# Thread:

|x|

2. ## Re: Uniform Continuity

Originally Posted by lovesmath
Prove that the function is uniformly continuous on the set of real numbers.
f(x)=|x|
I think I have to use the reverse triangle inequality, but I am not sure how to do it.
There you have it.

$\left| {\left| x \right| - \left| {x_0 } \right|} \right| \leqslant \left| {x - x_0 } \right| < \delta = \varepsilon$

3. ## Re: Uniform Continuity

Here is an additional question.

Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.

4. ## Re: Uniform Continuity

Originally Posted by lovesmath
Here is an additional question.

Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.
What have you tried so far?

-Dan

5. ## Re: Uniform Continuity

I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.

Not sure what to do next.

6. ## Re: Uniform Continuity

Originally Posted by lovesmath
I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.
The key to this is the fact that $D$ is bounded.
That means that $\exists M>0$ such that $D\subset [-M,M]$.

In your text material you should have a proof that a continuous function is bounded on a closed interval of real numbers. This new proof follows that proof very closely.
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