Proving the mean value inequality in higher dimensions for a differentiable function (rather than $C^1$) For a function $f: [a,b] \to \mathbb R$, the mean value theorem is an equality. Further, the differentiability imposed on $f$ is fairly weak: $f$ need only be continuous on $[a,b]$ and differentiable on $(a,b)$. Continuity of the derivative, or absolute continuity, or bounded variation are not required. When the codomain is an arbitrary Banach space (or just $\mathbb R^n$ with $n>1$), the best you can do is an inequality: if $f$ is $C^1$, you can show $\|f(b) - f(a)\| \leq \int_a^b \|f'(t)\|dt$ using the fundamental theorem of calculus. This seems like an unsatisfying generalization to me, not because it is an inequality, but because of the stronger differentiability requirement. I've been tinkering with trying to relax the $C^1$ hypothesis with something like the one-dimensional approach (Rolle's theorem and all that), but I doubt that will work. Does the above inequality hold if $f$ is only taken to be continuous on $[a,b]$ and differentiable on $(a,b)$? How does one prove this?
 A: Ok, let me start pointing out (although you probably know that) that the classical mean value theorem
$$\frac{f(b) - f(a)}{b-a} = f'(\xi) \text{ for some } \xi \in (a,b)$$
does not hold in the Banach setting; the classical counterexample is $[0,1] \rightarrow \Bbb{C}, f(x) = e^{2\pi i x}$.
This is basically because (the proof of) Rolle's theorem uses the ordering on $\Bbb{R}$ in an essential way.
For the other question, there is one simple solution that does involve the supremum of the derivative rather than the integral. We have
$$\Vert f(b) - f(a) \Vert \leq (b - a) \cdot \sup_{\xi \in (a,b)} \Vert f'(\xi) \Vert.$$
This is proved by choosing (using Hahn-Banach) a bounded functional $\phi \in B'$ with $\Vert \phi \Vert \leq 1$ and $\phi(f(b) - f(a)) = \Vert f(b) - f(a) \Vert$ and then applying the classical mean-value theorem to $\phi \circ f$.
For the question concerning the integral, let us assume that $\int_a^b \Vert f'(t) \Vert dt < \infty$ (otherwise, the estimate is trivial). Note that $f' : (a,b) \rightarrow B$ is measurable as it is the pointwise limit of the measurable functions $x \mapsto n \cdot (f(x + \frac{1}{n}) - f(x)) \cdot \chi_{(a, b-\frac{1}{n})}$. I only added the indicator function to avoid the problem $x + \frac{1}{n} \notin (a,b)$ although $x \in (a,b)$.
Let us again take $\phi \in B'$ arbitrary. Then the hypothesis also holds for $f_\phi := \phi \circ f$ and we also have $(\phi \circ f)' = \phi \circ f' \in L^1((a,b))$.
Now Theorem 7.21 in Rudin, Real and Complex Analysis implies
$$\phi(f(b) - f(a)) = f_\phi(b) - f_\phi(a) = \int_a^b f_\phi'(x) dx = \phi\left(\int_a^b f'(x) dx \right).$$
Actually, Rudin requires $f_\phi$ to be differentiable on $[a,b$] instead of $(a,b)$, but then the above applies to $a+\frac{1}{n}$ and $b-\frac{1}{n}$ instead of $a,b$ and we can take the limit as $f$ is continuous and $f'$ is integrable.
As $\phi \in B'$ was arbitrary (a corollary of) Hahn Banach implies
$$f(b) - f(a) = \int_a^b f'(x) dx.$$
Taking norms, you arrive at the desired conclusion.
In short, we only needed to assume (Lebesgue) integrability of the derivative, not $f \in C^1$. Note that this is always satisfied if $f'$ is bounded on $(a,b)$.
