Meaning of l.i.m.?

Question

I've been reading Akhiezer's "The Classical Moment Problem" recently and came across the expression (in Chapter 2): $$f(u) = Ux = \operatorname*{l.i.m.}_{\sigma(u)}\sum_0^n x_kP_k(u)$$ Does anyone know what l.i.m. means? I assume it means limit but he also uses "lim" throughout the text, so it must not mean a typical limit. I am also not sure what he'd be taking the limit of.

In case you want some context, I believe U is supposed to be an isometric operator that is generated by some generalised Fourier coefficients $x_k$ and orthogonal polynomials $P_k$, $\sigma(u)$ is a solution to the classical moment problem.

Answer 1

In the same page of the book, it is written that

$$ f(u) = \operatorname*{l.i.m.}_{\sigma(u)} f_n(u) $$

simply denotes the $L^2(\sigma)$-limit of the sequence $(f_n)$. Probably a more illuminating notation would be by

$$ f = L^2(\sigma)\text{-}\lim_{n\to\infty} f_n, \hspace{3em} f = \lim_{n\to\infty} f_n \quad \text{in } L^2(\sigma), \hspace{3em} f \xrightarrow[ \ n\to\infty \ ]{L^2(\sigma)} f $$

to match the usual limit notation as well as clarify which topological vector space we are working on. Then the statement simply tells that

$$ \sum_{k=0}^n x_k P_k(u) \xrightarrow[ \ n\to\infty \ ]{L^2(\sigma)} Ux $$

as $n\to\infty$

Answer 2

After searching the web, googling the book, and looking at articles related to the book, I'm pretty sure it's either going to be defined in the text somewhere, or it's a typo. My guess is the latter, because I've never seen that weird notation before.

Answer 3

A reasonable interpretation would be limit in mean.