Analysis continuity

**Jason Bourne** · November 23rd 2007, 10:35 PM

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

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

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

**kalagota** · November 24th 2007, 12:05 AM

Originally Posted by Jason Bourne

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

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

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

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

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

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

Thread: Analysis continuity

LinkBack

Thread Tools

Display

Analysis continuity

Similar Math Help Forum Discussions

complex analysis continuity

Continuity of functions - Real Analysis

Real Analysis, continuity on [0,1]

Real Analysis: continuity

continuity - analysis

Search Tags