When and how is the map $ zmapsto int |x|^{-z}phi(x) dx $ analytic?
$begingroup$
In Wolff's Lectures on Harmonic Analysis (see also the AMS published version here), the author claims without proofs in details the following:
for $phiinmathcal{S}(mathbb{R}^n)$, where $mathcal{S}$ denotes the Schwartz space, the map
$$
zmapsto int |x|^{-z}phi(x) dx
$$
is analytic in the "indicated regime", which may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.
I have the following questions:
- The "indicated regime" is not indicated in Wolff's notes. Where should it be?
- in order to apply the dominated convergence theorem, it seems (doesn't it?) that one needs to estimate the integral
$$
int bigg||x|^{-z}bigg|cdot bigg|log |x|bigg|cdot |phi(x)| dx.
$$
What dominated function should one use?
real-analysis complex-analysis harmonic-analysis
$endgroup$
add a comment |
$begingroup$
In Wolff's Lectures on Harmonic Analysis (see also the AMS published version here), the author claims without proofs in details the following:
for $phiinmathcal{S}(mathbb{R}^n)$, where $mathcal{S}$ denotes the Schwartz space, the map
$$
zmapsto int |x|^{-z}phi(x) dx
$$
is analytic in the "indicated regime", which may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.
I have the following questions:
- The "indicated regime" is not indicated in Wolff's notes. Where should it be?
- in order to apply the dominated convergence theorem, it seems (doesn't it?) that one needs to estimate the integral
$$
int bigg||x|^{-z}bigg|cdot bigg|log |x|bigg|cdot |phi(x)| dx.
$$
What dominated function should one use?
real-analysis complex-analysis harmonic-analysis
$endgroup$
add a comment |
$begingroup$
In Wolff's Lectures on Harmonic Analysis (see also the AMS published version here), the author claims without proofs in details the following:
for $phiinmathcal{S}(mathbb{R}^n)$, where $mathcal{S}$ denotes the Schwartz space, the map
$$
zmapsto int |x|^{-z}phi(x) dx
$$
is analytic in the "indicated regime", which may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.
I have the following questions:
- The "indicated regime" is not indicated in Wolff's notes. Where should it be?
- in order to apply the dominated convergence theorem, it seems (doesn't it?) that one needs to estimate the integral
$$
int bigg||x|^{-z}bigg|cdot bigg|log |x|bigg|cdot |phi(x)| dx.
$$
What dominated function should one use?
real-analysis complex-analysis harmonic-analysis
$endgroup$
In Wolff's Lectures on Harmonic Analysis (see also the AMS published version here), the author claims without proofs in details the following:
for $phiinmathcal{S}(mathbb{R}^n)$, where $mathcal{S}$ denotes the Schwartz space, the map
$$
zmapsto int |x|^{-z}phi(x) dx
$$
is analytic in the "indicated regime", which may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.
I have the following questions:
- The "indicated regime" is not indicated in Wolff's notes. Where should it be?
- in order to apply the dominated convergence theorem, it seems (doesn't it?) that one needs to estimate the integral
$$
int bigg||x|^{-z}bigg|cdot bigg|log |x|bigg|cdot |phi(x)| dx.
$$
What dominated function should one use?
real-analysis complex-analysis harmonic-analysis
real-analysis complex-analysis harmonic-analysis
edited Dec 19 '18 at 19:40
asked Dec 19 '18 at 19:33
user587192
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The basic theorem is that $g(z) = int_{mathbb R^n} f(x,z); dx$ is analytic in $z in U subseteq mathbb C$ if
$f(x,z)$ is analytic in $U$ for every $xin mathbb R^n$,
$f(x,z)$ is jointly measurable in $x$ and $z$,
$int_{mathbb R^n} |f(x,z)|; dx$ is uniformly bounded on every compact subset of $U$.
Then we can use Fubini's theorem to say
if $Gamma$ is any closed triangle in $U$,
$$ oint_Gamma g(z); dz = int_{mathbb R^n} oint_Gamma f(x,z); dz; dz = 0$$
and then Morera's theorem says $g(z)$ is analytic in $U$.
In your case, the only problem is the singularity of $|x|^{-z}$ at the origin: we have $$left| |x|^{-z} right| = |x|^{-text{Re}; z}$$
and we need $n-1 - text{Re}; z > -1$, i.e. $text{Re}; z < n$.
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The basic theorem is that $g(z) = int_{mathbb R^n} f(x,z); dx$ is analytic in $z in U subseteq mathbb C$ if
$f(x,z)$ is analytic in $U$ for every $xin mathbb R^n$,
$f(x,z)$ is jointly measurable in $x$ and $z$,
$int_{mathbb R^n} |f(x,z)|; dx$ is uniformly bounded on every compact subset of $U$.
