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 $n$.

The answer is C.

-sg-

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.

ReplyDeletethe 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..

ReplyDeleteThe 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]

ReplyDelete