Aufgaben:Problem 8

Concerning part a)

First thing: It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \).

Wouldn't it be much easier to just:

$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$

and then use Bernoulli-L'Hopital to get:

$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$

Second thing: Nevermind, confused myself...

Greetings A.

I didn't write that: I meant that the limit of the different quotient is the derivative. I was a bit inconsistent with the variable x (now it's fixed, I've used a y). If you prefer Bernoulli it's the same for me. Nick

Okay, I see what you meant now. I personally find it a bit confusing. But I guess it's a matter of taste.

A.

To Part b)

Question:

Since we had an whole exercise sheet on Lebesgue-convergence and i would assume they would ask of us to find an upper bound function g to prove that we're allowed to use that theorem. Why don't you do that? (It might be obvious but those just learning it by heart won't notice.)

Answer:

The upper bound function is for dominated convergence, not for monotone convergence. But I agree it is not so clearly stated, I'll fix it, thanks

Again to part b) As I remember (and according to Wikipedia http://de.wikipedia.org/wiki/Satz_von_der_monotonen_Konvergenz) the function to which \(f_k\) converges, has to be \(\in L^1 \). That should maybe shown too?

Trubo-Warrior (talk) 12:20, 5 January 2015 (CET)

Since we all like alternative versions here on the wiki, and I've encountered some confusion among my friends about the limit-thing we discussed above, I felt so free to put the alternative version in. Best, A.

As for the alternative version: yes, it was a good idea, so everybody can choose the one she/he prefers. For \( L^1 \): it is, since \( e^{-x} \) is in Schwartz space (over positive real numbers), but I've added it to clarify. I didn't have this requirement in the formulation of the exercise class, but I think you're right. Cheers, 'Nick (talk) 17:24, 12 January 2015 (CET)'

At the very end, don't think f(x) is in Schwartz Space, because t can take other values than integrers. I would prove that f(x) is in L1 separately: $$ \int_0^\infty \left| x^{t-1}e^{-x} \right| \, \mathrm{d}x = \int_0^\infty x^{t-1}e^{-x} \, \mathrm{d}x = \int_0^1 x^{t-1}e^{-x} \, \mathrm{d}x + \int_1^\infty x^{t-1}e^{-x} \, \mathrm{d}x $$

$$ \int_0^1 \! x^{t-1}e^{-x} \, \mathrm{d}x \leq \int_0^1 \! x^{t-1} \, \mathrm{d}x = \frac{1}{t}$$ because \( t>0 \), the integrant is positive and \(e^{-x} \leq 1\). and for the second integral, if \( 0 < t < 1 \) $$ \int_1^\infty x^{t-1}e^{-x} \, \mathrm{d}x \leq \int_1^\infty e^{-x} = \frac{1}{e}$$ and if \(t > 1 \) we can partially integrate until we get an integral of the above form multiplied by a constant (the additional terms don't go to zero but are less than infinity)

Which all together means that $$ \int_0^\infty \left| x^{t-1}e^{-x} \right| \, \mathrm{d}x < \infty$$ or maybe it is enough to say that the gamma function is analytic for \(t>0\) , but I wouldn't count on it "Carl (talk) 16:31, 15 January 2015 (CET)"

I agree that the story with Schwartz space doesn't really make sense, since we haven't even defined it for sets other than R^n; despite that, the version of the monotone convergence theorem that I had on the notes actually agrees with Lebesgue monotone convergence theorem, which doesn't require f to be in L1 (in fact, it states that it follows from the rest: http://en.wikipedia.org/wiki/Monotone_convergence_theorem). In any case, thanks for the proof :) 'Nick (talk) 21:38, 15 January 2015 (CET)'