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Uniformly continuous mapping

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose X = [0,1] x [0,1] and d is the metric on X induced from the Euclidean metric on R^2. Suppose also that Y = R^2 and d' is the Euclidean metric. Is the mapping

    T: [0,1] x [0,1] [tex]\rightarrow[/tex] R^2, T(x,y) = (xy, e^(x.y))

    uniformly continuous? Explain your answer.

    2. Relevant equations

    3. The attempt at a solution
    Hi everyone,
    So I know the definition for uniformly continuous, but am wondering if it's necessary to use it? We have in our notes that continuous linear maps on normed vecotr spaces are unifomrly continuous, and (Y,d') is a normed vector space.
    So by looking at the graph of the map, there is a discontinuity between the line on the x-axis and the exponential function. So can you say it is not continuous and thus not uniformly continuous?

    Thanks for any help
  2. jcsd
  3. Apr 20, 2010 #2
    I'm guessing that you mean [tex]T(x, y) = (xy, e^{xy})[/tex] here.

    To use this result, you would need [tex]X[/tex], not [tex]Y[/tex], to be a normed vector space ("on" refers to the domain), and you would need [tex]T[/tex] to be a linear map. Both of these are false; why?

    If [tex]T[/tex] is not continuous, it is certainly not uniformly continuous, but I don't understand the argument you offer; the words "there is a discontinuity between the line on the x-axis and the exponential function" don't make sense without elaboration.

    There is another topological property of [tex]X[/tex] which is relevant to uniform continuity of functions with domain [tex]X[/tex].
  4. Apr 21, 2010 #3


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    Homework Helper

    another idea might be to examine if the function is uniformly continuous on a limited domain, say the line y = 1
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