If f is infinitely differentiable and analytic on a dense set is f analytic?

  • Context: Graduate 
  • Thread starter Thread starter deluks917
  • Start date Start date
  • Tags Tags
    Differentiable Set
Click For Summary
SUMMARY

If a function f: R->R is infinitely differentiable and analytic on a dense set, it does not guarantee that f is analytic everywhere. The bump function, defined as f(x) = e^{-1/x^2} for x > 0 and f(0) = 0, serves as a counterexample, being analytic on R \ {0} but not on the entire real line. Additionally, smooth functions with compact support, such as those defined on the interval [a,b], are not analytic at the endpoints a and b due to differing limits of their Taylor series from either side.

PREREQUISITES
  • Understanding of infinitely differentiable functions
  • Familiarity with analytic functions and Taylor series
  • Knowledge of dense sets in real analysis
  • Concept of compact support in functions
NEXT STEPS
  • Study the properties of analytic functions in real analysis
  • Explore the concept of dense sets and their implications in function analysis
  • Investigate the characteristics of smooth functions with compact support
  • Learn about counterexamples in mathematical analysis, particularly in the context of differentiability and analyticity
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in the properties of differentiable and analytic functions.

deluks917
Messages
379
Reaction score
4
Let f: R->R. If f is infinitely differentiable and analytic on a dense set is f analytic? Is this true if we restric f to [0,1]?

note: by analytic I mean the radius of convergence of the taylor expansion is non-zero about every point.

Maybe this is simple but I was thinking about it and can't figure it out
 
Physics news on Phys.org
Depending on what you want, the bump function

[tex]\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{cc} e^{-1/x^2} & \text{if}~x\geq 0\\ 0 & \text{if}~x=0\end{array}\right.[/tex]

is a counterexample. It is analytic on [itex]\mathbb{R}\setminus \{0\}[/itex] but it is not analytic on entire [itex]\mathbb{R}[/itex].
 
Right, any smooth function of compact support , say in [a,b], is not analytic at either of the endpoints a,b , because the series must approach different values from the left and right.
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
23
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 13 ·
Replies
13
Views
10K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K