As you know, I’m not much for shortening already published and working regexes here, but in this case I think it is worthwhile as it makes what is being done much clearer.

After some tests, I think that the regex may even be shortened :

SEARCH (\w)(\w*)

REPLACE $1\L$2

Indeed, the unique ASCII character ( so, < \x{0080} ), which is, either, a word character and a punctuation character is the Low Line symbol _ ( \x{005F} ). But, as the regex \w* will catch the greatest amount of word characters, it will include all possible _ symbols, anyway ! Thus, the punctuation character, after a word, will be, necessarily, a character different from _ :-))

Moreover, as the \w and \s sets of characters have no common element, neither, the ending part (?:\s|[[:punct:]])* is useless !

Cheers,

guy038

Then hit “Replace all”

Teardown:
(\<.) Selects the first Character following a word beginning and makes it group $1
(\w*) Selects all following word characters and makes it group $2
((?:\s|[[:punct:]])*) Selects all following space or punctuation characters and makes it group $3

Replace by Group $1 as is followed by group $2 converted to lowercase followed by group $3as is. Repeat for all subsequent words.