Results 1 to 3 of 3

Math Help - Fréchet derivative and Gâteaux derivative

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    154

    Fréchet derivative and Gâteaux derivative

    Lets say we have the scalar field  f(x,y) = \begin{cases} \frac{xy}{\sqrt{x^2+y^2}} \ \text{if} \ (x,y) \neq (0,0) \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ (x,y) = (0,0) \end{cases}

    So by inspection, it seems that  f(x,y) is continuous at  (0,0) . Now the partial derivatives of  f are not continuous at  (0,0) . So  f is not differentiable at  (0,0) . So this is not Fréchet differentiable? But could it be Gâteaux differentiable?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by heathrowjohnny View Post
    Lets say we have the scalar field  f(x,y) = \begin{cases} \frac{xy}{\sqrt{x^2+y^2}} \ \text{if} \ (x,y) \neq (0,0) \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ (x,y) = (0,0) \end{cases}

    So by inspection, it seems that  f(x,y) is continuous at  (0,0) . Now the partial derivatives of  f are not continuous at  (0,0) . So  f is not differentiable at  (0,0) . So this is not Fréchet differentiable? But could it be Gâteaux differentiable?
    The Frechet derivative is for functions f:\mathbb{R}^n\mapsto \mathbb{R}^m, and it is a matrix (or a coloumn or row matrix). If it exists it is expressed in terms of its partial derivatives. However, existence of its partial derivatives does not gaurenntee Frechet differenciablility. There is a theorem which says if the partial derivatives are continous then the function is differenciable. I am not sure about the converse. I think it is wrong. You need to justify non-Frechet differenciability in another way. I am not familar with Gateaux derivative.

    Note: The function is continous because \left| \frac{xy}{\sqrt{x^2+y^2}} \right| \leq \frac{ \frac{1}{2}(x^2+y^2) }{\sqrt{x^2+y^2}} = \frac{1}{2}\sqrt{x^2+y^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2008
    Posts
    154
    The continuity of partial derivatives is not a necessary condition for differentiability.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Frechet Derivative
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 11th 2011, 02:52 PM
  2. Frechet Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2010, 06:36 PM
  3. Frechet Derivative Problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 30th 2009, 02:24 PM
  4. Finding the Frechet Derivative of a Map
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 24th 2009, 02:40 PM
  5. Frechet Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 19th 2009, 07:38 AM

Search Tags


/mathhelpforum @mathhelpforum