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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 2 years |
| seen | yesterday | |
| stats | profile views | 64 |
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May 23 |
awarded | Yearling |
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May 9 |
awarded | Caucus |
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Apr 16 |
comment |
Bernoulli shift on $S^\mathbb{Z}$ First prove it for cylinders. Then prove it for finite unions of cylinders. I think then you can use an approximation argument: approximate (in symmetric difference) an arbitrary measurable set with finite union of cylinders. |
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Apr 16 |
comment |
Nonconstant linear functional on a topological vector space is an open mapping You are assuming $f$ is continuous, which is not part of Rudin's statement. |
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Apr 16 |
revised |
Nonconstant linear functional on a topological vector space is an open mapping added 21 characters in body |
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Apr 16 |
revised |
Nonconstant linear functional on a topological vector space is an open mapping Fixed some typos. |
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Apr 16 |
suggested | suggested edit on Nonconstant linear functional on a topological vector space is an open mapping |
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Apr 16 |
awarded | Benefactor |
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Apr 15 |
revised |
Nonconstant linear functional on a topological vector space is an open mapping added 7 characters in body; edited title |
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Apr 13 |
accepted | Weakest hypothesis for integration by parts |
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Apr 13 |
comment |
Weakest hypothesis for integration by parts Can you provide the proof of $G$ being absolutely continuous? Also what assumptions do you need to integrate the right side of equation (2)? |
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Apr 12 |
comment |
Why is this function Lipschitz? $f(x)=x^{1/2}$ on $[0,1]$ satisfies your inequality with $C=1/2$, but it's not Lipschitz. |
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Apr 12 |
comment |
A Van der Corput style inequality for highly oscillatory integrals What if instead of $f'>0$, one had $|f'|>0$ as in the original van der Corput lemma? |
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Apr 12 |
awarded | Promoter |
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Apr 12 |
revised |
Weakest hypothesis for integration by parts added 29 characters in body; edited title |
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Apr 11 |
comment |
An equivalent condition for strong-mixing The answer seems to be No. The question is answered here: mathoverflow.net/questions/125245/silly-question-about-mixing |
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Apr 5 |
comment |
Identity involving partial sums of Fourier series Then sum over $k$ noting that if $y=x-x'$: $$\sum_{k=0}^{N-1} e^{i (k+1/2)y} - e^{-i (k+1/2) y} = \frac{e^{iNy}-1}{e^{iy/2}-e^{-iy/2}} - \frac{e^{-iNy}-1}{-e^{iy/2}+e^{-iy/2}} = \frac{e^{iNy}-2+e^{-iNy}}{e^{iy/2}-e^{-iy/2}} $$ Then we put the expression back into the integral and convert exponentials into sin functions. |
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Apr 5 |
comment |
Identity involving partial sums of Fourier series There are no telescopic sums. But, one could write \begin{split} S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \sum_{n=-k}^k e^{i k (x-x')} \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1)(x-x')} - e^{-i k (x-x')}}{e^{i (x-x')} -1}\\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1/2)(x-x')} - e^{-i (k+1/2) (x-x')}}{e^{i (x-x')/2} -e^{-i (x-x')/2}}, \end{split} where in the last step we multiplied and divided by $e^{-i (x-x')/2}$ to make exponentials symmetric. [continued in the next comment] |
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Apr 5 |
accepted | Identity involving partial sums of Fourier series |
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Apr 5 |
comment |
Identity involving partial sums of Fourier series Thank you Ron. I think there is a bit of redundancy in your solution. It seems that you turn exponential into sin and then sin back into exponential. |
