Talk:Nachbin's theorem

Borel transform

Why is Borel transform redirected to this page, which has some discussion of generalized Borel transforms? Surely there should be a page on the ordinary Borel transform, referred to here. Unfortunately I'm not qualified to write such a page. --Dylan Thurston 04:01, 15 March 2006 (UTC)

The ordinary Borel transform is the same thing as the Laplace transform, which this article explains. Since no one calls the Laplace transform the Borel transform any more, it seemed more appropriate to redirect here, where the usage is still in vogue. But maybe I'm mistaken. linas 02:10, 16 March 2006 (UTC)

Question

Can Borel and Borel-generalized transforms be applied when the function f(t) has no exponential order for example...$f(x)=e^{x^{2}}$ or simply if $|f(t)|>Ce^{Mt}$ for real C and M ?.. (unsigned, August 2006)

Uhh, does it satisfy conditions of psi-type?linas (talk) 04:37, 21 November 2010 (UTC)

"Direction" misleading.

The text "... to emphasize that the limit must hold in all directions θ." disconcerts me. In general, 2D limits are much much stronger than simply having 1D limits hold in a collection of directions. To wit: A function can have a complete set of directional derivatives through the origin, and yet be discontinuous at the origin. An example specific to this case, we could define the following $f$ with support only along a hyperbola on the complex plane, in terms of the Kronecker delta:

$f(a+bi)=\delta _{a,{\frac {1}{b}}}e^{\sqrt {a^{2}+b^{2}}}$

Observe that the limit of $f(re^{i\theta })$ when $r\to \infty$ for any given fixed $\theta$ is zero, because the parameterized path only hits at most a single point of $f$'s support! We would therefore conclude that $f$ has exponential type $-\infty$, or something else. However, if we take the true 2D limit with $\theta$ included and not fixed, then the limit is $\infty$, and the true exponential type of $f$ is correctly determined to actually be one.

Now, from my reading of Exponential_type#Formal_definition, the 2D limit is what was intended, please correct me if I'm wrong. One fix is to change the definition over to the more formal limit of $|z|\to \infty$. Alternatively, we could just change it to "... the limit must hold in all directions θ at once." which hints at the 2D limit. Thoughts? 18:38, 3 November 2014 (UTC)