Talk:Aufgaben:Problem 8

There are two things which I'm not sure about: whether in lemma 2 we can use Lebesgue (we need a discrete sequence of dominated functions, whereas h is a continuous variable; can we change h to 1/n and let \( n \rightarrow \infty \) instead of \( h \rightarrow 0 \) without any problem?) and how lemma 2 exactly justifies the passage with \( \color{red}{*} \)

Some derivative signs should be partial derivatives, but I guess it's not such a problem...

Best, Nick

So, should be okay now... I think the induction step at \( \color{red}{*} \) is clear enough.

Best, A.

Yes, we can. If the limit of f(h) exists for h -> 0, then for every discrete sequence hn f(hn) will have the same limit for n -> infinity. So I guess you can just substitute them and use Lebesgue at will. And \( \color{red}{*} \) isn't even an induction step... that's just applying the lemma we just proved. However, I believe in Lemma 2 the 1/sqrt(2*pi) on the right hand side of the equality is superfluous. It's not being used in the proof, as far as I can see, and it comes back to bite you in \( \color{red}{*} \).

Cheers, Nathalie

Ok, for the lemma and for the corollary it's all clear. I still can't see the induction step from the first derivative to the n-th derivative, though: the proof is for a specific function and doesn't work anymore when you have a derivative of this function inside the integral, instead of the function itself. In fact, maybe we could forget all about lemma 2 and its corollary by using something of this kind: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign But the problem is that we have \( \infty \) as integration bounds... I wonder whether we really have to justify all about this passage or whether we can just do it "the physician's way" (with no offense for anyone ;), because we're dealing with continuous functions with continuous derivatives and so on.

Cheers, Nick