Let $\;f:\mathbb R \rightarrow \mathbb R^m\;$ , $\;g:\mathbb R^m \rightarrow \mathbb R\;$ two functions and consider $\;\varepsilon \gt 0\;$. If $\;g\;$ is uniform continuous in $\;\{ \vert f \vert \le M\}\;$, where $\;\vert \bullet \vert \;$ is the Euclidean norm and $\;M\;$ some positive constant, then there is $\;δ\gt 0\;$ such that:
$\vert g(f_1)-g(f_2) \vert \le \varepsilon\;$ for $\;\vert f_1-f_2 \vert \le δ\; ,f_1,f_2 \in \{ \vert f \vert \le M\}\;$
My question:
Could I use the above condition using the same $\;f\in\{ \vert f \vert \le M\}\;$ but with different variables?
To be more specific, as I understand , it holds $\vert g(f_1(x))-g(f_2(x)) \vert \le \varepsilon\;$ for $\;\vert f_1(x)-f_2(x) \vert \le δ\; ,f_1,f_2 \in \{ \vert f \vert \le M\}\;$. I wonder if it's still true to claim $\vert g(f(x))-g(f(y)) \vert \le \varepsilon\;$ for $\;\vert f(x)-f(y) \vert \le δ\; ,f \in \{ \vert f \vert \le M\}\;$. Intuitively it seems possible, but I don't know how to prove it...
Any help would be valuable. Thanks in advance!
Yes. If $g$ is uniformly continuous on $\{|f|\le M\}$, then for every $\varepsilon>0$ there is $\delta>0$ such that $|g(y_1)-g(y_2)|\le \varepsilon$ for all $y_1,y_2\in \{|f|\le M\}$ with $|y_1-y_2|\le\delta$. So you can take $y_1=f(x_1)$ and $y_2=f(x_2)$ provided $|f(x_1)-f(x_2)|\le\delta$ and $|f(x_1)|\le M$, $|f(x_2)|\le M$.