Showing holomorphy of integral
$begingroup$
Let $f in L^1$ with compact support.
I want to show that $hat f: mathbb R to mathbb C, f(z):=int_{mathbb R}f(x)e^{-ixz}dlambda(x), $ can be extended to a holomorphic function on $mathbb C$ and I was wondering if I can do it like this:
Let $U_a:={z in mathbb C: -a<Im(z)<a}$ and $g(x,z):=f(x)e^{-ixz}$
1) $xmapsto g(x,z)$ is integrable for all $z in U_a$: $f$ is integrable and vanishes outside of its support. On the (compact) support $xmapsto e^{-ixz}$ is integrable because it is continuous. Therefore $xmapsto g(x,z)$ is integrable for all $z in U_a$.
2) $z mapsto g(x,z)$ is holomorphic because for a fixed $x$ $f(x)$ is just a constant and $z mapsto e^{-ixz}$ is holomorphic.
3) $|g(x,z)|=|f(x)e^{-ixz}|=|f(x)e^{Im(z)}x|<|f(x)|e^{a|x|}$
which is a integrable function $mathbb R to overline {mathbb R}$
Therefore $int_{mathbb R} g(.,z) dlambda(.)$ is holomorphic on $mathbb C$ (because $U_a$ was arbitrary) according to the 'holomorphy under the integral' lemma
real-analysis analysis fourier-analysis
asked Jan 20 at 19:27
user636610user636610
82
$endgroup$
add a comment |
$begingroup$
Let $f in L^1$ with compact support.
I want to show that $hat f: mathbb R to mathbb C, f(z):=int_{mathbb R}f(x)e^{-ixz}dlambda(x), $ can be extended to a holomorphic function on $mathbb C$ and I was wondering if I can do it like this:
Let $U_a:={z in mathbb C: -a<Im(z)<a}$ and $g(x,z):=f(x)e^{-ixz}$
1) $xmapsto g(x,z)$ is integrable for all $z in U_a$: $f$ is integrable and vanishes outside of its support. On the (compact) support $xmapsto e^{-ixz}$ is integrable because it is continuous. Therefore $xmapsto g(x,z)$ is integrable for all $z in U_a$.
2) $z mapsto g(x,z)$ is holomorphic because for a fixed $x$ $f(x)$ is just a constant and $z mapsto e^{-ixz}$ is holomorphic.
3) $|g(x,z)|=|f(x)e^{-ixz}|=|f(x)e^{Im(z)}x|<|f(x)|e^{a|x|}$
which is a integrable function $mathbb R to overline {mathbb R}$
Therefore $int_{mathbb R} g(.,z) dlambda(.)$ is holomorphic on $mathbb C$ (because $U_a$ was arbitrary) according to the 'holomorphy under the integral' lemma
real-analysis analysis fourier-analysis
asked Jan 20 at 19:27
user636610user636610
82
$endgroup$
$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49
add a comment |
$begingroup$
Let $f in L^1$ with compact support.
I want to show that $hat f: mathbb R to mathbb C, f(z):=int_{mathbb R}f(x)e^{-ixz}dlambda(x), $ can be extended to a holomorphic function on $mathbb C$ and I was wondering if I can do it like this:
Let $U_a:={z in mathbb C: -a<Im(z)<a}$ and $g(x,z):=f(x)e^{-ixz}$
1) $xmapsto g(x,z)$ is integrable for all $z in U_a$: $f$ is integrable and vanishes outside of its support. On the (compact) support $xmapsto e^{-ixz}$ is integrable because it is continuous. Therefore $xmapsto g(x,z)$ is integrable for all $z in U_a$.
2) $z mapsto g(x,z)$ is holomorphic because for a fixed $x$ $f(x)$ is just a constant and $z mapsto e^{-ixz}$ is holomorphic.
3) $|g(x,z)|=|f(x)e^{-ixz}|=|f(x)e^{Im(z)}x|<|f(x)|e^{a|x|}$
which is a integrable function $mathbb R to overline {mathbb R}$
Therefore $int_{mathbb R} g(.,z) dlambda(.)$ is holomorphic on $mathbb C$ (because $U_a$ was arbitrary) according to the 'holomorphy under the integral' lemma
real-analysis analysis fourier-analysis
asked Jan 20 at 19:27
user636610user636610
82
$endgroup$
Let $f in L^1$ with compact support.
I want to show that $hat f: mathbb R to mathbb C, f(z):=int_{mathbb R}f(x)e^{-ixz}dlambda(x), $ can be extended to a holomorphic function on $mathbb C$ and I was wondering if I can do it like this:
Let $U_a:={z in mathbb C: -a<Im(z)<a}$ and $g(x,z):=f(x)e^{-ixz}$
1) $xmapsto g(x,z)$ is integrable for all $z in U_a$: $f$ is integrable and vanishes outside of its support. On the (compact) support $xmapsto e^{-ixz}$ is integrable because it is continuous. Therefore $xmapsto g(x,z)$ is integrable for all $z in U_a$.
2) $z mapsto g(x,z)$ is holomorphic because for a fixed $x$ $f(x)$ is just a constant and $z mapsto e^{-ixz}$ is holomorphic.
