Talk:Aufgaben:Problem 8

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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 (apparently no one has done it in the other groups) 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

Thanks a lot for completing it! Now I think there's really everything, and lemma 2 in this more general form could also be useful for some other exercise. As far as I'm concerned (because I have to correct it), now it's proof-read. Shall I make the tick in the main page or is somebody in particular responsible for all the final checks?

'Nick (talk) 00:11, 5 January 2015 (CET)'

Have you guys considered just using induction to show that \( \hat \Phi_n(k) = (-i)^n\Phi_n(k)\) ? If not i could write an alternative solution...

Best,

"Carl (talk) 14:47, 8 January 2015 (CET)"


Yes, I found some solution on the internet similar to the one of your group at the time I had to solve the exercise. At the end I decided to extend the solution in the Felder-Script. Your solution looks right and elegant to me, I'm only not quite sure whether one can just use properties (2) and (3) of your solution without proof. But I'd be glad if you shared your solution here on the wiki.

Best, A.

thanks. what properties 2,3 do you mean? "Carl (talk) 14:19, 9 January 2015 (CET)"

I've heard there should be a really short version of the exercise which makes use of the proposition at page 96 in the script Fourier_ScahwartzAdded which states \( (x \phi) \hat \: (k) = i d_k \hat \phi (k) \) and \( (d_x \phi) \hat \: (k)=ik \hat \phi (k) \) (under some conditions), but I couldn't get hold of it so far... Has anyone seen it? 'Nick (talk) 21:58, 15 January 2015 (CET)'


Good, that you're bringing this up. I actually was just thinking of making the boring technical Lemma 2 easier by just using that property and comparing. I'm putting that in the proof. And I think you could like the solution of group Y in the dropbox (that's by the way the one I was referring to you, Carl). - A.