Given that |√ x-√ y|≤√|x-y| for x,y≥0, which δ proves √ x is uniformly continuous on [0,∞)?

Exercise · ZW-UNIFORM-CONTINUITY-009 · ★★★ Advanced

Given that \(|\sqrt x-\sqrt y|\le\sqrt{|x-y|}\) for \(x,y\ge0\), which \(\delta\) proves \(\sqrt x\) is uniformly continuous on \([0,\infty)\)?

Choose one option.

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B. \(\delta=\varepsilon^2\)

Explanation

With \(|x-y|<\varepsilon^2\) we get \(|\sqrt x-\sqrt y|\le\sqrt{|x-y|}<\varepsilon\). With \(\delta=\varepsilon\) we only reach \(\sqrt\varepsilon\), which exceeds \(\varepsilon\) when \(\varepsilon<1\). Note \(\sqrt x\) is not Lipschitz near \(0\), yet it is uniformly continuous.

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Review the concept: Uniform continuity

A function \(f\) is uniformly continuous on a set \(A\) if for each \(\varepsilon>0\) there is a \(\delta>0\), valid for all points at once, that controls \(|f(x)-f(y)|\) by \(|x-y|\) in \(A\).

\[\forall\varepsilon>0\ \exists\delta>0:\ x,y\in A,\ |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon\]

Given that \(|\sqrt x-\sqrt y|\le\sqrt{|x-y|}\) for \(x,y\ge0\), which \(\delta\) proves \(\sqrt x\) is uniformly continuous on \([0,\infty)\)?

Explanation

Review the concept: Uniform continuity

More practice problems: uniform continuity