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: 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 $$I$$.
 * 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").

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  (t &#183; c)  buidhe  12:07, 11 January 2021 (UTC)

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)