3) $|g(x,z)|=|f(x)e^{-ixz}|=|f(x)e^{Im(z)}x|<|f(x)|e^{a|x|}$
which is a integrable function $mathbb R to overline {mathbb R}$
Therefore $int_{mathbb R} g(.,z) dlambda(.)$ is holomorphic on $mathbb C$ (because $U_a$ was arbitrary) according to the 'holomorphy under the integral' lemma
real-analysis analysis fourier-analysis
real-analysis analysis fourier-analysis
asked Jan 20 at 19:27
user636610user636610
82
asked Jan 20 at 19:27
user636610user636610
82
asked Jan 20 at 19:27
user636610user636610
82
asked Jan 20 at 19:27
user636610user636610
82
asked Jan 20 at 19:27
user636610user636610
82
82
$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49
add a comment |
$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49
$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49
add a comment |
1 Answer
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You can start by arguing that $hat{f}(lambda)=int_{mathbb{R}}f(x)e^{-ilambda x}dx$ is continuous everywhere on $mathbb{C}$, which follows from the dominated convergence theorem. Then you can apply Morera's theorem by showing that $int_{Delta}hat{f}(lambda)dlambda=0$ for all triangles in $mathbb{C}$, which follows by interchanging the order of integration using Fubini's Theorem:
$$
int_{Delta}hat{f}(lambda)dlambda=int_{mathbb{R}}f(x)int_{Delta}e^{-ixlambda} dlambda dx = 0.
$$
The conclusion of Morera's theorem is that $hat{f}$ is an entire function.
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
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$begingroup$
You can start by arguing that $hat{f}(lambda)=int_{mathbb{R}}f(x)e^{-ilambda x}dx$ is continuous everywhere on $mathbb{C}$, which follows from the dominated convergence theorem. Then you can apply Morera's theorem by showing that $int_{Delta}hat{f}(lambda)dlambda=0$ for all triangles in $mathbb{C}$, which follows by interchanging the order of integration using Fubini's Theorem:
$$
int_{Delta}hat{f}(lambda)dlambda=int_{mathbb{R}}f(x)int_{Delta}e^{-ixlambda} dlambda dx = 0.
$$
The conclusion of Morera's theorem is that $hat{f}$ is an entire function.
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
You can start by arguing that $hat{f}(lambda)=int_{mathbb{R}}f(x)e^{-ilambda x}dx$ is continuous everywhere on $mathbb{C}$, which follows from the dominated convergence theorem. Then you can apply Morera's theorem by showing that $int_{Delta}hat{f}(lambda)dlambda=0$ for all triangles in $mathbb{C}$, which follows by interchanging the order of integration using Fubini's Theorem:
$$
int_{Delta}hat{f}(lambda)dlambda=int_{mathbb{R}}f(x)int_{Delta}e^{-ixlambda} dlambda dx = 0.
$$
The conclusion of Morera's theorem is that $hat{f}$ is an entire function.
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
You can start by arguing that $hat{f}(lambda)=int_{mathbb{R}}f(x)e^{-ilambda x}dx$ is continuous everywhere on $mathbb{C}$, which follows from the dominated convergence theorem. Then you can apply Morera's theorem by showing that $int_{Delta}hat{f}(lambda)dlambda=0$ for all triangles in $mathbb{C}$, which follows by interchanging the order of integration using Fubini's Theorem:
$$
int_{Delta}hat{f}(lambda)dlambda=int_{mathbb{R}}f(x)int_{Delta}e^{-ixlambda} dlambda dx = 0.
$$
The conclusion of Morera's theorem is that $hat{f}$ is an entire function.
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
You can start by arguing that $hat{f}(lambda)=int_{mathbb{R}}f(x)e^{-ilambda x}dx$ is continuous everywhere on $mathbb{C}$, which follows from the dominated convergence theorem. Then you can apply Morera's theorem by showing that $int_{Delta}hat{f}(lambda)dlambda=0$ for all triangles in $mathbb{C}$, which follows by interchanging the order of integration using Fubini's Theorem:
$$
int_{Delta}hat{f}(lambda)dlambda=int_{mathbb{R}}f(x)int_{Delta}e^{-ixlambda} dlambda dx = 0.
$$
The conclusion of Morera's theorem is that $hat{f}$ is an entire function.
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 24 at 0:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
add a comment |
add a comment |
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$begingroup$
Perhaps the fastest way is to show, using the dominated convergence theorem, that $$int f(x)e^{izx},mathrm{d}x=sum_{n=0}^{infty}left(int frac{(ix)^n}{n!}f(x),mathrm{d}xright)z^n,$$ which defines an entire function.
$endgroup$
– Sangchul Lee
Jan 20 at 19:37
$begingroup$
@SangchulLee I also thought about that but how exactly does one use the dct to argue the interchange of summation and integration? We always used the dct only for a series $f_n$ of functions
$endgroup$
– user636610
Jan 20 at 19:43
$begingroup$
For each fixed $z$, you can consider $s_n(x)=sum_{k=0}^{n}frac{(izx)^k}{k!}f(x)$. If $f(x)=0$ a.e. outside $[-R,R]$, then $s_n$'s are dominated by the integrable function $e^{R|z|}|f(x)|$ and converges pointwise to $e^{izx}f(x)$. So, one can apply the dominated convergence theorem.
$endgroup$
– Sangchul Lee
Jan 20 at 19:49