Exercise 4.6.10

Question

Let $\alpha>0$ be given. Show that if $f$ is continuous at $x$, then it is $\alpha$-continuous at $x$ as well. Explain how it follows that $D_{f}^{\alpha} \subseteq D_{f}$.

Ulisse Mini · Accepted Answer

Let $\epsilon = \alpha/2$ and use continuity to get $\delta>0$ with $0<|x-y|<\delta$ implying $|f(x)-f(y)|<\alpha/2$, which shows every $y,z \in V_{\delta}(x)$ satisfies $|f(y)-f(z)| < \alpha$ by the triangle inequality. Thus $f$ is $\alpha$-continuous at $x$.

The negation of ``continuous at $x$ implies $\alpha$-continuous at $x$'' is ``not $\alpha$-continuous at $x$ implies not continuous at $x$'', hence $D_f^\alpha \subseteq D_f$.

Exercise 4.6.10

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Exercise 4.6.10

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