Intuition behind the proof of the Inverse Fourier Transform?
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I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.
The proof I have read asks you to consider an auxiliary function $g_a(x)$ defined as
$$g_a(x) = int_-infty^infty e^-a lvert trverte^2pi i t xoperatornamed!t = frac2a4pi^2x^2+a^2.$$
It then proceeds to prove that $lim_arightarrow 0 f*g_a(x) = f(x)$ and therefore shows that $hathatf(x) = f(-x)$ by the dominated convergence theorem.
My question is as follows: Where did $g$ come from?
So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function?
I've also noticed that $g_a(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?
That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_a(t)$ "naturally"?
As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $epsilon>0$ be given and consider $delta = f(epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $delta$ if you never show how you arrived at that $delta$. In a similar vein, how do I reasonably arrive at $g_a(x)$?
fourier-analysis intuition
edited Apr 2 '16 at 11:55
Watson
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
add a comment |Â
up vote
7
down vote
favorite
I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.
The proof I have read asks you to consider an auxiliary function $g_a(x)$ defined as
$$g_a(x) = int_-infty^infty e^-a lvert trverte^2pi i t xoperatornamed!t = frac2a4pi^2x^2+a^2.$$
It then proceeds to prove that $lim_arightarrow 0 f*g_a(x) = f(x)$ and therefore shows that $hathatf(x) = f(-x)$ by the dominated convergence theorem.
My question is as follows: Where did $g$ come from?
So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function?
I've also noticed that $g_a(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?
That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_a(t)$ "naturally"?
As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $epsilon>0$ be given and consider $delta = f(epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $delta$ if you never show how you arrived at that $delta$. In a similar vein, how do I reasonably arrive at $g_a(x)$?
fourier-analysis intuition
edited Apr 2 '16 at 11:55
Watson
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
2
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
1
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.
The proof I have read asks you to consider an auxiliary function $g_a(x)$ defined as
$$g_a(x) = int_-infty^infty e^-a lvert trverte^2pi i t xoperatornamed!t = frac2a4pi^2x^2+a^2.$$
It then proceeds to prove that $lim_arightarrow 0 f*g_a(x) = f(x)$ and therefore shows that $hathatf(x) = f(-x)$ by the dominated convergence theorem.
My question is as follows: Where did $g$ come from?
So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function?
I've also noticed that $g_a(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?
That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_a(t)$ "naturally"?
As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $epsilon>0$ be given and consider $delta = f(epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $delta$ if you never show how you arrived at that $delta$. In a similar vein, how do I reasonably arrive at $g_a(x)$?
fourier-analysis intuition
edited Apr 2 '16 at 11:55
Watson
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.
The proof I have read asks you to consider an auxiliary function $g_a(x)$ defined as
$$g_a(x) = int_-infty^infty e^-a lvert trverte^2pi i t xoperatornamed!t = frac2a4pi^2x^2+a^2.$$
It then proceeds to prove that $lim_arightarrow 0 f*g_a(x) = f(x)$ and therefore shows that $hathatf(x) = f(-x)$ by the dominated convergence theorem.
My question is as follows: Where did $g$ come from?
So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function?
I've also noticed that $g_a(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?
That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_a(t)$ "naturally"?
As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $epsilon>0$ be given and consider $delta = f(epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $delta$ if you never show how you arrived at that $delta$. In a similar vein, how do I reasonably arrive at $g_a(x)$?
