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This is taken from Conway's A course in functional Analysis (p. 98, Exercise 9):

If $(S,d)$ is a metric space and $X$ is a normed space, show that if $f:S\rightarrow X$ is a function such that for all $L\in X^*$ (the continuous dual of $X$) $L\circ f:S\rightarrow\mathbb C$ is Lipschitz, then so is $f$.

I tried deriving the claim using the Uniform Boundedness Principle but it seems the wrong tool, for to apply it I somehow need to be looking at a Banach space and bounded operators from it. The natural choice seems to consider $X^*$ and $A_s\in\mathcal B(X^*,\mathbb C)$ given by $A_s(L)=L(f(s))$ for any $s\in S$, but both the fact that the Lipschitz constants can depend on $L$ and no obvious way of relating a bound on $||A_s||$ to $|f|$ make me a little bit puzzled.

Any pointer would be greatly appreciated, thanks!

user3157436
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1 Answers1

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Define $$A=\left\lbrace\frac{f(x)-f(y)}{d(x,y)}\colon x,y\in S \,, \, x\neq y \right\rbrace \subseteq X $$ then $A$ is weakly bounded.

user90803
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