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I would like to show that $f\colon\mathbb{R}\to\mathbb{R}, x\longmapsto\lvert x\rvert^{1/4}$ is uniformly continuous.

My idea is the following. Let $\varepsilon>0$ and set $\delta=\varepsilon^4$. I need to show that $\lvert x-y\rvert\leq\delta$ implies $\lvert f(x)-f(y)\rvert\leq\varepsilon$.

Without loss of generality, assume that $\lvert x\rvert\leq\lvert y\rvert$. Then, in particular, $\lvert x\rvert^{1/4}\leq\lvert y\rvert^{1/4}$ and $$ \lvert f(x)-f(y)\rvert=\lvert y\rvert^{1/4}-\lvert x\rvert^{1/4}\tag{1}. $$

Due to $\lvert~\lvert x\rvert - \lvert y\rvert~\rvert\leq\lvert x-y\rvert<\delta$, we get by assumption that $\lvert y\rvert - \lvert x\rvert < \delta=\varepsilon^4$ and thus $$ \lvert y\rvert\leq \lvert x\rvert + \varepsilon^4\leq(\lvert x\rvert^{1/4} + \varepsilon)^4 $$ meaning that $$ \lvert y\rvert^{1/4}\leq \lvert x\rvert^{1/4}+\varepsilon. $$

Consequently, for $(1)$, we get that $$ \lvert f(x)-f(y)\rvert\leq\varepsilon. $$

Please tell me if my proof is okay, thanks a lot!

BridgeTYH
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1 Answers1

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This is a slight modification of https://math.stackexchange.com/a/728364. When $x,y\ge0$ we have a chain of inequalities: $$\begin{align*}|\sqrt[4]x-\sqrt[4]y|^4&\le|\sqrt[4]x-\sqrt[4]y|^2\cdot |\sqrt[4]x+\sqrt[4]y|^2\\ &=|\sqrt x-\sqrt y|^2\\ &\le|\sqrt x-\sqrt y|\cdot |\sqrt x+\sqrt y|\\ &=|x-y|\le\delta=\epsilon^4.\end{align*}$$ For the general case note that $|f(x)-f(y)|^4\le ||x|-|y||\le |x-y|\le\delta=\epsilon^4$.

Kenta S
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