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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.

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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 $\|\phi\|=1$.

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  • $\begingroup$ So $L$ is the span in $F$ of the fixed vector $z = f(y) - f(x)$, but how is $h$ defined? Are you using the fact that $L$ has a basis consisting of a single element $z$, assigning $h:z \mapsto \| f(y) - f(x) \| \in \mathbb{R}$ and then extending $h$ to a map $L \to \mathbb{R}$ linearly? $\endgroup$ – joeb Feb 1 '17 at 2:38
  • $\begingroup$ exactly, I am using this fact. $\endgroup$ – Tsemo Aristide Feb 1 '17 at 2:39
  • $\begingroup$ I think I get it now. If $\tilde{z} = tz \in L$ is such that $1 =\| \tilde{z} \| = |t| \| z \|$ then $| h(\tilde{z}) | = |th(z)| = |t||z| = 1$ so that $\| h \| = \sup \{|h(\tilde{z})| : \| \tilde{z} \| = 1 \} = \sup\{1\} = 1$. Then apply the Hahn-Banach Theorem as you said. Thanks so much $\endgroup$ – joeb Feb 1 '17 at 2:48

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