Exercise · ZW-UNIFORM-CONTINUITY-001 · ★☆☆ Basic
For \(f(x)=2x\) on \(\mathbb R\), which \(\delta\) (in terms of \(\varepsilon\)) guarantees \(|f(x)-f(y)|<\varepsilon\) whenever \(|x-y|<\delta\)?
Choose one option.
A \(\delta=\varepsilon\) B \(\delta=\varepsilon/2\) C \(\delta=2\varepsilon\) D \(\delta=\varepsilon^2\) Show answer B. \(\delta=\varepsilon/2\)← Back to the question
Explanation Since \(|f(x)-f(y)|=2|x-y|\), it suffices that \(|x-y|<\varepsilon/2\). This \(\delta\) does not depend on the point, so \(f\) is uniformly continuous . With \(\delta=\varepsilon\) one would only get \(2|x-y|<2\varepsilon\), which is not enough.
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 \(A\).
\[\forall\varepsilon>0\ \exists\delta>0:\ x,y\in A,\ |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon\]
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