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Continuity of two-variable function

  1. Oct 16, 2005 #1
    Hi, I have some troubles understanding the basic facts about investigating the continuity of two-variable functions.
    Our professor gave us very simple example to show us the basic facts:

    Very important is that projections are continuous, it means

    \pi_1 :[x,y] \longmapsto x

    \pi_2 :[x,y] \longmapsto y

    are continuous.
    Now let's have this function:

    f(x,y) = \sqrt{x^2+y^2}

    It's continuous on the whole [itex]\mathbb{R}^2[/itex]. Why?

    1. Step:
    Projections are continuous

    2. Step:

    x,y \longmapsto x^2 + y^2[/tex]...continuous

    3. Step:
    Square root is continuous on [itex][0,\infty)[/itex]
    What I don't fully understand is the second step. I can't see why, from the fact that projections are continuous, we can say that

    x^2 + y^2

    is continous. Of course I didn't expect any other result, when you look at it it's obvious that it will be continuous, but I just can't see the rigorous mathematical background.

    You know, I see it is some equivalent of the limit of product of single variable function (? is it the right expression ?), but I don't think it's sufficient. I would need some equivalent for two-variable functions...

    Hope you understand my problem :)

    Thank you very much.
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 18, 2005 #2


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    I'll give it a try, but I'm not a mathematician, so I might get it wrong ; feel free to correct me if that's the case.
    Consider the function [tex] a \rightarrow a^2[/tex]. That's a polynomial and hence continuous.
    Now, consider the first function:
    [tex] (x,y) \rightarrow x \rightarrow x^2 [/tex]
    This is a projection, followed by our squaring, so that's a continuous function from (x,y) to x^2.
    Next, consider the second function:
    [tex] (x,y) \rightarrow y \rightarrow y^2 [/tex]
    This is also a projection, followed by squaring, so that's a continuous function too, from (x,y) to y^2.
    The sum of two continuous functions, is continuous, so if we add our two functions, we have [tex] (x,y) \rightarrow x^2 + y^2[/tex]
    should be continous.
    Did I sneak in something ?
  4. Dec 14, 2005 #3


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    Let the point [itex] M_0 (x_0,y_0) [/itex] belong to the domain of the function [itex] f(x,y) [/itex]. Then the function [itex] z=f(x,y) [/itex] is continous at the point [itex] M_0 (x_0,y_0) [/itex] if we have

    [tex] \lim_{\substack{x\rightarrow x_0\\y\rightarrow y_0}} f(x,y) = f(x_0,y_0) [/tex]

    and (x,y) approaches [itex] M_0 (x_0,y_0) [/itex] in arbitary fashion all the while remaining in the domain of the function.

    Now if [itex] x=x_0 + \Delta x , y = y_0 + \Delta y [/itex] then the above condition can be rewritten as

    [tex] \lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} f(x_0 + \Delta x,y_0 + \Delta y) = f(x_0,y_0) [/tex]

    [tex] \lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} [f(x_0 + \Delta x,y_0 + \Delta y) - f(x_0,y_0)] = 0 [/tex]

    [tex] \lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z = 0 [/tex]

    So for [tex] z=\sqrt{x^2+y^2} [/tex],
    what is
    [tex] \lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z [/tex] for any point in the domain?
    Last edited: Dec 14, 2005
  5. Dec 14, 2005 #4


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    Doesn't get any easier than this.

    It's very simliar, if not the same, as continuity for a single variable.
  6. Dec 15, 2005 #5


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    Actually, it does- put it in polar coordinates and it IS continuity for a single variable!

    The point made earlier, by vanesch, is that you cannot be certain of getting the correct limit of
    [tex] \lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z [/tex]

    By simply taking [itex]\Delta y[/itex] going to zero and then taking [itex]\Delta x[/itex] going to 0. However, you can find the limit by seeing what happens when [itex](\Delta x, \Delta y)[/itex] is very close to (0,0). In polar coordinates that is measured by the single variable r. What do you get if you put this problem in polar coordinates?
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