fourier-analysis intuition
fourier-analysis intuition
edited Apr 2 '16 at 11:55
Watson
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
edited Apr 2 '16 at 11:55
Watson
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
edited Apr 2 '16 at 11:55
Watson
14.9k92866
edited Apr 2 '16 at 11:55
Watson
14.9k92866
edited Apr 2 '16 at 11:55
Watson
14.9k92866
14.9k92866
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
asked Mar 9 '15 at 7:27
JessicaK
4,38651633
4,38651633
2
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
1
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25
add a comment |Â
2
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
1
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25
2
2
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
1
1
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If you consider the operator $A=frac1ifracddx$ on the domain consisting of periodic and differentiable functions on $[-pi,pi]$, then you can consider, as Cauchy did, $(lambda I-A)^-1f$. The led to a expression with poles at $lambda=0,pm 1,pm 2,cdots$ and residue at $N$ equal to $frac12piint_-pi^pif(t)e^-intdt e^inx$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $infty$, which would appear to be
$$
lim_lambdarightarrowinftylambda(lambda I-A)^-1f=lim_lambdarightarrowinftyfraclambdalambda I-Af = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=frac1ifracddx$ on $(-infty,infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
frac12pi iint_-infty^infty ((t+iepsilon)I-A)^-1f-((t-iepsilon)I-A)^-1f dt
$$
This integral gives your integral with $a=epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you consider the operator $A=frac1ifracddx$ on the domain consisting of periodic and differentiable functions on $[-pi,pi]$, then you can consider, as Cauchy did, $(lambda I-A)^-1f$. The led to a expression with poles at $lambda=0,pm 1,pm 2,cdots$ and residue at $N$ equal to $frac12piint_-pi^pif(t)e^-intdt e^inx$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $infty$, which would appear to be
$$
lim_lambdarightarrowinftylambda(lambda I-A)^-1f=lim_lambdarightarrowinftyfraclambdalambda I-Af = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=frac1ifracddx$ on $(-infty,infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
frac12pi iint_-infty^infty ((t+iepsilon)I-A)^-1f-((t-iepsilon)I-A)^-1f dt
$$
This integral gives your integral with $a=epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
add a comment |Â
up vote
1
down vote
accepted
If you consider the operator $A=frac1ifracddx$ on the domain consisting of periodic and differentiable functions on $[-pi,pi]$, then you can consider, as Cauchy did, $(lambda I-A)^-1f$. The led to a expression with poles at $lambda=0,pm 1,pm 2,cdots$ and residue at $N$ equal to $frac12piint_-pi^pif(t)e^-intdt e^inx$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $infty$, which would appear to be
$$
lim_lambdarightarrowinftylambda(lambda I-A)^-1f=lim_lambdarightarrowinftyfraclambdalambda I-Af = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=frac1ifracddx$ on $(-infty,infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
frac12pi iint_-infty^infty ((t+iepsilon)I-A)^-1f-((t-iepsilon)I-A)^-1f dt
$$
This integral gives your integral with $a=epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you consider the operator $A=frac1ifracddx$ on the domain consisting of periodic and differentiable functions on $[-pi,pi]$, then you can consider, as Cauchy did, $(lambda I-A)^-1f$. The led to a expression with poles at $lambda=0,pm 1,pm 2,cdots$ and residue at $N$ equal to $frac12piint_-pi^pif(t)e^-intdt e^inx$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $infty$, which would appear to be
$$
lim_lambdarightarrowinftylambda(lambda I-A)^-1f=lim_lambdarightarrowinftyfraclambdalambda I-Af = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=frac1ifracddx$ on $(-infty,infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
frac12pi iint_-infty^infty ((t+iepsilon)I-A)^-1f-((t-iepsilon)I-A)^-1f dt
$$
This integral gives your integral with $a=epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
If you consider the operator $A=frac1ifracddx$ on the domain consisting of periodic and differentiable functions on $[-pi,pi]$, then you can consider, as Cauchy did, $(lambda I-A)^-1f$. The led to a expression with poles at $lambda=0,pm 1,pm 2,cdots$ and residue at $N$ equal to $frac12piint_-pi^pif(t)e^-intdt e^inx$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $infty$, which would appear to be
$$
lim_lambdarightarrowinftylambda(lambda I-A)^-1f=lim_lambdarightarrowinftyfraclambdalambda I-Af = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=frac1ifracddx$ on $(-infty,infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
frac12pi iint_-infty^infty ((t+iepsilon)I-A)^-1f-((t-iepsilon)I-A)^-1f dt
$$
This integral gives your integral with $a=epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
edited Sep 7 at 6:54
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
answered Sep 7 at 6:39
DisintegratingByParts
56.3k42373
56.3k42373
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
add a comment |Â
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
I find this answer interesting thank you. Do you happen to have any references that discusses this in more detail?
– JessicaK
Sep 10 at 21:45
1
1
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
@JessicaK : E. C. Titchmarsh, who was a prominent student of G. H. Hardy, adopted and developed this approach to give some of the earliest general and rigorous pointwise convergence results for general Fourier transforms and series associated with Sturm-Liouville problems. The Titchmarsh classic is Eigenfunction Expansions Associated with 2nd Order Differential Equations.
– DisintegratingByParts
Sep 10 at 23:54
add a comment |Â
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2
The "intuition" essentially boils down to $int e^ik(x-y) dk = delta(x-y) (2pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
1
Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f ast g_a to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $widehatg_a to 1$ pointwise.
– PhoemueX
Mar 9 '15 at 21:25