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Def. Continuity in terms of sequences: How do I generalize to multivariate fcns?

  1. Dec 25, 2005 #1

    benorin

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    Working from "Principles of Mathematical Analysis", by Walter Rudin I have gleaned the following definition of continuity of a function (which maps a subset one metric space into another):

    Suppose [itex]f:E\rightarrow Y[/itex], where [itex]\left( X, d_{X}\right) \mbox{ and } \left( Y, d_{Y}\right) [/itex] are metric spaces, and [itex]E\subset X[/itex]. Let [itex]x\in E[/itex] be a limit point of E.

    f is continuous at x if, and only if, for every sequence [itex]\left\{ x_{n}\right\} \rightarrow x[/itex] such that [itex]x_{n}\in E\forall n\in\mathbb{N}[/itex], we have [itex]f(x_{n})\rightarrow f(x)\mbox{ as }n\rightarrow\infty[/itex].

    My question is, when generalizing the above definition to multivariate functions, the last line of the definition would include which of the following:

    a. [tex]f(x_{n},y_{n})\rightarrow f(x)\mbox{ as }n\rightarrow\infty[/tex],

    or

    b. [tex]f(x_{n},y_{m})\rightarrow f(x)\mbox{ as }n,m\rightarrow\infty[/tex] ?

    I am uncertian if I need the double limit. :rofl: DUH! I get: X is a metric space, it could be of an arbitrary dimension if need be. But suppose that it were a product of metric spaces (with different metrics,) would the question then merit investigation?
     
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  3. Dec 25, 2005 #2

    matt grime

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    You need a sequence v_n=(x_n,y_n,z_n,....) tending to v=(x,y,z,...) where v is a vector and the metric (ie the notion of tending to) is the euclidean one, so the first of your two options, though I suspect that the second would be equivalent in some sense, since the first has to be for *all* possible sequences.
     
  4. Dec 25, 2005 #3

    HallsofIvy

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    Neither a) nor b). Since f is a function of two variables, for f to be continuous at (x,y), it must be true that for every sequence xn converging to x and every sequence ym converging to y, the (double) sequence f(xn,ym) (with m and n going to infinity independently) must converge to f(x,y).

    If you have a product of metric spaces with different metrics, then you would use the appropriate metric on each component.
     
  5. Dec 25, 2005 #4

    Hurkyl

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    Ack! That's backwards! It took me a while to figure out what you meant. :tongue2: Symbolically, that's grammatically incorrect! (Though it's okay in English)

    Sure, but I suspect that you could always do a problem by falling back on the product metric.
     
  6. Dec 25, 2005 #5

    mathwonk

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    the whole point of giving the definition for a metric space is to make you realize that dimension has nothing to do with it, but only the metric.

    so in several variables, i.e. in the product of several copies of the erals, you just use any of your favorite product metrics, eucldiean, sum, max.


    there is no difference which one you use either, until you get to infinite dimensions.

    so several variables is the same as one, unless it means infinitely many variables.

    then you have to decide whether two functions are close if they are close everywhere, or their difference has small integral, or the square of their difference has small integral, or what....
     
  7. Dec 26, 2005 #6

    benorin

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    The context I was looking to apply it to has, in fact, infinite dimensions, namely Hilbert spaces. The particular application was proving the following:

    Let H denote a Hilbert space with the inner product [itex] ( \cdot , \cdot ) : H\times H \rightarrow \mathbb{C}[/itex]. Let [itex]\parallel \cdot\parallel [/itex] denote the norm induced by the inner product. Fix [itex]z\in H[/itex]. Then the following functions are continuous mappings [itex]\forall x,y\in H[/itex]:

    [tex] x\rightarrow (x,z), \\ y\rightarrow (z,y) , \\ x\rightarrow \parallel x \parallel , \mbox{ and }x,y\rightarrow (x,y) [/tex]

    The first three are easy consequences of the triangle and Schartz inequalities. The proof of the fourth is from whence my question arose.

    Let [itex]x_{n}\rightarrow x \mbox{ and } y_{n}\rightarrow y\mbox{ as }n\rightarrow\infty[/itex]. Then

    [tex] \left| \left( x_{n}, y_{n}\right) - \left( x, y\right) \right| = \left| \left( x_{n}, y_{n}\right) - \left( x, y_{n}\right) + \left( x, y_{n}\right) - \left( x, y\right) \right| [/tex]
    [tex] \leq \left| \left( x_{n}-x, y_{n}\right) \right| + \left| \left( x, y_{n}-y\right) \right| [/tex]
    [tex]\leq \parallel x_{n}-x \parallel\cdot\parallel y_{n}\parallel + \parallel x \parallel\cdot\parallel y_{n}-y\parallel\rightarrow 0\mbox{ as } n\rightarrow\infty[/tex]

    :uhh: so everything is continuous now, right?
     
    Last edited: Dec 26, 2005
  8. Dec 27, 2005 #7

    mathwonk

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    hence you are using essentially the square root of the infinite sum, or integral, of the squares of the differences of the coordinates, among the examples i gave above.

    and yes your calculation looks ok to me.

    to connect this to my explicit examples, note that your source is using an axiomatic approach to "hilbert space" whereas all hilbert spaces are actually isomorphic to concrete function spaces, either "separable", i.e. given by square summable sequences (functions defined on Z witha discrete measure); or not, given by square integrable functions defined on a measure space.
     
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