Suppose $E,F$ are real Banach spaces, $U \subset E$ is open and $f: U \to F$ is a differentiable. Further suppose that two points $x,y \in U$ are such that the line segment between them is contained in $U$, so that if $h$ denotes the map $\mathbb{R} \to E \, ; \, t \mapsto (1-t)x + ty$ then $h([0,1]) \subset U$.
I'm reading a proof that claims, under the hypotheses written above, there exists some $t_0 \in (0,1)$ for which the point $x_0 = h(t_0)$ satisfies
$ \| f(y) - f(x) \| \leqslant \| Df(x_0)(y - x) \| \leqslant \| Df(x_0) \| \| y - x \|$.
The proof starts off as follows: "Using the Hahn-Banach Theorem, choose $\phi$ in the conjugate space $F^*$ of $F$ so that $\| \phi \| = 1$ and $\|f(y) - f(x) \| = \phi( f(y) - f(x) )$."
From there the proof simply applies the mean value theorem to $\phi \circ f \circ h : [0,1] \to \mathbb{R}$, using the chain rule along the way.
I'm assuming by "conjugate space" he means the space $F^* = L(F,\mathbb{R})$ of continuous linear functionals $F \to \mathbb{R}$.
How does the Hahn-Banach Theorem allow the author to choose $\phi$ as they did? It might be a stupid question, but I've never used the Hahn-Banach Theorem, and frankly only heard of it a week or so ago. I've seen a proof of a formulation, which I can follow, but from what I hear there are different formulations of the theorem.
I assume $f(y)\neq f(x)$. If $f(y)=f(x)$ the inequality is trivially true. Consider the linear function $h$ define on the line $L=Vect(f(y)-f(x))$, such that $h(f(y)-f(x))=\|f(y)-f(x)\|$, it is bounded and its norm is $1$ since $h({{f(y)-f(x)}\over{\|f(y)-f(x)\|}})=1$. Hahn Banach implies you can extend $h$ to $\phi:F\rightarrow R$ such that $\phi(f(y)-f(x))=\|f(y)-f(x)\|$ and $\|\phi\|=1$.