Suppose to have two Banach spaces $E$ and $F$, with $E$ reflexive. Suppose to have a continuous map $T:E \to F$ which maps bounded subsets into precompact subsets. $T$ is not assumed to be linear.
Finally, suppose to have a sequence $(e_n)_n\subset E$ weakly convergent to $e\in E$.
Then, up to subsequences, $T(e_n)$ converges strongly to $T(e)$.
Is it correct? How can I prove it? Can I drop the reflexivity assumption?
Best Answer
This is not true I'm afraid. Weak convergence and nonlinearity are fundamentally incompatible(*).
Let $F$ be finite-dimensional. Then any continuous (nonlinear) operator is bounded and thus compact. But it will not generally translate weakly convergent sequences into (weakly) convergent sequences.
Consider, e.g., the map $T \colon E \to \mathbb R$ with $T(x) = \|x\|$. Even if $E$ is the nicest of all spaces, the sequence space $\ell^2$, you will have $e_n \xrightarrow{\sigma} 0$ as $n \to \infty$ but $T(e_n) = 1$ for every $n$.
Update: To motivate the claim (*) a bit: The straightforward way you prove that bounded linear operators $T \colon E \to F$ map weakly convergent sequences in $E$ to weakly convergent sequences in $F$ is by observing that $f(Tx_n) = (f \circ T)(x_n)$ and that $f \circ T \in F'$. This approach takes you nowhere if $T$ is nonlinear.