Let $E$ and $F$ be Banach spaces and $T: E → F$ a function weakly Lipschitzian then $T$ is Lipschitzian

Question

Help me with this problem:

Let $E$ and $F$ be Banach spaces and $T: E → F$ a function weakly Lipschitzian (not necessarily linear), means that $f \circ T$ is a Lipschitz function for all $f \in F^*$, where $F^*$ it is the dual of $F$. Prove that $T$ is Lipschitzian.

I tried the following:

If $f \circ T$ is a Lipschitz function, then $$\left| (f \circ T)(x)-(f \circ T)(y)\right|\leq K_f \left\|x-y \right\| \Rightarrow $$ $$\left| f (T(x)- T(y))\right|\leq K_f \left\| x-y \right\|$$

Take $f \neq0$.

Suppose that T is not Lipschitz, then exist $(x_n), (y_n) \in E$ $\ $ such that $$\left\|T(x_n)- T(y_n)\right\| > n \left\| x_n-y_n \right\| \ \forall \ n \in \mathbb{N}$$

Since $f$ is bounded, we have $$\left| f(T(x)- T(y))\right| \leq \left\| f \right\| \left\| T(x)- T(y) \right\|$$

Therefore $$\left| f (T(x_n)- T(y_n))\right|\leq K_f \left\| x_n-y_n \right\| < \frac{K_f}{n} \left\|T(x_n)- T(y_n)\right\|$$ $\Rightarrow$ $\left\|f\right\|= \inf{\left\{C>0 \ ; \ |f(x)|\leq C\left\|x\right\|, x \in F \right\}}< \frac{K_f}{n} \ \ \ \forall \ n \in \mathbb{N}$

Then $\left\|f\right\|=0 \Rightarrow f=0$, Contradiction, therefore $T$ is Lipschitzian.

Answer 1

The bound $K_f/n$ holds only for the sequence $v_n:=(T(x_n)-T(y_n))$ and not for all $x\in F$. This is why the argument does not hold.