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It is known that we may choose smooth $f:\mathbb R \to [0,1]$ such that $f(x)=1$ if $x\geq \frac{3}{4} $ and $f(x)=0$ if $x\leq -\frac{3}{4}+1.$

Define $h(x)= \sin (\frac{\pi}{2} f(x+1))$ if $x\leq 0$ and $h(x)= \cos (\frac{\pi}{2} f(x))$ if $x\geq 0.$

We note that support of $h$ is $[-3/4, 3/4]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ $\|h\|_{L^2}=1,$ and this $h:\mathbb R \to \mathbb C$ is smooth.

My Question: Can we expect to choose $h:\mathbb R \to \mathbb C$ such that support of $h$ is $[-3/4, 3/4]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ and $$\|h\|_{L^2}^2=\int_{-3/4}^{3/4} |h(x)|^2 dx =3/2?$$

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  • $\begingroup$ 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? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 5, 2018 at 14:28
  • $\begingroup$ @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? $\endgroup$
    – Math Learner
    Commented Aug 5, 2018 at 14:47
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    $\begingroup$ 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 $[1, \infty)$). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 5, 2018 at 14:55
  • $\begingroup$ @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. $\endgroup$
    – Math Learner
    Commented Aug 5, 2018 at 16:04
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    $\begingroup$ 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 $x > 0$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 5, 2018 at 22:46

1 Answer 1

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This is impossible to do. Indeed, suppose that the support of $h$ is $[-1/3,1/3]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1]$. Then for all $x\in[0,1/3]$ we have $x-1<-1/3$, whence $h(x-1)=0$ and $h^2(x)=1-h^2(x-1)=1$. Similarly, for all $x\in[2/3,1]$ we have $h(x)=0$, whence $h^2(x-1)=1-h^2(x)=1$, so that $h^2(y)=1$ for all $y\in[2/3-1,1-1]=[-1/3,0]$. So, $h^2=1$ on $[-1/3,1/3]$ and hence $\|h\|_{L^2}^2=\int_{-1/3}^{1/3} h^2(x)\, dx =2/3\ne3/2$.

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  • $\begingroup$ Thanks. But we need to start with support of $h$ is $[-3/4, 3/4].$ Can we say anything for this. PS: If support of $h$ is $[-1/3, 1/3]$ then $h^2(x)+h^{2}(x-1)=1$ for all $x\in [0,1]$ is NOT possible. $\endgroup$
    – Math Learner
    Commented Aug 5, 2018 at 2:27
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    $\begingroup$ In your question, it was assumed that the support of $h$ is $[-1/3,1/3]$, and this is what is assumed in my answer as well. $\endgroup$
    – Iosif Pinelis
    Commented Aug 5, 2018 at 2:38
  • $\begingroup$ Oh, Thanks. It was typo. I have edited the question. $\endgroup$
    – Math Learner
    Commented Aug 5, 2018 at 14:08
  • $\begingroup$ Can we say anything for the $[-3/4, 3/4 ]$ supported function. $\endgroup$
    – Math Learner
    Commented Aug 5, 2018 at 14:09

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