# Why do we call this theorem in vector valued functions “mean value theorem”?

I didn't understand why we call this theorem below mean value theorem. I didn't understand the relationship with the mean value theorem we learn in the first semester Calculus.

Mean value for vector valued functions: Let $f:[a,b]\to \mathbb R^n$ a continuous function, differentiable in the open interval $(a,b)$. If $|f'(t)|\le M$ for every $t\in (a,b)$, then $|f(b)-f(a)|\le M(b-a)$.

• I know it as mean value inequality, and it is an extension of the mean value theorem, for real valued functions, in which equality holds. – Fimpellizieri Dec 31 '16 at 20:24
• See this link. – Fimpellizieri Dec 31 '16 at 20:26
• – littleO Dec 31 '16 at 21:32

In one-variable calculus, you get an equality $$f(b)-f(a)=f'(c)(b-a)$$ for some $c\in [a,b]$. Taking absolute values, this gives $$|f(b)-f(a)|=|f'(c)|(b-a)\leq M(b-a).$$
This means that the inequality $|f(b)-f(a)|\leq M(b-a)$ can be thought of as a weak version of the standard mean value theorem. A stronger theorem doesn't hold in general, so we give this name to the above inequality.
One can show that if $x$ is such that $|h^Tx| \le M \|h\|$ for all $h$, then $\|x\| \le M$.
Now pick some $h$, then note that $\phi_h(t) = h^T f(t)$ satisfies $h^T(f(b)-f(a) ) = \phi_h(b) -\phi_h(a) = \phi'_h(\xi) (b-a) = h^T f'(\xi) (b-a)$ for some $\xi$. This is where the usual mean value theorem is used. If we have $\|f'(t)\| \le M$, then the above becomes $|h^T(f(b)-f(a))| \le M \|h\|\|b-a\|$ and since this is true for all $h$ we have $\|f(b)-f(a)\| \le M \|b-a\|$.