I need help with a particular question:
A function $f(x)$ is Lipschitz on $\mathbb R$ if $\exists C$ that is a constant such that $\forall x, y \in \mathbb R,|f(x)-f(y)| \leq C|x-y|$.
Show that if $f$ is Lipschitz on $\mathbb R$, and if $(x_n)$ is a Cauchy sequence, then $(f(x_n))$ is also Cauchy.
We are required to show that $\forall \varepsilon > 0, \exists K \in \mathbb N$ such that $\forall x_n, x_m > K, |f(x_n)-f(x_m)| < \varepsilon$.
Since $(x_n)$ is Cauchy, $\forall \varepsilon > 0, \exists K' \in \mathbb N$ such that $\forall n, m > K', |x_n-x_m| < \varepsilon$.
What could be right/wrong.
Since $f$ is Lipschitz, would it imply that $f(x_n)$ is Lipschitz?
If it is, then we have $|f(x_n)-f(x_m)| \leq C|x_n-x_m|$.
and thus $\frac{|f(x_n)-f(x_m)|}{C} \leq |x_n-x_m|$.
Since $(x_n)$ is Cauchy, then $\frac{|f(x_n)-f(x_m)|}{C} \leq |x_n-x_m| < \varepsilon$ and
$\frac{|f(x_n)-f(x_m)|}{C} < \varepsilon$.
Best Answer
Hints:
$$\{x_n\}\;\;\text{Cauchy means}\;\; \forall\;\epsilon>0\;\exists\;N_\epsilon\in\Bbb N\;\;s.t.\;\;n,m>N_\epsilon\Longrightarrow |x_n-x_m|<\frac{\epsilon}{K}$$
so now, again for $\,n,m>N_\epsilon\,$ :
$$|f(x_n)-f(x_m)|<K|x_n-x_m|<K\frac{\epsilon}{K}=\epsilon$$