# Uniform Continuity

• Nov 17th 2012, 02:44 PM
lovesmath
Uniform Continuity
|x|
• Nov 17th 2012, 02:50 PM
Plato
Re: Uniform Continuity
Quote:

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.

$\displaystyle \left| {\left| x \right| - \left| {x_0 } \right|} \right| \leqslant \left| {x - x_0 } \right| < \delta = \varepsilon$
• Nov 17th 2012, 04:16 PM
lovesmath
Re: Uniform Continuity

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.
• Nov 17th 2012, 05:15 PM
topsquark
Re: Uniform Continuity
Quote:

Originally Posted by lovesmath

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
• Nov 18th 2012, 05:52 PM
lovesmath
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.
• Nov 19th 2012, 04:12 AM
Plato
Re: Uniform Continuity
Quote:

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 $\displaystyle D$ is bounded.
That means that $\displaystyle \exists M>0$ such that $\displaystyle D\subset [-M,M]$.

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.