Let \(\Omega\) denote a domain on \(\mathbb{R}^n\).
Case: multi-dimensional domain and one-dimensional codomain Link to heading
If \(\Omega\) is convex, a differentiable function \(f\colon \Omega\to \mathbb{R}\), also satisfies a mean value theorem. To see this, let \(x,y\in \Omega\). Consider \(g\colon [0,1] \to \mathbb{R}\) defined by
\[ g(t)=f((1-t)x + ty). \]Then, there exists \(t_0\in (0,1)\), such that
\[ f(y)-f(x)=g(1)-g(0)=g'(t_0)=\nabla f((1-t_0)x + t_0 y) (x-y). \]Case: multi-dimensional domain and multidimensional-dimensional codomain Link to heading
For a differentiable function \(F\colon \Omega\to \mathbb{R}^m\), such a statement does not exist. The main reason, is that for each component we may find a point satisfying the mean value theorem. But the points do not coincide tin general.
But we have
\[ \lVert F(x+h) - F(x)\rVert = \lVert DF(x)h + \mathcal{o}(h)\rVert. \]If \(F\) is Lipschitz, we find \(\sup_{x\in \Omega}\lVert DF(x)\rVert\) is a Lipschitz constant of \(F\) (see (0x695deae7) ).