Mean value inequality. Let U⊆RkU\subseteq\mathbb R^k be open, let f:U→Rmf:U\to\mathbb R^m be continuously differentiable, and suppose the line segment [x,y][x,y] lies in UU. If M≥0M\ge0 satisfies

∥Df(z)∥≤Mfor all z on the segment from x to y,\|Df(z)\|\le M \quad \text{for all } z \text{ on the segment from } x \text{ to } y,

where ∥Df(z)∥\lVert Df(z)\rVert is the , then

∥f(y)−f(x)∥≤M∥y−x∥.\|f(y)-f(x)\|\le M\|y-x\|.
Remarks

For k=m=1k=m=1, this gives ∣f(y)−f(x)∣≤M∣y−x∣|f(y)-f(x)|\le M|y-x|.