# Analysis continuity

• Nov 23rd 2007, 10:35 PM
Jason Bourne
Analysis continuity
Let $\displaystyle \varphi : R \rightarrow [0,\inf)$ be a real valued function and assume that $\displaystyle \varphi$ is continuous at 0 with $\displaystyle \varphi(0) = 0$. Let $\displaystyle f : R \rightarrow R$ be a real valued function and assume that there exists a constant c >0 such that

$\displaystyle \mid f(x) - f(y) \mid \leq c\varphi(x-y)$

for all $\displaystyle x,y \in R$. Prove that f is continuous.
• Nov 24th 2007, 12:05 AM
kalagota
Quote:

Originally Posted by Jason Bourne
Let $\displaystyle \varphi : R \rightarrow [0,\inf)$ be a real valued function and assume that $\displaystyle \varphi$ is continuous at 0 with $\displaystyle \varphi(0) = 0$. Let $\displaystyle f : R \rightarrow R$ be a real valued function and assume that there exists a constant c >0 such that

$\displaystyle \mid f(x) - f(y) \mid \leq c\varphi(x-y)$

for all $\displaystyle x,y \in R$. Prove that f is continuous.

let $\displaystyle \varepsilon ~ > ~ 0$
let w=x-y. then
$\displaystyle \varphi (w)$ continuous at 0 $\displaystyle \implies \exists \delta$ such that
if $\displaystyle |w - 0| < \delta \implies |\varphi(w) - \varphi(0)| = |\varphi(w)| = \varphi(w) < \frac{\varepsilon}{c}$

$\displaystyle \implies c\varphi(w)$ is continuous.. (why?)

now, $\displaystyle \mid f(x) - f(y) \mid \leq c\varphi(x-y) = c\varphi(w) < \varepsilon$. QED