound like functional analysis to me. Functional analysis is basically linear algebra and calculus on infinite dimensional vector spaces. For example the Hilbert space of quantum physics is studied in functional analysis.

Sorry, what you described does not seem like a definition of functional analysis to me, that is closer to abstract algebra which functional analysis can be a part of to verify properties of an abstraction.

If you solve for a functional equation algebraically, it seems as though it is the the same as solving for x a constant, but when solving for an unknown function, you would more than just a constant if those solutions exist, so it would be more useful to know of actual functions containing variables that would satisfy these properties, not just constants. 

For instance,

f(x)^{1/2}=f(x)^2

would show

1^{2/3}=f(x)

I suppose a less ideal but relevant question is: how does one know for sure that they've found all conceivable solutions that satisfy a functional equation? For instance, there's the functional equation

f(x)=f(1-x) satisfied by the Riemann Xi function, \xi(x). But, it could also be solved by f(x)=c_1+c_2x^2-c_2x which is much simpler in comparison. However, how would one determine that there isn't an even more complicated function than the Riemann Xi function that would also satisfy this functional equation? How does one know there isn't an unknown complicated function that satisfies f(x)^{1/2}=f(x)^2? How does one know there isn't an undiscovered operator that provides a solution? Solving for it algebraically certainly doesn't help.


Edited by SFNQuestions, 24 January 2017 - 02:20 AM.

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#4 studiot

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Posted 24 January 2017 - 02:34 PM

Hello,

As I understand, one typically derives properties of functions that satisfy certain functional equations by substituting x for various values like 0, 1, y, -x, x^2 and so on. But what I am wondering is why algebraically solving for the unknown function doesn't work, or at least what it means when you do. 

 

 

Perhaps some clarification of terminology is in order.

 

Some authors use the term 'functional analysis' to mean the theory of functions of a (real or complex) variable.

That is the word 'functional' is used as an adjective.

I think this is the meaning intended here.

 

 

Unfortunately the word functional is also used in mathematics as a noun with a particular meaning.

This meaning was introduced by Kantorovich, Banach  and Kerysig

 

A functional is a map from a space of test functions, \Im ,  (ie functions of interest) to its underlying field.

In your case I think the underlying field is the field of Real Numbers, R

 

So a definite integral is such a map from the space of integrable functions to the reals and outputs a real number for each definite integral.

 

Functionals can be non linear or linear.

 

A linear real functional, F, is a map  \Im  \mapsto R such that for any two functions, 

 
\left( {\varphi \,and\,\psi  \in \Im } \right) and scalars (real numbers) \left( {a\,and\,b \in R} \right)
 
 
F\left( {a\varphi \, + b\,\psi } \right) = aF\left( \varphi  \right) + bF\left( \psi  \right)
 
Some texts on the subject are
 
Kantorovich  Functional Analysis
 
Kreysig Introductory Functional Analysis with Applications
 
Griffel Applied functional Analysis
 
Functional Analysis grew out of Dirac's version of the theory of 'generalised functions' called Distribution Theory.

Edited by studiot, 24 January 2017 - 05:55 PM.

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#5 wtf

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Posted 24 January 2017 - 05:50 PM

I think OP is asking about functional equations.

https://en.wikipedia...tional_equation

I'm afraid I don't know anything about the general theory so I can't help the OP. I'm guessing that you'd have to show you've found all solutions by ad hoc methods on a case by case basis.

Edited by wtf, 24 January 2017 - 05:51 PM.

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