Talk:Strongly measurable function

Confusion with strong / uniform measurability vs. strong / uniform continuity of semigroups

I have some problems with interpreting these two statements in the Wikipedia article:

1. A semigroup of linear operators can be strongly measurable yet not strongly continuous.
2. It is uniformly measurable if and only if it is uniformly continuous, i.e., if and only if its generator is bounded.

In the following, I want to explain where the problems are hidden. In my eyes, when we speak of a one-parameter semigroup then this means a homomorphism $T:(0,\infty )\to M$ of the semigroup $((0,\infty ),+)$ (the OPEN interval) in some other semigroup $(M,\cdot )$ (usually a Banach algebra or a space of linear bounded operators on a Banach space with the strong topology). The semi-open interval $[0,\infty )$ has more structure: it is a monoid with identity $0$. If $M$ carries a topology then this distinction is important when we regard continuity properties of semigroups. Basically, there are three levels of definitions of what is meant by "continuity of a semigroup" in the literature:

1. continuity of the semigroup homomorphism $T:(0,\infty )\to M$
2. existence of the limit $L:=\lim _{t\to 0+}T(t)$ (in this case extend $T$ at $0$ by $T(0):=L$) and
3. when extending $T(0):=I$ (in case $M$ is also a monoid) then whether $L=I$ ("continuity at 0").

Hille and Phillips show in their monumental treatise "Functional Analysis and Semi-Groups" that if $X$ is a Banach space, $L(X)$ the space of bounded linear operators equipped with the strong operator topology then if the one-parameter semigroup $T:(0,\infty )\to L(X)$ is Bochner measurable (i.e. strongly measurable) then $T$ is strongly continuous (so this is "level 1"-continuity). The limit $L=\lim _{t\to 0+}T(t)$ need not exist ("level 2"-continuity) and even if it exist it need not be equal to $I$ ("level 3"-continuity). This is what Davies (in the reference of this Wikipedia article) shows in his Example 6.1.10. Similarly, if $B$ is a Banach algebra (e.g. $L(X)$ equipped with the operator norm (giving it the uniform topology)) then if $T:(0,\infty )\to B$ is Bochner measurable (i.e. uniformly measurable in case $B=L(X)$) then $T$ is also (uniformly) continuous on $(0,\infty )$. Again, the (uniform) limit $L=\lim _{t\to 0+}T(t)$ need not exist and even if it exists it need not be equal to $I$.

So, the two statements above should be read as follows:

1. A semigroup of linear operators can be strongly measurable (and thus strongly continuous in $(0,\infty )$) yet not strongly continuous in $0$.
2. It is uniformly measurable if and only if it is uniformly continuous in $(0,\infty )$. It is uniformly continuous in $0$ if and only if it is uniformly continuous in $[0,\infty )$ (and thus uniformly measurable) if and only if its generator is bounded.

Yadaddy ag (talk) 11:37, 30 August 2016 (UTC)

Requested move 4 January 2021

The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review after discussing it on the closer's talk page. No further edits should be made to this discussion.

The result of the move request was: Moved (non-admin closure) (t · c) buidhe 12:07, 11 January 2021 (UTC)

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Strongly measurable functions → Strongly measurable function – Per WP:PLURAL, there is no reason for the article to use the plural form. 𝟙𝟤𝟯𝟺𝐪𝑤𝒆𝓇𝟷𝟮𝟥𝟜𝓺𝔴𝕖𝖗𝟰 (𝗍𝗮𝘭𝙠) 01:23, 4 January 2021 (UTC)

Indeed; why did you open an RM instead of just moving it youself? --JBL (talk) 14:35, 4 January 2021 (UTC)

• Since it's already listed as a requested move, might as well formally support. kennethaw88 • talk 23:13, 8 January 2021 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.