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  1. Hello
    Im trying to prove that a function h: R^n -> R^m is differentialbe if and only if each of the m components hi: R^n -> R is differentiable.
    I know that i have to use the coordinate projection function and the chain rule for one implication, but im having lots of trouble starting the problem off.
  2. jcsd
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  4. EnumaElish

    EnumaElish 2,332
    Science Advisor
    Homework Helper

    If all components are differentiable then my guess is that you can use the triangle inequality to show that the function itself is dif'able.

    For the only if part, you could assume that one of the components is not differentiable while the function is dif'able and hopefully a contradiction will be obvious.
  5. HallsofIvy

    HallsofIvy 41,260
    Staff Emeritus
    Science Advisor

    Exactly what is your definition of "differentiable"?
  6. Ok i used the triangle inequality to show that the function itself is differentiable; my work looks like
    assume (h1, h2, ...... , hm) is differentiable
    therefore by the triangle inequality,
    = ||(h1 .... hm)(x) - ( - D(||/ ||x-Xo||
    and since h= (h1,h2, ....., hm)
    ||h(x) - h(Xo) - Dh(Xo)(x-Xo)|| / ||x-Xo||

    now my proof was that the function is differentiable if and only if each component is differentiable.
    and i assumed that one component was not differentable where the entire fn was differentiable. To hopefully find that it is a contradiction.
    my work looks like: assume hj is not differentiable

    since h= (hi,h2,..hj...,hm)
    = ||(hi, - (hi,....hj,....,hm)(Xo) - D(hi,...hj,...,hm)(Xo)(x-Xo)||
    / ||x-Xo||

    and i went on to say that this is
    <= || hi(x) - hi(Xo) - Dhi(x-Xo) + ....+ hj(x) - hj(Xo) - Dhj(x-Xo) +....
    but i assumed hj was not differentiable
    therefor this is a contradiction, therefor if h is differentiable
    then the,
    sum (i=1 to m) hi must be differentiable

    is this a good proof showing that all components are differentiable.. or am i doing somthing wrong?
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