# Analysis continuity

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• November 23rd 2007, 10:35 PM
Jason Bourne
Analysis continuity
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.
• November 24th 2007, 12:05 AM
kalagota
Quote:

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