Understanding a step in the proof of Fourier Inversion Theorem by Stein
$begingroup$
I'm reading Stein and Shakarchi's Fourier Analysis, and have a question about the proof of Fourier inversion for Schwartz class functions.
In the proof below, I have two questions. Here, $K_delta(x)=delta^{-1/2}e^{-pi x^2/delta}$.
First, why does the integral $int f(x)K_delta (x)dx$ goes to $f(0)$ as $delta$ tends to $0$ since $K_delta$ is a good kernel? I've attached all the previous propositions in the text as well, but I can't see why it must follow because $K_delta$ is a good kernel. My guess is, using Corollary 1.7, we have $(f*K_delta) (0)=int f(x)K_delta (0-x) dx to f(0)$ uniformly as $delta to 0$ and $K_delta$ is even so $K_delta (-x)=K_delta (x)$ and we can get the result. But this requires $K_delta$ being even and not just a good kernel.
Finally, how does the second integral, $int hat{f}(xi)G_delta (xi)dxi$ converges to $int hat{f}(xi)dxi$ as $delta to 0$? I think we are supposed to change the order of $lim_{delta to 0}$ and the integration, but how are we guaranteed to do this?
functional-analysis analysis fourier-analysis fourier-transform
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
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add a comment |
$begingroup$
I'm reading Stein and Shakarchi's Fourier Analysis, and have a question about the proof of Fourier inversion for Schwartz class functions.
In the proof below, I have two questions. Here, $K_delta(x)=delta^{-1/2}e^{-pi x^2/delta}$.
First, why does the integral $int f(x)K_delta (x)dx$ goes to $f(0)$ as $delta$ tends to $0$ since $K_delta$ is a good kernel? I've attached all the previous propositions in the text as well, but I can't see why it must follow because $K_delta$ is a good kernel. My guess is, using Corollary 1.7, we have $(f*K_delta) (0)=int f(x)K_delta (0-x) dx to f(0)$ uniformly as $delta to 0$ and $K_delta$ is even so $K_delta (-x)=K_delta (x)$ and we can get the result. But this requires $K_delta$ being even and not just a good kernel.
Finally, how does the second integral, $int hat{f}(xi)G_delta (xi)dxi$ converges to $int hat{f}(xi)dxi$ as $delta to 0$? I think we are supposed to change the order of $lim_{delta to 0}$ and the integration, but how are we guaranteed to do this?
functional-analysis analysis fourier-analysis fourier-transform
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
$endgroup$
add a comment |
$begingroup$
I'm reading Stein and Shakarchi's Fourier Analysis, and have a question about the proof of Fourier inversion for Schwartz class functions.
In the proof below, I have two questions. Here, $K_delta(x)=delta^{-1/2}e^{-pi x^2/delta}$.
First, why does the integral $int f(x)K_delta (x)dx$ goes to $f(0)$ as $delta$ tends to $0$ since $K_delta$ is a good kernel? I've attached all the previous propositions in the text as well, but I can't see why it must follow because $K_delta$ is a good kernel. My guess is, using Corollary 1.7, we have $(f*K_delta) (0)=int f(x)K_delta (0-x) dx to f(0)$ uniformly as $delta to 0$ and $K_delta$ is even so $K_delta (-x)=K_delta (x)$ and we can get the result. But this requires $K_delta$ being even and not just a good kernel.
Finally, how does the second integral, $int hat{f}(xi)G_delta (xi)dxi$ converges to $int hat{f}(xi)dxi$ as $delta to 0$? I think we are supposed to change the order of $lim_{delta to 0}$ and the integration, but how are we guaranteed to do this?
functional-analysis analysis fourier-analysis fourier-transform
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
$endgroup$
I'm reading Stein and Shakarchi's Fourier Analysis, and have a question about the proof of Fourier inversion for Schwartz class functions.
In the proof below, I have two questions. Here, $K_delta(x)=delta^{-1/2}e^{-pi x^2/delta}$.
