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Math Help - If f is differentiable on a connected open set S..

  1. #1
    Senior Member Pinkk's Avatar
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    If f is differentiable on a connected open set S..

    and \partial_{x_{1}}f(x) = 0 for all x\in S, must f be independent of x_{1}, that is, given a,b where a_{j} = b_{j} for all j \ne 1, then f(a)=f(b).


    So I know how to show this is true if S is convex (because then I can use the Mean Value Theorem), but I am having a hard time coming up with such a function, say on a non-convex subset of \mathbb{R}^{2}, where this does not hold. Any hints or ideas would be appreciated, thanks.
    Last edited by Pinkk; September 22nd 2010 at 07:01 PM.
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    Quote Originally Posted by Pinkk View Post
    and \partial_{x_{1}}f(x) = 0 for all x\in S, must f be independent of x_{1}, that is, given a,b where a_{j} = b_{j} for all j \ne 1, then f(a)=f(b).


    So I know how to show this is true if S is convex (because then I can use the Mean Value Theorem), but I am having a hard time coming up with such a function, say on a non-convex subset of \mathbb{R}^{2}, where this does not hold. Any hints or ideas would be appreciated, thanks.
    An open connected set in \mathbb{R}^n is path-connected. Given two points with a continuous path connecting them, you can split the path into segments each of which lies in a convex subset of S (e.g. an open ball contained in S). So f will be independent of x_1 on each segment, and therefore along the whole path.
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  3. #3
    Senior Member Pinkk's Avatar
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    Hmm, that makes sense, but my textbook (without actually showing the example) says you can find such an example if you have an open connected set in \mathbb{R}^{2} with a hole in it. But if you can cover an open connected set with overlapping convex sets, then how can such a counterexample exist?
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  4. #4
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    Quote Originally Posted by Pinkk View Post
    Hmm, that makes sense, but my textbook (without actually showing the example) says you can find such an example if you have an open connected set in \mathbb{R}^{2} with a hole in it. But if you can cover an open connected set with overlapping convex sets, then how can such a counterexample exist?
    Cancel my previous comment. Here's an example of what the book claims (and it doesn't even need a hole in the domain, just a slit). Let Y be the negative y-axis together with the origin, Y = \{(0,y):y\leqslant0\}, and let S be the complement of Y, S = \mathbb{R}^2\setminus Y. Define f on S by

    f(x,y) = \begin{cases}y^3&(\text{if }x\geqslant0),\\ |y^3|&(\text{if }x<0).\end{cases}

    If y>0 then f(x,y) = y^3 regardless of whether x is positive or negative. So \frac{\partial f}{\partial x} = 0 for all x (and f is independent of x).

    If y<0 then \frac{\partial f}{\partial x} = 0 for x<0 and also for x>0. But f(x,y) will be y^3 if x is positive, and -y^3 if x is negative. So it depends on x.

    Interesting result!
    Last edited by Opalg; September 24th 2010 at 12:46 AM. Reason: modified function to make it differentiable
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    If we define sign(x)
    <br />
sign(x) = \begin{cases}1&(\text{if }x\geqslant0),\\ -1&(\text{if }x<0).\end{cases}<br />
    then
    <br />
f(x,y)=g(y) \; sign(x)<br />
    and
    <br />
\frac{\partial f}{\partial x} = g(y) \; 2 \delta (x)<br />

    and
    f(x,y) depends on x.
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