Lipschitz map between metric and normed spaces Let be $F:(X,d)\to V$ a map between $(X,d)$ metric space and $V$ normed space, such that for each $f\in V'$ (linear and continuous), $f\circ F$ is lipschitz map. Show that $F$ is a Lipschitz map.
I try something like that:
Proof: For every $f\in V'$, there are $C_f, L_f >0$ such that, for every $x,y\in X$
$$|f(x)|\le C_f |x|$$
and
$$|f(F(x)-F(y))|=|f(F(x))-f(F(y))|\le L_f d(x,y).$$
If $F(x)-F(y)\neq 0$, by Hahn-Banach theorem, there is $f_{xy}\in V'$ such that
$$f_{xy}(F(x)-F(y))=|F(x)-F(y)|.$$
So,
$$|f_{xy}(F(x))-f_{xy}(F(y))|=|F(x)-F(y)|\le L_{f_{xy}}d(x,y).$$
I stopped here. How can I to standardize the constant $L_{f_{xy}}$? Is it the good way to solve the problem? Any help?
I appreciate.
 A: Lets assume $F$ is not Lipschitz. I also want to use the notation $F(x,y)=F(x)-F(y)$ because it is very much more convenient.
If $F$ is not Lipschitz, $F(x,y)/d(x,y)$ is unbounded on $X\times X - D$ where $D$ is the diagonal. This means that we can find a sequence $(x_n,y_n)$ so that $\|F(x_n,y_n)\|≥ 2^{2n}d(x_n,y_n)$. (And $x_n\neq y_n$.)
With Hahn Banach get a functional $f_n$ so that $f_n(F(x_n,y_n))=2^{-n}\|F(x_n,y_n)\|=2^n d(x_n,y_n)$ and $\|f_n\|=2^{-n}$. Then
$$f:=\sum_{k=1}^\infty g(k)\ f_k$$
with $g(k) \in S^1$ (the complex unit circle) is a linear functional with $\|f\|≤1$ from the geometric series. I want to choose $g(n)$ so that $\sum_k^{n-1} g(k)\ (f_k\circ F)(x_n,y_n)$ has the same argument as $g(n)\ (f_n \circ F)(x_n,y_n)$. This consideration allows me to get an upper bound on the following expression:
$$|(f\circ F)(x_n,y_n)|≥\left|\ \left|\sum_{k=1}^{n}g(k)\ (f_k\circ F)(x_n,y_n)\right| -\left|\sum_{k=n+1}^{\infty}g(k)\ (f_k\circ F)(x_n,y_n)\right|\ \right|$$
Now the first expression has absolute value $≥(f_n \circ F)(x_n,y_n)=2^n d(x_n,y_n)$ since $g(k)$ has been chosen so that the first $n-1$ parts of the sum have the same argument as the $n$-th part, and thus the addition of the two does not make the modulus smaller.
The second expression has absolute value $≤\sum_{k=n+1}^\infty 2^{-k}\|F(x_n,y_n)\|=2^{n-2}d(x_n,y_n)$, which is smaller than the first expression, so we can leave the outermost $|\cdot|$ out and the entire thing is larger than:
$$|(f\circ F)(x_n,y_n)|≥2^n(1-2^{-2})d(x_n,y_n)≥2^{n-1}d(x_n,y_n)$$
So $(f\circ F)(x,y)/d(x,y)$ is not bounded on $X \times X - D$, which is a contradiction to it being Lipschitz.
