Exercise 1801Calculus › Multivariable calculus › Partial derivativesM1Added 3 Nov 2026

Mean value inequality

#mean value#inequality#gradient

Statement

Let $f:\mathbb{R}^n\to\mathbb{R}$ be $C^1$ and $\| \nabla f\|\le M$ . Show that $|f(b)-f(a)|\le M\|b-a\|$ .

Hint

Apply the univariate MVT to $t\mapsto f(a+t(b-a))$ .

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Restriction. $g(t)=f(a+t(b-a))$ on $[0,1]$ . Then $g(0)=f(a)$ , $g(1)=f(b)$ , and $g'(t)=\nabla f(a+t(b-a))\cdot(b-a)$ .

MVT. $f(b)-f(a)=g'(c)$ for some $c\in(0,1)$ .

Bound. $|g'(c)|\le\|\nabla f\|\cdot\|b-a\|\le M\|b-a\|$ .

Conclusion. Lipschitz bound $|f(b)-f(a)|\le M\|b-a\|$ .