# A subcontinuous function, which is not continuous

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

• Link to the Math SE post (from 9 months ago): math.stackexchange.com/q/3393342/279515
– user82537
Jul 18, 2020 at 17:19
• I delete it, since i didn't have any interaction on Math SE.. Jul 20, 2020 at 11:06
• Maybe you could have instead provided a community wiki answer there linking to the accepted answer here?
– user82537
Jul 20, 2020 at 12:28

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., $$e_0^*$$.
• That means if $x\in[(n+1]^{-1},n^{-1}]$ then $$f(x) = f(\frac1n) +(f(\frac1{n+1}) -f(\frac1n))\frac{x-\frac1n}{\frac1{n+1} - \frac1n}$$ Jul 18, 2020 at 9:35
• The sequence $u_k$ must be general (and converges to $0$), I don't know why you assume that $0\leq u_k\leq 1$? Jul 27, 2020 at 15:42
• What I called $u_k$ obviously wasn't what was called $u_k$ in the question: I changed this to $t_k$ to avoid confusion. Terms with $u_k<0$ are unproblematic, all terms will be $\leq 1$ eventually, and terms between $0$ and $1$ can be written $t/n+(1-t)/(n+1)$ with $0\leq t\leq 1$ so their image by $f is$te_n+(1-t)e_{n+1}\$. Jul 27, 2020 at 16:10