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Is this derivative Lipschitz?

  1. Jul 21, 2009 #1
    Here is a tough one:

    Say we have a multivariable function f:R^n -> R
    and for any x, and direction u, the function g:R->R defined as g(t)=f(x+tu) has that g'(t) is Lipschitz with the same Lipschitz constant (say M). For special cases, taking u to be any basis element we see that every partial restricted to the direction of its axis is Lipschitz

    Can I then prove that the derivative Df(x) (or equivalently any partial derivative) is Lipschitz as well.

    the 1-dimensional case is trivial:

    |(Df(y)-Df(x))\cdot e_i| = |g'(y)-g'(x)|<M|y-x| Lipschitz. (g is defined above with x=0 and u=e_1)

    It is the n>1 cases that poses a problem, because I find that if y-x is not parallel with e_i we run into problems.

    I have been working on this problem for a while and find I am working in circles now.
  2. jcsd
  3. Jul 21, 2009 #2
    You know sometimes people think that if partial derivatives exist, then the function would be differentiable, but that's not always true, because it can happen that the expression

    \frac{f(x+tu) - f(x)}{t}

    has a limit when [itex]t\to 0[/itex] for all fixed [itex]u\in\mathbb{R}^n[/itex], but for some suitable differentiable function [itex]u(t)=a_1t + \frac{1}{2}a_2t^2 + \cdots[/itex], a limit of an expression

    \frac{f(x+u(t)) - f(x)}{t}

    (if existing) does not agree with the partial derivative in direction [itex]a_1[/itex].

    Perhaps a similar problem could ruin your hypothesis? There might exist a differentiable function, for which the derivatives in any fixed direction are always Lipschitz, but for which derivatives along some curves are not?
  4. Jul 22, 2009 #3
    Thanks for your reply.

    I know that the existence of partial derivatives does not guarantee a differentiable function.

    I have been unable to construct a counter example, the condition that all those function have a uniform Lipschitz constant is a very strong condition.

    Also, on the note of partial derivatives, it is true however that if all the partials are continuous that the function is differentiable (in fact it is [tex]C^1[/tex])

    I guess I'll just think about this problem some more.

    PS i read how to imbed latex into posts.
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