## 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

Solution:
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 $n$.
-sg-

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]