A subcontinuous function, which is not continuous

Question

Let $E$ be a Banach space and $T: E\rightarrow E$ be a mapping. $T$ is said to be subcontinuous if for any sequences $(u_n)_{n\in\mathbb{N}}$ in $E$ that converge strongly to $u$ the sequence $(T(u_n))_{n\in\mathbb{N}}$ converges weakly to $T(u)$.

I am looking for a subcontinuous function which is not continuous. I thought I can find something in $l^2$, but I didn't.

I already posted it on math.stackexchange, but to no avail

Thank you

Answer 1

Let $(e_n)$ be the standard orthonormal basis of $\ell^2$: recall that, as a sequence, $(e_n)$ converges weakly to $0$. Now define a map $f\colon\mathbb{R} \to \ell^2$ by $f(\frac{1}{n})=e_n$ and $f(t)=0$ if $t\leq 0$, and interpolating linearly between $\frac{1}{n}$ and $\frac{1}{n+1}$: this is continuous at every point except at $0$ where it is subcontinuous (because any sequence of the form $t_k\, e_{n_k} + (1-t_k)\, e_{n_k+1}$ with $0\leq t_k\leq 1$ and $n_k\to+\infty$, converges weakly to $0$ when $k\to+\infty$). If you want a subcontinuous function $\ell^2\to\ell^2$, just right-compose $f$ with a nonzero continuous linear map $\ell^2\to\mathbb{R}$, e.g., $e_0^*$.