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How to choose compaclty supported smooth complex valued function satisfying $|h|^2(x)+|h|^2(x-1)=1$?
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We know the following trigonometric identities
$sin^2x + cos^2 x =1$ for all $xin mathbb R,$ and $sin^2 z + cos^2 z =1$ for all $zin mathbb C.$
It is known that we may choose smooth $f:mathbb R to [0,1]$ such that $f(x)=1$ if $xgeq 1 $ and $f(x)=0$ if $xleq 0.$ Define $h(x)= sin (fracpi2 f(x+1))$ if $xleq 0$ and $h(x)= cos (fracpi2 f(x))$ if $xgeq 0.$ We note that support of $h$ is $[-1, 1]$ and $h^2(x)+ h^2(x-1)=1$ for all $xin [0,1],$ and $h:mathbb R to mathbb R$ is smooth.
We note that this $h$ is real valued, my curiosity is, what happens if we want $h$ to be indeed complex valued.
My Question: Can we choose complex valued function $h:mathbb R to mathbb C$ such that the support of
$h$ is $[-1, 1]$ (or say $[-a, a]$ for some $a>0$) and $|h|^2(x)+ |h|^2(x-1)=1$ (or say $=b$ for some $b>0$) for all $xin [0,1],$ and $h:mathbb R to mathbb C$ is smooth.
real-analysis complex-analysis functions examples-counterexamples
asked Aug 6 at 16:02
Math Learner
1619
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0
down vote
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We know the following trigonometric identities
$sin^2x + cos^2 x =1$ for all $xin mathbb R,$ and $sin^2 z + cos^2 z =1$ for all $zin mathbb C.$
It is known that we may choose smooth $f:mathbb R to [0,1]$ such that $f(x)=1$ if $xgeq 1 $ and $f(x)=0$ if $xleq 0.$ Define $h(x)= sin (fracpi2 f(x+1))$ if $xleq 0$ and $h(x)= cos (fracpi2 f(x))$ if $xgeq 0.$ We note that support of $h$ is $[-1, 1]$ and $h^2(x)+ h^2(x-1)=1$ for all $xin [0,1],$ and $h:mathbb R to mathbb R$ is smooth.
We note that this $h$ is real valued, my curiosity is, what happens if we want $h$ to be indeed complex valued.
My Question: Can we choose complex valued function $h:mathbb R to mathbb C$ such that the support of
$h$ is $[-1, 1]$ (or say $[-a, a]$ for some $a>0$) and $|h|^2(x)+ |h|^2(x-1)=1$ (or say $=b$ for some $b>0$) for all $xin [0,1],$ and $h:mathbb R to mathbb C$ is smooth.
real-analysis complex-analysis functions examples-counterexamples
asked Aug 6 at 16:02
Math Learner
1619
I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We know the following trigonometric identities
$sin^2x + cos^2 x =1$ for all $xin mathbb R,$ and $sin^2 z + cos^2 z =1$ for all $zin mathbb C.$
It is known that we may choose smooth $f:mathbb R to [0,1]$ such that $f(x)=1$ if $xgeq 1 $ and $f(x)=0$ if $xleq 0.$ Define $h(x)= sin (fracpi2 f(x+1))$ if $xleq 0$ and $h(x)= cos (fracpi2 f(x))$ if $xgeq 0.$ We note that support of $h$ is $[-1, 1]$ and $h^2(x)+ h^2(x-1)=1$ for all $xin [0,1],$ and $h:mathbb R to mathbb R$ is smooth.
We note that this $h$ is real valued, my curiosity is, what happens if we want $h$ to be indeed complex valued.
My Question: Can we choose complex valued function $h:mathbb R to mathbb C$ such that the support of
$h$ is $[-1, 1]$ (or say $[-a, a]$ for some $a>0$) and $|h|^2(x)+ |h|^2(x-1)=1$ (or say $=b$ for some $b>0$) for all $xin [0,1],$ and $h:mathbb R to mathbb C$ is smooth.
real-analysis complex-analysis functions examples-counterexamples
asked Aug 6 at 16:02
Math Learner
1619
We know the following trigonometric identities
$sin^2x + cos^2 x =1$ for all $xin mathbb R,$ and $sin^2 z + cos^2 z =1$ for all $zin mathbb C.$
It is known that we may choose smooth $f:mathbb R to [0,1]$ such that $f(x)=1$ if $xgeq 1 $ and $f(x)=0$ if $xleq 0.$ Define $h(x)= sin (fracpi2 f(x+1))$ if $xleq 0$ and $h(x)= cos (fracpi2 f(x))$ if $xgeq 0.$ We note that support of $h$ is $[-1, 1]$ and $h^2(x)+ h^2(x-1)=1$ for all $xin [0,1],$ and $h:mathbb R to mathbb R$ is smooth.
We note that this $h$ is real valued, my curiosity is, what happens if we want $h$ to be indeed complex valued.
My Question: Can we choose complex valued function $h:mathbb R to mathbb C$ such that the support of
$h$ is $[-1, 1]$ (or say $[-a, a]$ for some $a>0$) and $|h|^2(x)+ |h|^2(x-1)=1$ (or say $=b$ for some $b>0$) for all $xin [0,1],$ and $h:mathbb R to mathbb C$ is smooth.
real-analysis complex-analysis functions examples-counterexamples
asked Aug 6 at 16:02
Math Learner
1619
asked Aug 6 at 16:02
Math Learner
1619
asked Aug 6 at 16:02
Math Learner
1619
asked Aug 6 at 16:02
Math Learner
1619
1619
I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32
add a comment |Â
I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32
I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32
I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32
add a comment |Â
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I suppose you mean a non-constant function?
– Jacob Mazor
Aug 6 at 17:32