# Uniformly continuous mapping

## 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] $$\rightarrow$$ R^2, T(x,y) = (xy, e^(x.y))

## 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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T: [0,1] x [0,1] $$\rightarrow$$ R^2, T(x,y) = (xy, e^(x.y))
I'm guessing that you mean $$T(x, y) = (xy, e^{xy})$$ here.

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 $$X$$, not $$Y$$, to be a normed vector space ("on" refers to the domain), and you would need $$T$$ to be a linear map. Both of these are false; why?

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 $$T$$ 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 $$X$$ which is relevant to uniform continuity of functions with domain $$X$$.

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