# uniform continuity

• Aug 11th 2009, 06:14 PM
Kat-M
uniform continuity
Let $f$ be continuous on $[a,b)$. Prove that $lim_{x \rightarrow b^-}f(x)$ exists $iff$ $f$ is uniformly continuous on $[a,b)$.
• Aug 11th 2009, 08:48 PM
siclar
If we know $\displaystyle\lim_{x\rightarrow b-}f(x)$ exists and we want to show f is actually uniformly continuous, it helps to invoke a nice theorem.

Theorem: If $f$ is a continuous real function on a closed and bounded set of real numbers, then $f$ is in fact uniformly continuous.

Then if we can define a value of $f$ at $b$ which maintains continuity, we've got it!

Spoiler:
Use the sequential definition of continuity, i.e. $f$ is continuous if and only if $x_n \rightarrow x$ implies $f(x_n)\rightarrow f(x)$
.

Conversely if the function is uniformly continuous then a straightforward use of the definition shows the limit exists.

Spoiler:
Let $\epsilon>0$ be given and let $x_n$ be any sequence converging to $x$. Then there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Use the Cauchy criterion.