First, why does the integral $int f(x)K_delta (x)dx$ goes to $f(0)$ as $delta$ tends to $0$ since $K_delta$ is a good kernel? I've attached all the previous propositions in the text as well, but I can't see why it must follow because $K_delta$ is a good kernel. My guess is, using Corollary 1.7, we have $(f*K_delta) (0)=int f(x)K_delta (0-x) dx to f(0)$ uniformly as $delta to 0$ and $K_delta$ is even so $K_delta (-x)=K_delta (x)$ and we can get the result. But this requires $K_delta$ being even and not just a good kernel.
Finally, how does the second integral, $int hat{f}(xi)G_delta (xi)dxi$ converges to $int hat{f}(xi)dxi$ as $delta to 0$? I think we are supposed to change the order of $lim_{delta to 0}$ and the integration, but how are we guaranteed to do this?
functional-analysis analysis fourier-analysis fourier-transform
functional-analysis analysis fourier-analysis fourier-transform
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
asked Sep 25 '16 at 7:32
takecaretakecare
2,33221437
2,33221437
add a comment |
add a comment |
1 Answer
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For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
$$ lim_{deltarightarrow 0} frac{1}{sqrt{delta}} int_{Bbb R} left(f(x)-f(0)right) e^{-pi x^2/delta} dx = 0 $$
Given $epsilon>0$ find $eta$ so that $|x|<eta Rightarrow |f(x)-f(0)|<epsilon/2$ and then $delta$ so that the contribution from the integral over $|x|geq epsilon$ (the condition (iii)) is smaller than $epsilon/2$
For the second part the easiest is to note that $hat{f}$ is $L^1$ and use Dominated convergence (since $G_delta(x)$ goes pointwise to 1). But you may also give an $epsilon,delta$ - proof by hand:
The function $hat{f}$ is integrable, say $I=int|hat{f}|<+infty$ so given $epsilon>0$ first find $M$ so that
$$int_{|xi|>M} |hat{f}(xi)| dxi < epsilon/2$$
Now find $delta>0$ so that
$$sup_{|xi|leq M} (1-G_delta(xi)) < frac{epsilon}{2 I}$$
Then $$ left|int_{|xi|leq M} hat{f}(xi)(1-G_delta(xi)) dxi right|<epsilon/2 $$ Combining with the above, noting that $0< G_deltaleq 1$ we obtain the result.
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
$$ lim_{deltarightarrow 0} frac{1}{sqrt{delta}} int_{Bbb R} left(f(x)-f(0)right) e^{-pi x^2/delta} dx = 0 $$
Given $epsilon>0$ find $eta$ so that $|x|<eta Rightarrow |f(x)-f(0)|<epsilon/2$ and then $delta$ so that the contribution from the integral over $|x|geq epsilon$ (the condition (iii)) is smaller than $epsilon/2$
For the second part the easiest is to note that $hat{f}$ is $L^1$ and use Dominated convergence (since $G_delta(x)$ goes pointwise to 1). But you may also give an $epsilon,delta$ - proof by hand:
The function $hat{f}$ is integrable, say $I=int|hat{f}|<+infty$ so given $epsilon>0$ first find $M$ so that
$$int_{|xi|>M} |hat{f}(xi)| dxi < epsilon/2$$
Now find $delta>0$ so that
$$sup_{|xi|leq M} (1-G_delta(xi)) < frac{epsilon}{2 I}$$
Then $$ left|int_{|xi|leq M} hat{f}(xi)(1-G_delta(xi)) dxi right|<epsilon/2 $$ Combining with the above, noting that $0< G_deltaleq 1$ we obtain the result.
