Exercise 1273Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

Bounded derivative implies Lipschitz

#Lipschitz#bounded derivative#MVT

Statement

Show that if $f$ is differentiable on $I$ and $|f'|\le M$ on $I$ , then $f$ is $M$ -Lipschitz.

Hint

Apply the MVT to $x<y$ .

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MVT. For $x<y$ , there is $c\in]x,y[$ with $|f(y)-f(x)|=|f'(c)||y-x|\le M|y-x|$ .

Conclusion. $f$ is $M$ -Lipschitz.