# Inequality involving infinite norm.

Define $$\omega_f(\delta) = \sup \{|f(x)-f(y)|: (x,y)\in [0,1]^{2}\text{ and } |x-y|\leq \delta\}.$$ Let $$F(f)(\delta) = \omega_f(\delta)$$ for arbitrary $$\delta\geq 0$$ and $$f\in \mathcal{C}([0,1]).$$ Then show that $$||F(f)-F(g)||_{\infty}\leq 2||f-g||_{\infty}.$$

Proof Attempt: First we observe that for $$\delta\geq 0$$ $$\omega_f(\delta) = \max\{|f(x)-f(y)|:(x,y)\in[0,1]^2 \text{ and } |x-y|\leq \delta\}.$$ This is simply because $$f$$ is continuous function over a compact set and thus there exists $$x_1,x_2$$ such that $$|x_1-x_2|\leq \delta$$ and $$\omega_f(\delta) = |f(x_1)-f(x_2)|.$$ Thus we have the following inequality for $$\delta\geq 0,$$ $$|F(f)(\delta) - F(g)(\delta)| = | |f(t_1) -f(t_2)| - |g(t_3)-g(t_4)||\leq |f(t_1)-f(t_2) - g(t_3)+g(t_4)|$$ $$\leq |f(t_1)-g(t_3)| +|f(t_2)-g(t_4)|\leq 2||f-g||_{\infty}$$ where $$(t_1,t_2), (t_3,t_4)\in [0,1]^2$$ with $$|t_1-t_2|\leq \delta$$ and $$|t_3-t_4|\leq \delta$$ such that $$F(f)(\delta) = \omega_f(\delta) = |f(t_1)-f(t_2)|$$ and $$F(g)(\delta) = \omega_f(\delta) = |g(t_3)-g(t_4)|.$$ Since the above inequality holds for each $$\delta\geq 0$$ we have that $$\newcommand\norm[1]{\left\lVert#1\right\rVert}$$ $$\norm{F(f)-F(g)}_{\infty}\leq 2\norm{f-g}_{\infty}.$$

Is this reasoning correct?

• Yes, good job sir. – Ben W Jan 7 at 16:08

Your proof is not correct, because you assume that $$|f(t_1)-g(t_3)| + |f(t_2)-g(t_4)| \leq 2 \Vert f-g \Vert_{\infty}$$ for arbitrary $$t_1, t_2, t_3, t_4$$. Choosing $$f(x) = g(x) = x$$ shows that this can not be true in general.
I would proceed as follows: For $$\delta \ge 0$$ and all $$(x, y) \in [0, 1]^2$$ with $$|x-y| \le \delta$$ $$|f(x) - f(y)| \le |f(x) - g(x)| + |g(x) - g(y)| + |g(y) - f(y) | \\ \le \Vert f-g \Vert_{\infty} + \omega_g(\delta) + \Vert f-g \Vert_{\infty}$$ which means that the right-hand side is an upper bound for the set $$\{|f(x)-f(y)|: (x,y)\in [0,1]^{2}\text{ and } |x-y|\leq \delta\}$$ and therefore $$\omega_f(\delta) \le \omega_g(\delta) + 2 \Vert f-g \Vert_{\infty} \, .$$ The same relationship holds with $$f, g$$ interchanged, so that $$|\omega_f(\delta) - \omega_g(\delta)| \le 2 \Vert f-g \Vert_{\infty} \, .$$ This holds for all $$\delta \ge 0$$, therefore $$\Vert F(f)-F(g) \Vert_{\infty} \leq 2 \Vert f-g \Vert_{\infty}\, .$$
Remark: We did not use the fact that the functions are continuous, or that the domain is a compact interval. The same conclusion holds for arbitrary bounded functions defined on arbitrary subsets of $$\Bbb R$$ (or even between metric spaces, if the differences are measured in the corresponding metrics).