Does one need the Hahn-Banach theorem to prove the mean value inequality for maps into a normed space? Consider the following mean value theorem:
If $f$ is a continuous mapping of $\,[a,b]$ into a normed linear space $X$, whose norm doesn't derive from an inner product, and $f$ is differentiable on $]a,b[$, there exists $c \in \,]a,b[$ such that $$\left\| \frac {f(b)-f(a)}{b-a} \right\| \le \|f'(c)\|$$
Now I ask if one can prove it by contradiction starting from the auxiliary function $$\varphi(t)=\left\| \frac {f(b)-f(a)}{b-a} \right\|(t-a)-\|f(t)-f(a)\| \qquad (\,t \in [a,b]\,)$$ My contribution is rather poor.
I note that $\varphi$ is continuous on $[a,b]\,$ and $\,\varphi(a)=\varphi(b)=0$.
Supposed $$\left\| \frac {f(b)-f(a)}{b-a} \right\| > \|f'(c)\| \qquad \forall \,c \in \,]a,b[$$ I think to prove that $\varphi$ is strictly monotonic on $[a,b]$, so that $\varphi(a) \neq \varphi(b)$, a contradiction.
Unfortunately I stop when I begin to differentiate $\|f(t)-f(a)\|$ to find $\varphi'$.
Does someone know more ? Thanks in advance.
meaning of the title:
Typically the proof uses an inner product if it is available.
If it doesn't exist, one appeals to the Hahn-Banach theorem to find a functional as a substitute of the inner product (see theorem 1.2.1, p. 3 in this book).
 A: A proof without using a norming linear functional for $f(b)-f(a)$ can be found in the 2nd edition of Rudin's Principles of Mathematical Analysis. It's stated for Euclidean space, but it works the same for all normed spaces. (Trivia: when the 2nd  edition was translated into Russian, the translator added a footnote with the linear-functional proof; subsequently, Rudin replaced  the proof in the 3rd edition).
The idea is to use the triangle inequality to show that $f$ must have certain rate of change on smaller and smaller subintervals; then take intersection over a chain of subintervals. Care must be taken to ensure the intersection is an interior point of $[a,b]$.
Following Rudin's notation, let $L=(b-a)/3$ and $M=\|f(b)-f(a)\|$. The function $g(x)=\|f(x+L)-f(x)\|$ satisfies $g(a)+g(a+L)+g(a+2L)\ge M$ by the triangle inequality. It follows  that $g(x)\ge M/3$ for some $x_1\in (a,a+2L)$, for otherwise we would have
$$g(a)\le \frac{M}{3}, \quad  g(a+L)<\frac{M}{3}, \quad  g(a+2L) \le \frac{M}{3}$$
Replace out original interval $I_0=[a,b]$ with $I_1=[x_1,x_1+L]$ and repeat. (Note that $I_1\subset (a,b)$.)
Let $x_*=\bigcap_n I_n$. The endpoints of $I_n$ form two sequences $s_n,t_n$ converging to $x_*$ from opposite directions. Since $$\frac{\|f(t_n)-f(s_n)\|}{|t_n-s_n|}\ge \frac{\|f(b)-f(a)\|}{|b-a|}$$   it follows that $$\|f'(x_*)\|\ge \frac{\|f(b)-f(a)\|}{|b-a|}$$

Note
The more intuitive approach of repeated bisection does not guarantee we get an interior point. However, after $I_1$ was obtained above, subsequent steps could be bisection.