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
add a comment |
$begingroup$
For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
$$ lim_{deltarightarrow 0} frac{1}{sqrt{delta}} int_{Bbb R} left(f(x)-f(0)right) e^{-pi x^2/delta} dx = 0 $$
Given $epsilon>0$ find $eta$ so that $|x|<eta Rightarrow |f(x)-f(0)|<epsilon/2$ and then $delta$ so that the contribution from the integral over $|x|geq epsilon$ (the condition (iii)) is smaller than $epsilon/2$
For the second part the easiest is to note that $hat{f}$ is $L^1$ and use Dominated convergence (since $G_delta(x)$ goes pointwise to 1). But you may also give an $epsilon,delta$ - proof by hand:
The function $hat{f}$ is integrable, say $I=int|hat{f}|<+infty$ so given $epsilon>0$ first find $M$ so that
$$int_{|xi|>M} |hat{f}(xi)| dxi < epsilon/2$$
Now find $delta>0$ so that
$$sup_{|xi|leq M} (1-G_delta(xi)) < frac{epsilon}{2 I}$$
Then $$ left|int_{|xi|leq M} hat{f}(xi)(1-G_delta(xi)) dxi right|<epsilon/2 $$ Combining with the above, noting that $0< G_deltaleq 1$ we obtain the result.
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
add a comment |
$begingroup$
For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
$$ lim_{deltarightarrow 0} frac{1}{sqrt{delta}} int_{Bbb R} left(f(x)-f(0)right) e^{-pi x^2/delta} dx = 0 $$
Given $epsilon>0$ find $eta$ so that $|x|<eta Rightarrow |f(x)-f(0)|<epsilon/2$ and then $delta$ so that the contribution from the integral over $|x|geq epsilon$ (the condition (iii)) is smaller than $epsilon/2$
For the second part the easiest is to note that $hat{f}$ is $L^1$ and use Dominated convergence (since $G_delta(x)$ goes pointwise to 1). But you may also give an $epsilon,delta$ - proof by hand:
The function $hat{f}$ is integrable, say $I=int|hat{f}|<+infty$ so given $epsilon>0$ first find $M$ so that
$$int_{|xi|>M} |hat{f}(xi)| dxi < epsilon/2$$
Now find $delta>0$ so that
$$sup_{|xi|leq M} (1-G_delta(xi)) < frac{epsilon}{2 I}$$
Then $$ left|int_{|xi|leq M} hat{f}(xi)(1-G_delta(xi)) dxi right|<epsilon/2 $$ Combining with the above, noting that $0< G_deltaleq 1$ we obtain the result.
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
$$ lim_{deltarightarrow 0} frac{1}{sqrt{delta}} int_{Bbb R} left(f(x)-f(0)right) e^{-pi x^2/delta} dx = 0 $$
Given $epsilon>0$ find $eta$ so that $|x|<eta Rightarrow |f(x)-f(0)|<epsilon/2$ and then $delta$ so that the contribution from the integral over $|x|geq epsilon$ (the condition (iii)) is smaller than $epsilon/2$
For the second part the easiest is to note that $hat{f}$ is $L^1$ and use Dominated convergence (since $G_delta(x)$ goes pointwise to 1). But you may also give an $epsilon,delta$ - proof by hand:
The function $hat{f}$ is integrable, say $I=int|hat{f}|<+infty$ so given $epsilon>0$ first find $M$ so that
$$int_{|xi|>M} |hat{f}(xi)| dxi < epsilon/2$$
Now find $delta>0$ so that
$$sup_{|xi|leq M} (1-G_delta(xi)) < frac{epsilon}{2 I}$$
Then $$ left|int_{|xi|leq M} hat{f}(xi)(1-G_delta(xi)) dxi right|<epsilon/2 $$ Combining with the above, noting that $0< G_deltaleq 1$ we obtain the result.
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
edited Sep 25 '16 at 8:21
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
answered Sep 25 '16 at 7:46
H. H. RughH. H. Rugh
23.2k11134
23.2k11134
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
add a comment |
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
$begingroup$
We haven't learned Lebesgue integrals in this book, and only treat Riemann integration, so is there another easy way to guarantee the change of limit and integration other than using DCT?
$endgroup$
– takecare
Sep 25 '16 at 8:00
add a comment |
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