What Does it Mean for a Function to be Smooth on the Complex Plane?

  • Context: Graduate 
  • Thread starter Thread starter AxiomOfChoice
  • Start date Start date
  • Tags Tags
    Function Smooth
Click For Summary
SUMMARY

A "smooth" function on the complex plane refers to a function that satisfies specific regularity conditions, typically denoted as C^2 or C^{infinity}. This means that the partial derivatives of the real and imaginary components of the function exist and are continuous. In mathematical discussions, "smooth" is often used interchangeably with these classifications, but it does not imply that the function is holomorphic. The distinction between these definitions is crucial for understanding the behavior of functions in complex analysis.

PREREQUISITES
  • Understanding of C^2 and C^{infinity} function classes
  • Familiarity with partial derivatives in multivariable calculus
  • Basic knowledge of complex analysis
  • Concept of holomorphic functions
NEXT STEPS
  • Research the properties of C^2 and C^{infinity} functions
  • Study the implications of mixed partial derivatives in multivariable calculus
  • Explore the definition and properties of holomorphic functions
  • Learn about the applications of smooth functions in complex analysis
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in the properties of smooth functions in mathematical contexts.

AxiomOfChoice
Messages
531
Reaction score
1
What is a "smooth" function?

If someone tells you to consider a "smooth" function on the complex plane, what does that mean, exactly? Does it mean that the partial derivatives of the real and imaginary parts exist and are continuous? Does it mean something else?
 
Physics news on Phys.org


Usually, "let f be a smooth function" is an expression people use to be "let f be a function satisfying certain regularity conditions".

Ex: for a smooth enough map f:R^2-->R, we have equality of the mixed partial derivative. Here, smooth means C^2.

Sometimes also, people equate smooth with C^{infinity}.

I've never seen smooth to mean holomorphic though...
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K