Question #170335

Show that the function f:R2 to R defined by f(x,y)= x2+y2 is not uniformly continuous


1
Expert's answer
2021-03-24T14:04:13-0400

Inorder to show that f is uniformly continuous, we show that given ϵ>0δ>0(x1,y1)(x2,y2)<δ    f(x1,y1)f(x2,y2)<ϵ\epsilon >0 \exist \, \delta>0 \ni |(x_1,y_1)-(x_2,y_2)| < \delta \implies |f(x_1,y_1) -f(x_2,y_2)|<\epsilon

Suppose that this is true and consider two points in R2\mathbb{R^2} say (a,b) and (a+δ2,b+δ2)(a + \frac{\delta}{2}, b + \frac{\delta}{2})

Now consider

f(a+δ2,b+δ2)f(a,b)=(a+δ2)2+(b+δ2)2(a2+b2)=(a+b)δ+δ22|f(a + \frac{\delta}{2}, b + \frac{\delta}{2})-f(a,b)| = | (a + \frac{\delta}{2})^2 +(b+ \frac{\delta}{2})^2 - (a^2+b^2)| = |(a+b) \delta + \frac{\delta^2}{2}|

And as a+ba+b gets larger, the quantity above is greater than ϵ\epsilon which is sufficiently small.

So, we have arrived at a contradiction. Hence, we can conclude that f is not uniformly continuous.


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