Exercise · ZW-UNIFORM-CONTINUITY-006 · ★★☆ Intermediate

For \(f(x)=x^2\) on \([0,3]\) we have \(|f(x)-f(y)|=|x+y|\,|x-y|\). What is the smallest constant \(L\) with \(|f(x)-f(y)|\le L|x-y|\) on \([0,3]\)?

Choose one option.

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  • B. \(L=6\)

Explanation

On \([0,3]\) the factor \(|x+y|\) is at most \(6\), so \(L=6\) works and is the smallest possible (approached with \(x,y\) near \(3\)). Hence \(\delta=\varepsilon/6\) works. \(L=3\) and \(L=5\) fail; \(L=9\) works but is not the smallest.

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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 \(A\).

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

More practice problems: uniform continuity

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