Sunday, May 23, 2010

GR9768.64: Continous Functions

64.  Suppose that f  is a continuous real-valued function defined on the closed interval [0,1]. Which of the following must be true?
  •   I.  There is a constant C > 0 such that | f(x) - f(y) | ≤ C for all x and y in [0,1].
  •  II.  There is a constant D > 0 such that | f(x) - f(y) | ≤ 1 for all x and y in [0,1] that satisfy |x-y| ≤ D.
  • III.  There is a constant E > 0 such that | f(x) - f(y) | E |x-y| for all x and y in [0,1].  
       (A)  I only
       (B)  III only
       (C)  I and II only
       (D)  II and III only
       (E)  I, II, and III

The definition of continuity: $f$ is continuous at $x = a$ if for all $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |x-a| < \delta $ implies $|f(x) - f(a)| < \epsilon$.
Since $f$ is continuous on a closed interval $[0,1]$, we see that $f$ is also uniformly continuous on $[0,1]$.
The definition of uniformly continuous: $f$ is uniformly continuous on $E = [a,b]$ if given $\epsilon > 0$ and $x, y \in E$, there exists $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < \delta$.
I is true, with $C = \epsilon > 0$.
II must be true, with $\delta = D > 0$ and $\epsilon = 1$.
III is actually false. Yes, if we have this, then it implies $f$ is uniformly continuous. But what we have is actually the opposite. And we know the converse of a true statement is not necessarily still true. As for a counter example, take $f(x) = x^n$ for really large $n$.
The answer is C.


  1. Another way to see that III is false is to note that it is the definition of Lipschitz continuity, which is stronger than uniform continuity. sqrt(x) is an example of a uniformly continuous function that is not Lipschitz continuous.

  2. the explanation about uniform continuity is wrong. Every continuous function on a compact set in uniform continuous and because [a,b] is compact... also the function x^n is Lipschtitz for all n(on compacts)... so we are looking for a uniformly continuous function which is not Lipschtitz i think one that fits is f(x)=xlnx and f(0)=0..

  3. The property in three is possessed by very simple functions. All you need is that the derivative is infinite at a point. X^1/3 at x=0 is such an example. Specifically ( x^ 1/3-0)/(x-0) is unbounded as x goes to zero even though the function is uniformly continuous on [0,1]