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

  • $\begingroup$ Link to the Math SE post (from 9 months ago): math.stackexchange.com/q/3393342/279515 $\endgroup$
    – user82537
    Jul 18, 2020 at 17:19
  • $\begingroup$ I delete it, since i didn't have any interaction on Math SE.. $\endgroup$
    – Motaka
    Jul 20, 2020 at 11:06
  • 1
    $\begingroup$ Maybe you could have instead provided a community wiki answer there linking to the accepted answer here? $\endgroup$
    – user82537
    Jul 20, 2020 at 12:28

1 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., $e_0^*$.

  • $\begingroup$ What the you mean exactly by : interpolating linearly between... $\endgroup$
    – Motaka
    Jul 18, 2020 at 9:28
  • 2
    $\begingroup$ That means if $x\in[(n+1]^{-1},n^{-1}]$ then $$f(x) = f(\frac1n) +(f(\frac1{n+1}) -f(\frac1n))\frac{x-\frac1n}{\frac1{n+1} - \frac1n} $$ $\endgroup$
    – username
    Jul 18, 2020 at 9:35
  • $\begingroup$ The sequence $u_k$ must be general (and converges to $0$), I don't know why you assume that $0\leq u_k\leq 1$? $\endgroup$
    – Motaka
    Jul 27, 2020 at 15:42
  • $\begingroup$ 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 $te_n+(1-t)e_{n+1}$. $\endgroup$
    – Gro-Tsen
    Jul 27, 2020 at 16:10

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