general mean value theorem

Can anyone give me the intuitive explanation of the general mean value theorem stated in my notes as under:

Let $f:U\rightarrow \mathbb R$ and $U\subseteq \mathbb R^n$ and let $f$ is differentiable at $K\subseteq U$ which is convex.. If $\gamma(t)=(1-t)a+t(b)$ is a line segment joining $a,b$ and $t\in[0,1]$ Then there is a point $c$ on the line segment s.t.

$$f(b)-f(a)=\nabla f(c)(b-a).$$

I'm facing problem how to interpret this theorem.How is it similar to mean value theorem in one-dimension. Please help....

• I don't think it would say "then $f$ is differentiable". Maybe it said "and let $f$ be differentiable". – Michael Hardy Oct 4 '14 at 18:01

As you stated it, it is exactly the one dimensional theorem applied to the function $g:[0,1]\to\mathbb{R}$ given by $g(t)=f(\gamma(t))$. Indeed, $$f(b)-f(a)=f(\gamma(1))-f(\gamma(0))=g(1)-g(0)=g'(s)$$ for some $s\in(0,1)$, but $$g'(t)=\nabla f(\gamma(t))\cdot \gamma'(t)=\nabla f(\gamma(t))\cdot(b-a)$$ where $\cdot$ is the scalar product. So $$f(b)-f(a)=g'(s)=\nabla f(\gamma(s))\cdot (b-a)$$ if you set $c=\gamma(s)$, you have your formula.
If you set $\hat{e}=\dfrac{b-a}{\|b-a\|}$, that is the unit vector pointing from $a$ to $b$, you can rewrite the statement as $$\dfrac{f(b)-f(a)}{\|b-a\|}=\nabla f( c)\cdot \hat{e}$$ which is maybe closer to the one-dimensional one and could help you to grasp the meaning of this.
• You have $\nabla f(\gamma(t))$ where you presumably meant $\nabla (f\circ\gamma)(t)$. A reasonable reader could think you meant $(\nabla f)(\gamma(t))$, which would not be correct. – Michael Hardy Oct 4 '14 at 18:04
• Uhm, no, I think I really mean $\nabla f$ calculated at the point $\gamma(t)$ … I already have a derivative of $\gamma$ outside. Moreover $f\circ \gamma$ is a real function of one real variable, so it does not really make any sense to write its gradient instead of simply writing its derivative… or am I missing something? – wisefool Oct 4 '14 at 22:10