Question

A function $f: mathbb{R} ightarrow mathbb{R}$ is said to be Lipschitz if there exists some $M > 0$ such that egin{align} |f(x) - f(y)| le M|x - y|, end{align} for all $x, y in mathbb{R}$. (a) Show that if $f: mathbb{R} ightarrow mathbb{R}$ is Lipschitz, then it is uniformly continuous on $mathbb{R}$. (b) Are all Lipschitz functions uniformly continuous? Prove this or give a counterexample (with justification).

          A function $f: mathbb{R} 
ightarrow mathbb{R}$ is said to be Lipschitz if there exists some $M > 0$ such that egin{align*} |f(x) - f(y)| le M|x - y|, end{align*} for all $x, y in mathbb{R}$.

(a) Show that if $f: mathbb{R} 
ightarrow mathbb{R}$ is Lipschitz, then it is uniformly continuous on $mathbb{R}$.

(b) Are all Lipschitz functions uniformly continuous? Prove this or give a counterexample (with justification).

A function f: mathbbR
ightarrow mathbbR is said to be Lipschitz if there exists some M > 0 such that eginalign* |f(x) - f(y)| le M|x - y|, endalign* for all x, y in mathbbR.

(a) Show that if f: mathbbR
ightarrow mathbbR is Lipschitz, then it is uniformly continuous on mathbbR.

(b) Are all Lipschitz functions uniformly continuous? Prove this or give a counterexample (with justification).

Added by Rosa B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Penny Riley

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A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous on $\mathbb{R}$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, y \in \mathbb{R}$, if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon$. A Show more…

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A function f : R → R is said to be Lipschitz if there exists some M > 0 such that |f(x) − f(y)| ≤ M|x − y|, for all x, y ∈ R. (a) Show that if f : R → R is Lipschitz, then it is uniformly continuous on R. (b) Are all Lipschitz functions uniformly continuous? Prove this or give a counterexample (with justification).