Question

show x^1/3 is uniformly continuous on (-infinity,+infinity)

Answer 1

Let f : |R |R be defined by, f(x) = x^1/3 for all x in |R.

CASE 1 : Consider the restriction of f to [-1,1], i.e. consider the function g = f |_[-1,1]. Since, g is continuous on the
closed and bounded(compact) interval [-1,1], hence g is uniformly continuous on [-1,1].

CASE 2 : Suppose x |R \ [-1,1]. Then, | f'(x) | = | 1/(3x^2/3) | < 1/3 (as |x| > 1 ).
Suppose now that y > x > 1 or x < y < -1 . Since f is continuous and differentiable on [x,y], hence by
  the Mean Value Theorem, there is a z in (x,y) such that f(y) - f(x) = f'(z) (x-y) .
Since z (x,y), hence |z| > 1. Thus, |f'(z)| < 1/3. As a result, |f(y) - f(x)| = |f'(z)| |x-y| < 1/3 |x-y|.
Thus, f is lipschitz on |R \ [-1,1]. Since every lipschitz function is uniformly continuous, hence
f |_{|R \ [-1,1]} is uniformly continuous on |R \ [-1,1].

Thus, f is uniformly continuous on [-1,1] and on |R \ [-1,1].

Choose > 0 . Then there is a ₁ > 0 such that if x,y [-1,1] and |x-y| < ₁ , then |f(x) - f(y)| < /2 (by the uniform continuity of f |_[-1,1] ).
Also, there exists a ₂ > 0 such that if |x|,|y| > 1 and |x-y| < ₂ , then |f(x) - f(y)| < /2 (by the uniform continuity of
f |_{|R \ [-1,1]} ).

Let = min{ ₁ , ₂ } > 0. If |x-y| < , then :
1) x,y [-1,1] implies that |f(x) - f(y)| < /2 < .
2) x,y |R \ [-1,1] implies that |f(x) - f(y)| < /2 < .
3) x [-1,1] and y > 1 implies that there is a z in (x,y) such that |x-z| < |x-y| < and |z-y| < |x-y| < .
Thus, |f(x) - f(y)| |f(x) - f(z)| + |f(z) - f(y)| < /2 + /2 = .
4) Similarly, if x [-1,1] and y< -1, |f(x) - f(y)| < .

In other words, |x-y| < implies that |f(x) - f(y)| < .

Since > 0 is arbitrary, hence f is uniformly continuous on |R.