Mixing
Compactly supported $f in L^1$. Then $hat f$ can be extended to an holomorphic function on $mathbb C$
$begingroup$
Compactly supported $f in L^1$. Then $hat f$ can be extended to an holomorphic function on $mathbb C$, where $hat f$ is the fourier transform of f.
I tried the following
$hat f(x)=frac{1}{sqrt{2 pi}}int_{mathbb{R}}f(t)exp(-itx) dlambda(t)=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)exp(-itx) dt=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)sum_{n=0}^{infty}frac{(-itx)^n}{n!}dt=frac{1}{sqrt{2 pi}}sum_{n=0}^{infty}x^nint_{-infty}^{infty}f(t)frac{(-it)^n}{n!}dt=sum_{n=0}^{infty}frac{1}{sqrt{2 pi}}x^na_ndt$
with $a_n:=int_{-infty}^{infty}f(t)frac{(-it)^n}{n!}$
so $hat f(x)$ can be written as a power series.
For interchanging integration and summation I thought:
For $x in mathbb C$ with $|x|le T$ and $mathrm{supp} f subseteq [-R,R]$ we have
$int sum_{n=0}^{infty}|f(t)||frac{(-itx)^n}{n!}|le sum_{n=0}^{infty}frac{(RT)^n}{n!}|f|_1=e^{RT}|f|_1<infty$
and then using dominated convergence theorem
complex-analysis fourier-transform
asked Jan 15 at 19:53
user626880user626880
204
$endgroup$
add a comment |
$begingroup$
Compactly supported $f in L^1$. Then $hat f$ can be extended to an holomorphic function on $mathbb C$, where $hat f$ is the fourier transform of f.
I tried the following
$hat f(x)=frac{1}{sqrt{2 pi}}int_{mathbb{R}}f(t)exp(-itx) dlambda(t)=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)exp(-itx) dt=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)sum_{n=0}^{infty}frac{(-itx)^n}{n!}dt=frac{1}{sqrt{2 pi}}sum_{n=0}^{infty}x^nint_{-infty}^{infty}f(t)frac{(-it)^n}{n!}dt=sum_{n=0}^{infty}frac{1}{sqrt{2 pi}}x^na_ndt$
with $a_n:=int_{-infty}^{infty}f(t)frac{(-it)^n}{n!}$
so $hat f(x)$ can be written as a power series.
For interchanging integration and summation I thought:
For $x in mathbb C$ with $|x|le T$ and $mathrm{supp} f subseteq [-R,R]$ we have
$int sum_{n=0}^{infty}|f(t)||frac{(-itx)^n}{n!}|le sum_{n=0}^{infty}frac{(RT)^n}{n!}|f|_1=e^{RT}|f|_1<infty$
and then using dominated convergence theorem
complex-analysis fourier-transform
asked Jan 15 at 19:53
user626880user626880
204
$endgroup$
$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15
add a comment |
$begingroup$
Compactly supported $f in L^1$. Then $hat f$ can be extended to an holomorphic function on $mathbb C$, where $hat f$ is the fourier transform of f.
I tried the following
$hat f(x)=frac{1}{sqrt{2 pi}}int_{mathbb{R}}f(t)exp(-itx) dlambda(t)=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)exp(-itx) dt=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)sum_{n=0}^{infty}frac{(-itx)^n}{n!}dt=frac{1}{sqrt{2 pi}}sum_{n=0}^{infty}x^nint_{-infty}^{infty}f(t)frac{(-it)^n}{n!}dt=sum_{n=0}^{infty}frac{1}{sqrt{2 pi}}x^na_ndt$
with $a_n:=int_{-infty}^{infty}f(t)frac{(-it)^n}{n!}$
so $hat f(x)$ can be written as a power series.
For interchanging integration and summation I thought:
For $x in mathbb C$ with $|x|le T$ and $mathrm{supp} f subseteq [-R,R]$ we have
$int sum_{n=0}^{infty}|f(t)||frac{(-itx)^n}{n!}|le sum_{n=0}^{infty}frac{(RT)^n}{n!}|f|_1=e^{RT}|f|_1<infty$
and then using dominated convergence theorem
complex-analysis fourier-transform
asked Jan 15 at 19:53
user626880user626880
204
$endgroup$
Compactly supported $f in L^1$. Then $hat f$ can be extended to an holomorphic function on $mathbb C$, where $hat f$ is the fourier transform of f.
I tried the following
$hat f(x)=frac{1}{sqrt{2 pi}}int_{mathbb{R}}f(t)exp(-itx) dlambda(t)=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)exp(-itx) dt=frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(t)sum_{n=0}^{infty}frac{(-itx)^n}{n!}dt=frac{1}{sqrt{2 pi}}sum_{n=0}^{infty}x^nint_{-infty}^{infty}f(t)frac{(-it)^n}{n!}dt=sum_{n=0}^{infty}frac{1}{sqrt{2 pi}}x^na_ndt$
with $a_n:=int_{-infty}^{infty}f(t)frac{(-it)^n}{n!}$
so $hat f(x)$ can be written as a power series.
