Exercise 4.4.3

Question

Show that $f(x)=1 / x^{2}$ is uniformly continuous on the set $[1, \infty)$ but not on the set $(0,1]$.

Ulisse Mini · Accepted Answer

By Lipschitz over $[1, \infty)$
  $$
  \left|\frac{1/x^2 - 1/y^2}{x - y}ight|
  = \left|\frac{y^2 - x^2}{x^2y^2(x-y)}ight|
  = \left|\frac{(x - y)(x + y)}{x^2y^2(x-y)}ight|
  = \left|\frac{x + y}{x^2y^2}ight|
  = \left|\frac{1}{xy^2} + \frac{1}{x^2y}ight|
  \le 2
  $$
  For $(0, 1]$ consider $x_n = 1/n$ and $y_n = 1/2n$. we have $|x_n - y_n| 	o 0$ but
  $$|f(x_n) - f(y_n)| = |n^2 - 4n^2| = 3n^2$$
  is unbounded, hence $f$ is not uniformly continuous on $(0, 1]$ by Theorem 4.4.5.

Exercise 4.4.3

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Exercise 4.4.3

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