In H. Cartan's Differential Calculus, Theorem 3.1.1 is called the Mean Value Theorem and is stated as:

Theorem: Let $f:[a,b]\to\mathbf R^n$ and $g:[a,b]\to\mathbf R$ be two functions which are continuous on $[a,b]$ and differentiable on $(a,b)$. Assume that $\|f'(x)\|\leq g'(x)$ for all $x\in(a,b)$. Then $\|f(b)-f(a)\|\leq g(b)-g(a)$.

On the other hand, the universally known (Lagrange's) Mean Value Theorem states that

Theorem: Let $f:[a,b]\to\mathbf R$ be a function which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c\in(a,b)$ such that $(f(b)-f(a))/(b-a)=f'(c)$.

Since they are both being called Mean Value Theorems, I think may be one follows easily from the other.

But I fail to see the connection.

Can somebody please shed some light on this.



Cartan's mean value theorem follows from Lagrange's mean value theorem: Let $f,g$ be as in the theorem and assume for now that $g'(x)>0$ for all $x$. Consider the real-valued function (here the dot denotes the standard dot product on $\mathbb R^n$, which induces the euclidian norm; $y\cdot z=\sum_n y_nz_n$) \begin{align*} h(x)&=(f(b)-f(a))\cdot \left[(f(x)-f(a))(g(b)-g(x))-(f(b)-f(x))(g(x)-g(a))\right]. \end{align*} Lagrange's mean value theorem ensures the existence of $x\in(a,b)$ such that $$ 0=(f(b)-f(a))\cdot f'(x)(g(b)-g(a))-(f(b)-f(a))\cdot(f(b)-f(a))g'(x)). $$ The Cauchy-Schwarz inequality now yields $$ \|f(b)-f(a)\|g'(x)\leq (g(b)-g(a))\|f'(x)\|\leq(g(b)-g(a))g'(x), $$ which finishes the argument, since we assumed $g'(x)>0$.

It remains to remove the assumption $g(x)>0$. Observe that $g'(x)\geq \|f'(x)\|\geq 0$. Consider then $\tilde g(x)=g(x)+\epsilon(x-a)$, which yields $$ \|f(b)-f(a)\|\leq g(b)-g(a) + \epsilon (b-a). $$ Now let $\epsilon\rightarrow 0$.

  • $\begingroup$ Thank you for the response. I cannot see how $h$ is real valued since $f$ is vector valued. Can you please address this point? $\endgroup$ – caffeinemachine Oct 23 '14 at 15:38
  • $\begingroup$ The dot is a dot product (inner product). $\endgroup$ – Jonas Dahlbæk Oct 23 '14 at 19:54

"Lagrange's" version of the MVT guarantees the existence of a point $\tau\in\ ]a,b[\ $ such that $$f(b)-f(a)=f'(\tau)\>(b-a)\ ;$$ but for real-valued functions only. Its proof relays heavily on the order present in ${\mathbb R}$, but not in ${\mathbb R}^d$ for $d\geq2$. Take the function $${\bf f}(t):=(\cos t,\sin t)\qquad(0\leq t\leq 2\pi)$$ as an example: One has ${\bf f}(2\pi)-{\bf f}(0)=0$, but $|{\bf f}'(t)|\equiv1$; whence there is no such $\tau$ in this case.

For vector-valued functions there is still a weaker form of the MVT in form of an inequality: In the simplest version $$|{\bf f}'(t)|\leq M\qquad(a\leq t\leq b)$$ guarantees $$|{\bf f}(b)-{\bf f}(a)|\leq M\cdot(b-a)\ ,\tag{1}$$ but this estimate contains no information about the direction of the vector ${\bf f}(b)-{\bf f}(a)$ in terms of the vector-valued function $t\mapsto {\bf f}'(t)$. The estimate $(1)$ is in essence an application of the triangle inequality, and "passing to the limit".

"Cartan's" version of the MVT is a loaded up version of $(1)$.

  • $\begingroup$ So as I understand, there is no immediate connection between the two MVTs. That is, it is not immediate that Lagrange's, say, follows from Cartan's? $\endgroup$ – caffeinemachine Oct 23 '14 at 15:45
  • 1
    $\begingroup$ @caffeinemachine: You cannot get Lagrange's version of the MVT from $(1)$ or similar. $\endgroup$ – Christian Blatter Oct 23 '14 at 16:30

Suppose Lagrange Mean value theorem is true, under the assumption of your first theorem except $f:[a,b]\to \mathbb{R}$

We can be show there exists $c\in(a,b)$ such that $(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c)$. This is Cauchy Mean value theorem.

Then take absolute value on both sides, since $0\leq |f'(x)|\leq g'(x)$ for all $x\in(a,b)$, $g$ is increasing, we get $$|f(b)-f(a)||g'(c)|=(g(b)-g(a))|f'(c)| \leq (g(b)-g(a))|g'(c)|$$ So the simplified first theorem holds.


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