Then we can use Fubini's theorem to say
if $Gamma$ is any closed triangle in $U$,
$$ oint_Gamma g(z); dz = int_{mathbb R^n} oint_Gamma f(x,z); dz; dz = 0$$
and then Morera's theorem says $g(z)$ is analytic in $U$.
In your case, the only problem is the singularity of $|x|^{-z}$ at the origin: we have $$left| |x|^{-z} right| = |x|^{-text{Re}; z}$$
and we need $n-1 - text{Re}; z > -1$, i.e. $text{Re}; z < n$.
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
add a comment |
$begingroup$
The basic theorem is that $g(z) = int_{mathbb R^n} f(x,z); dx$ is analytic in $z in U subseteq mathbb C$ if
$f(x,z)$ is analytic in $U$ for every $xin mathbb R^n$,
$f(x,z)$ is jointly measurable in $x$ and $z$,
$int_{mathbb R^n} |f(x,z)|; dx$ is uniformly bounded on every compact subset of $U$.
Then we can use Fubini's theorem to say
if $Gamma$ is any closed triangle in $U$,
$$ oint_Gamma g(z); dz = int_{mathbb R^n} oint_Gamma f(x,z); dz; dz = 0$$
and then Morera's theorem says $g(z)$ is analytic in $U$.
In your case, the only problem is the singularity of $|x|^{-z}$ at the origin: we have $$left| |x|^{-z} right| = |x|^{-text{Re}; z}$$
and we need $n-1 - text{Re}; z > -1$, i.e. $text{Re}; z < n$.
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
add a comment |
$begingroup$
The basic theorem is that $g(z) = int_{mathbb R^n} f(x,z); dx$ is analytic in $z in U subseteq mathbb C$ if
$f(x,z)$ is analytic in $U$ for every $xin mathbb R^n$,
$f(x,z)$ is jointly measurable in $x$ and $z$,
$int_{mathbb R^n} |f(x,z)|; dx$ is uniformly bounded on every compact subset of $U$.
Then we can use Fubini's theorem to say
if $Gamma$ is any closed triangle in $U$,
$$ oint_Gamma g(z); dz = int_{mathbb R^n} oint_Gamma f(x,z); dz; dz = 0$$
and then Morera's theorem says $g(z)$ is analytic in $U$.
In your case, the only problem is the singularity of $|x|^{-z}$ at the origin: we have $$left| |x|^{-z} right| = |x|^{-text{Re}; z}$$
and we need $n-1 - text{Re}; z > -1$, i.e. $text{Re}; z < n$.
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
$endgroup$
The basic theorem is that $g(z) = int_{mathbb R^n} f(x,z); dx$ is analytic in $z in U subseteq mathbb C$ if
$f(x,z)$ is analytic in $U$ for every $xin mathbb R^n$,
$f(x,z)$ is jointly measurable in $x$ and $z$,
$int_{mathbb R^n} |f(x,z)|; dx$ is uniformly bounded on every compact subset of $U$.
Then we can use Fubini's theorem to say
if $Gamma$ is any closed triangle in $U$,
$$ oint_Gamma g(z); dz = int_{mathbb R^n} oint_Gamma f(x,z); dz; dz = 0$$
and then Morera's theorem says $g(z)$ is analytic in $U$.
In your case, the only problem is the singularity of $|x|^{-z}$ at the origin: we have $$left| |x|^{-z} right| = |x|^{-text{Re}; z}$$
and we need $n-1 - text{Re}; z > -1$, i.e. $text{Re}; z < n$.
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
answered Dec 19 '18 at 19:57
Robert IsraelRobert Israel
328k23216469
328k23216469
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
add a comment |
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
$begingroup$
Thanks for your answer. I'm interested in a bit more details. While Fubini and Morera's theorems together are indeed handy (thanks for pointing that out!) I'm still curious about Wolff's argument using dominated convergence: how would you find a dominated function? Would you elaborate on why one (only) needs $n-1-text{Re }z>-1$? I don't know how to use the decaying properties of the Schwartz function $phi$ to get this inequality.
$endgroup$
– user587192
Dec 19 '18 at 20:19
1
1
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
$begingroup$
It doesn't have to do with the Schwartz function, just with the radial factor $r^{n-1} ; dr$ in the volume element of $mathbb R^n$ and the fact that $int_0^1 r^d ; dr$ converges iff $d > -1$. The Schwartz function $phi(x)$, since it goes to $0$ at $infty$ faster than any polynomial, takes care of the integration over a neighbourhood of $infty$.
$endgroup$
– Robert Israel
Dec 19 '18 at 21:06
add a comment |
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