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Stuck on proof regarding partial derivatives

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose the function f:R^2→R has 1st order partial derivatives and that
    δf(x,y)/δx = δf(x,y)/δy = 0 for all (x,y) in R^2.
    Prove that f is constant; there exists c such that f(x,y) = c for all (x,y) in R.

    There's a hint as well:
    First show that the restriction of f:R^2→R to a line parallel to one of the coordinate axes is constant.

    2. Relevant equations
    The chapter defines partial derivatives in general:
    Let O be open in R^n and i be an index with 1≤i≤n. A function f:O→R has a partial derivative with respect to its ith component at the point x in O provided that
    lim_(t→0)(f(x + t*e_i) - f(x))/t exists.


    3. The attempt at a solution
    The chapter has no mention of integrals but I tried using integrals anyway since I have no idea what to do with the given hint.

    0 =
    y
    ∫δf(x,t)/δy dt +
    0
    x
    ∫δf(t,0)/δy dt
    0
    = f(x,y) - f(x,0) + f(x,0) - f(0,0)
    = f(x,y) - f(0,0)

    implying f(x,y) = f(0,0) for all (x,y).
    But then I realized that the partial derivatives being 0 means their antiderivatives are constant, so that wont work out.
    Now I'm looking at the hint again but, I'm really stumped on how to use it. The chapter mentions the Mean Value Theorem. I thought of trying that somehow, but it's not given that the function is continuous. Can anyone point me in the right direction?
     
  2. jcsd
  3. Feb 19, 2012 #2

    LCKurtz

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    Set y = c, constant, and consider h(x) = f(x,c). Can you apply the MVT to that?
     
  4. Feb 20, 2012 #3
    Well, another reason I'm having trouble with this problem is I'm having trouble believing it's true.
    What if f(x,y) =
    {2 if (x,y) = (0,0),
    {3 if (x,y) ≠ (0,0).

    Wouldn't the partial derivatives be 0 still? Again it's not given that the function is continuous. Do I have to assume it's continuous?
     
  5. Feb 20, 2012 #4

    LCKurtz

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    If the partial derivative with respect to x exists at (x,c) then the function of x given by f(x,c) must be continuous. Same for f(c,y). Assuming the derivative exists implies continuity.
     
  6. Feb 20, 2012 #5
    Ah, yes. But this section in the book specifically states that
    "For n > 1, a function f: R^n→R that has first order partial derivatives need not be continuous," and even provides an example:
    f(x,y) =
    {xy/(x^2+y^2) if (x,y) ≠ (0,0)
    {0 if (x,y) = (0,0)

    The partials at (0,0) are 0.

    Okay, I'm going try this:
    Suppose f is not continuous. Then f is not constant, since f being constant implies it's continuous. Then we are done?

    Then with that assertion that f should be continuous I can use MVT. Yes?
     
  7. Feb 20, 2012 #6

    LCKurtz

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    But I didn't say the function f(x,y) was continuous as a function of two variables. What you do have is that as a function of one variable with the other held constant it is continuous. And its derivative is given to exist and is 0.
     
  8. Feb 20, 2012 #7
    Oh, I think I see now. For a function of one variable, differentiability implies continuity. So that's where the hint fits in? Use the line parallel to an axis to equate it to a function with an interval domain to imply continuity?

    So I can do something like this:
    I = {(x,c) | c = some constant}, the line.
    g:I→R ; g(x) = f(x,c). Since g is differentiable on I it is continuous?
     
  9. Feb 20, 2012 #8

    LCKurtz

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    You have a function of one variable whose derivative is 0 and you are trying to show it is a constant. Perhaps if you show it is constant along the line where y = c that would get you started. Look in your calculus book to see how f'(x) identically 0 implies f(x) is a constant. I bet you will see a reference to the MVT.
     
  10. Feb 21, 2012 #9

    LCKurtz

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    And I might add that once you have f(x,y) is constant along lines (x,c) and (a,y) you need to show that f is the same constant everywhere.
     
  11. Feb 21, 2012 #10
    I think I've written out a satisfying proof. Thanks for your help.
     
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