Timeline for How to choose compactly supported smooth $h$ so $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ and $\int_{-3/4}^{3/4} |h(x)|^2 dx =3/2$? [closed]
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S Aug 8, 2018 at 6:25 | history | unlocked | CommunityBot | ||
S Aug 8, 2018 at 6:25 | history | locked | CommunityBot | ||
S Aug 8, 2018 at 6:25 | history | closed |
Mateusz Kwaśnicki Jan-Christoph Schlage-Puchta András Bátkai Yemon Choi Stefan Waldmann |
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Aug 5, 2018 at 22:46 | comment | added | Mateusz Kwaśnicki | Take $f$ as in the question, fix $c \geqslant 0$ and define $h(x) = \sin(\tfrac{\pi}{2} f(x+1)(1 + c i f(x+1) (1 - f(x+1))))$ for $x \leqslant 0$, $h(x) = \cos(\tfrac{\pi}{2} f(x)(1 + c i f(x) (1 - f(x))))$ for $x > 0$. | |
Aug 5, 2018 at 16:04 | comment | added | Math Learner | @MateuszKwaśnicki: Thanks a lot. I got your argument. But I'm unable to follow your last comment concerning complexed valued function $h.$ If $h$ is indeed complex valued with the desired property, why one can expect $\|h\|_{L^2}^2= 3/2$? Please can you explain bit more on this? Thanks. | |
Aug 5, 2018 at 14:55 | comment | added | Mateusz Kwaśnicki | It is the same question: simply note that $\int_{-3/4}^{3/4} |h(x)|^2 dx = \int_{-1}^1 |h(x)|^2 dx = \int_{-1}^0 |h(x)|^2 dx + \int_0^1 |h(x)|^2 dx = \int_0^1 (|h(x)|^2 + |h(x-1)|^2) dx = 1.$ (Unless $h$ is indeed complex-valued, in which case the integral can be any number in $[1, \infty)$). | |
Aug 5, 2018 at 14:47 | comment | added | Math Learner | @MateuszKwaśnicki: Thanks. This is slightly different. In the previous one change of variable does the trick. But in this I do not know how to handle the situation? | |
Aug 5, 2018 at 14:28 | comment | added | Mateusz Kwaśnicki | You asked essentially the same question yesterday or the day before yesterday. I gave you an answer in a comment and suggested Math.SE as a more appropriate site. Now I see that you deleted the previous question and asked it again. Why? | |
Aug 5, 2018 at 14:06 | history | edited | Math Learner | CC BY-SA 4.0 |
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Aug 5, 2018 at 2:12 | answer | added | Iosif Pinelis | timeline score: 2 | |
Aug 5, 2018 at 0:52 | history | edited | Math Learner | CC BY-SA 4.0 |
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Aug 5, 2018 at 0:38 | history | edited | Math Learner | CC BY-SA 4.0 |
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Aug 4, 2018 at 21:45 | review | Close votes | |||
Aug 8, 2018 at 6:25 | |||||
Aug 4, 2018 at 21:25 | history | asked | Math Learner | CC BY-SA 4.0 |