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> 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 $T(u)$.
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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



A subcontinuous function, which is not continuous
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., $e\_0^\*$.


