You may want to browse through the forum because I could swear I saw it recently
EDIT: http://www.mathhelpforum.com/math-he...ontinuity.html OK not exactly the same, but close enough/
You may want to browse through the forum because I could swear I saw it recently
EDIT: http://www.mathhelpforum.com/math-he...ontinuity.html OK not exactly the same, but close enough/
Here's one way to show its not uniformly continuous on
Proof by contradiction
Suppose f(x) is uniformly continuous on . Fix e > 0 Thus, there is a d such that:
lx-yl < d --> l f(x) - f(y) l < e
Choose an n so large such that n > max{1/ , 7e/3}.
Let x = 1/ , y = x + d/2
Clearly, lx-yl < d
But, lf(x) - f(y)l = l n - 1/(1/ +d/2)^2 l = l n + 4n/(4+n*d^2+2d* ) l = n* l 1 - 4/(4+n*d^2+2d* )l
but we said n > 1/ , thus
1 - 4/(4+n*d^2+2d* ) > 3/7 and
n*l(1 - 4/(4+n*d^2+2d* )l > n*3/7 > e
contradiction
Note, there are likely mistakes here, I rushed it a bit.