Mean Value Theorem Stated as an Inequality

Question

The mean value theorem is usually stated as follows:

Let $(F, \parallel \parallel)$ be a Banach space and $f : [a, b] \to F$ be a function that is continuous on $[a, b]$ and differentiable on $(a, b).$ Then $\exists\ c \in (a,b)$ s.t $$\| f(b) − f(a) \| = \| f'(c) \| \|(b − a)\| $$

My professor instead uses an inequality:

Let $(F, \parallel \parallel)$ be a Banach space and $f : [a, b] \to F$ be a function that is continuous on $[a, b]$ and differentiable on $(a, b).$ Then $$\| f(b) − f(a) \| \leq \sup_{c \in [a,b]} \|f'(c)\|\|(b − a)\| .$$

Isn't the inequality weaker than the equality? Why would you want to use an inequality and supremum?

Answer 1

The first version is simply false once you take $F$ to be an arbitrary Banach space which is not $\Bbb{R}$.

Here's a counterexample. Take $F = \Bbb{R}^2$, and let $f: [0, 2 \pi] \to \Bbb{R}^2$ be defined by \begin{align} f(t) = (\cos t, \sin t) \end{align} Then, for all $t \in [0, 2 \pi]$, we have $f'(t) = (-\sin t, \cos t)$, which lies on the unit circle; i.e $\lVert f'(t)\rVert = 1$. So, we have \begin{align} \lVert f(2\pi) - f(0)\rVert = 0, \end{align} whereas for any $t \in [0,2 \pi]$, we have \begin{align} \lVert f'(t)\rVert \cdot (2 \pi - 0) = 1 \cdot 2 \pi = 2 \pi \neq 0. \end{align}

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So, in higher dimensions, there is simply no equality in general. Roughly speaking, the claim fails because even if you assume finite-dimensionality of the Banach space $F$, and you have a map $f: [a,b] \to \Bbb{R}^n$, with component functions $(f_1, \dots, f_n)$, you can try applying the standard mean-value theorem of single variable calculus to each $f_i$. This will yield a certain $c_i \in (a,b)$, but there is no guarantee that $c_1 = \dots c_n$ are all equal.

The inequality version, while not as satisfying as an equality, is still extremely powerful. There are also significant generalizations one can make for the mean-value inequality.