A function that doesn't have a max/min in general, but does when restricted

  • Context: Graduate 
  • Thread starter Thread starter wisvuze
  • Start date Start date
  • Tags Tags
    Function General
Click For Summary
SUMMARY

The discussion centers on constructing a function f: R² → R that is everywhere differentiable but lacks a local minimum at the origin (0,0), while exhibiting a strict minimum along any line through the origin. A proposed function is f(x,y) = -e^(1/x²) if y = x² and 0 otherwise, although it is noted that this function is not continuous everywhere. The participant suggests that smoothing the function could lead to a viable solution, indicating a need for further refinement to achieve the desired properties.

PREREQUISITES
  • Understanding of differentiable functions in multivariable calculus
  • Familiarity with the concept of local minima and maxima
  • Knowledge of piecewise functions and their properties
  • Basic principles of continuity and smoothing techniques in mathematical functions
NEXT STEPS
  • Research techniques for constructing piecewise functions with specific differentiability properties
  • Explore the concept of bump functions and their applications in smoothing functions
  • Study the implications of local minima in multivariable calculus
  • Investigate examples of functions that exhibit different behaviors in restricted domains
USEFUL FOR

Mathematicians, calculus students, and anyone interested in advanced function properties and optimization in multivariable contexts.

wisvuze
Messages
372
Reaction score
1
Hello, I'm trying to construct a function f from R^2 -> R where f is everywhere differentiable, doesn't have a local min at 0 but if I restrict the domain of f to any line through the origin, 0 will be a (strict) min at that point..
I've been fooling around with this, but I don't really have any good strategies for figuring out how to make this function. Any tips? thank you
 
Physics news on Phys.org
Hmmm...

Intuitively, I think something like this could work:

[tex]f(x,y)=\left\{\begin{array}{cc} -e^{1/x^2} &\text{if}~y=x^2\\ 0 &\text{otherwise}\end{array}\right.[/tex]

This isn't continuous everywhere. But I do think that it is differentiable in 0 (because the function [itex]e^{1/x^2}[/itex] is part of a bump function.

Now I guess we should just smooth the function up. I have a feeling it might work.

I'll think of a better example.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K