For interchanging integration and summation I thought:
For $x in mathbb C$ with $|x|le T$ and $mathrm{supp} f subseteq [-R,R]$ we have
$int sum_{n=0}^{infty}|f(t)||frac{(-itx)^n}{n!}|le sum_{n=0}^{infty}frac{(RT)^n}{n!}|f|_1=e^{RT}|f|_1<infty$
and then using dominated convergence theorem
complex-analysis fourier-transform
complex-analysis fourier-transform
asked Jan 15 at 19:53
user626880user626880
204
asked Jan 15 at 19:53
user626880user626880
204
asked Jan 15 at 19:53
user626880user626880
204
asked Jan 15 at 19:53
user626880user626880
204
asked Jan 15 at 19:53
user626880user626880
204
204
$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15
add a comment |
$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15
$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15
add a comment |
1 Answer
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$begingroup$
Yes, this works.
More generally, suppose $(T, Sigma, mu)$ is a $sigma$-finite measure space, $g: T times U to mathbb C$ is a measurable function (where $U subseteq mathbb C$ is open) such that $g(t, cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T times K$ for all compact subsets $K$ of $U$, and
$f in L^1(mu)$. Then $F(z) = int_T f(t) g(t,z); dmu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ oint_C F(z); dz = oint_C int_T f(t) g(t,z); dmu(t) ; dz
= int_T oint_C f(t) g(t,z); dz ; dmu(t) = 0 $$
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, this works.
More generally, suppose $(T, Sigma, mu)$ is a $sigma$-finite measure space, $g: T times U to mathbb C$ is a measurable function (where $U subseteq mathbb C$ is open) such that $g(t, cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T times K$ for all compact subsets $K$ of $U$, and
$f in L^1(mu)$. Then $F(z) = int_T f(t) g(t,z); dmu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ oint_C F(z); dz = oint_C int_T f(t) g(t,z); dmu(t) ; dz
= int_T oint_C f(t) g(t,z); dz ; dmu(t) = 0 $$
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Yes, this works.
More generally, suppose $(T, Sigma, mu)$ is a $sigma$-finite measure space, $g: T times U to mathbb C$ is a measurable function (where $U subseteq mathbb C$ is open) such that $g(t, cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T times K$ for all compact subsets $K$ of $U$, and
$f in L^1(mu)$. Then $F(z) = int_T f(t) g(t,z); dmu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ oint_C F(z); dz = oint_C int_T f(t) g(t,z); dmu(t) ; dz
= int_T oint_C f(t) g(t,z); dz ; dmu(t) = 0 $$
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Yes, this works.
More generally, suppose $(T, Sigma, mu)$ is a $sigma$-finite measure space, $g: T times U to mathbb C$ is a measurable function (where $U subseteq mathbb C$ is open) such that $g(t, cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T times K$ for all compact subsets $K$ of $U$, and
$f in L^1(mu)$. Then $F(z) = int_T f(t) g(t,z); dmu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ oint_C F(z); dz = oint_C int_T f(t) g(t,z); dmu(t) ; dz
= int_T oint_C f(t) g(t,z); dz ; dmu(t) = 0 $$
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
$endgroup$
Yes, this works.
More generally, suppose $(T, Sigma, mu)$ is a $sigma$-finite measure space, $g: T times U to mathbb C$ is a measurable function (where $U subseteq mathbb C$ is open) such that $g(t, cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T times K$ for all compact subsets $K$ of $U$, and
$f in L^1(mu)$. Then $F(z) = int_T f(t) g(t,z); dmu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ oint_C F(z); dz = oint_C int_T f(t) g(t,z); dmu(t) ; dz
= int_T oint_C f(t) g(t,z); dz ; dmu(t) = 0 $$
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 20:29
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
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$begingroup$
If you are not sure it probably means you should try to replace at the end the DCT by a direct bound (using that the series converges in $L^1([c,d])$ with $[c,d]$ containing $supp(f)$ and that $int_c^d$ is continuous in $L^1([c,d])$)
$endgroup$
– reuns
Jan 16 at 3:14
$begingroup$
Why did you delete your question about Fourier series? I put some time into answering it.
$endgroup$
– zhw.
Jan 19 at 20:01
$begingroup$
@zhw. I revoked it
$endgroup$
– user626880
Jan 19 at 20:13
$begingroup$
I know you did. I was wondering why.
$endgroup$
– zhw.
Jan 19 at 20:15