Is the Mapping T Uniformly Continuous on [0,1] x [0,1]?

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SUMMARY

The mapping T: [0,1] x [0,1] → R^2 defined by T(x,y) = (xy, e^(xy)) is not uniformly continuous. The discussion highlights that while continuous linear maps on normed vector spaces are uniformly continuous, T is neither linear nor defined on a normed vector space, as X is not a normed vector space. The presence of a discontinuity between the x-axis and the exponential function further confirms that T is not continuous, and thus cannot be uniformly continuous.

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Homework Statement


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.


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

I'm guessing that you mean [tex]T(x, y) = (xy, e^{xy})[/tex] here.

Pyroadept said:
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.

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?

Pyroadept said:
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?

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].
 
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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