What can we say about the Begaman transform of $fast g (t_2)- fast g(t_1)$?
up vote
0
down vote
favorite
Let $f, gin mathcal{S}(mathbb R)$. Then the convolution of two Schwartz class function is again Schwartz class function, that is, $fast g in mathcal{S}(mathbb R).$
Now we define
$$ H(t)= H(t_1,t_2)= int_{t_1}^{t_2} frac{d}{dr}(fast g)(r) dr = fast g (t_2)- fast g(t_1), (t_1, t_2 in mathbb R)$$
Since $(fast g)'= f'ast g,$ we may notice that, by Holder inequality, $|H|_{L^{infty}(mathbb R^{2})} leq |f'|_{L^{p}} |g|_{L^{p'}} < infty$ ($p'$ is the Holder conjugate).
We consider the Bergaman transform of $H$ as follows:
$$ BH(z)= int_{mathbb R^{2}} H(t) e^{2pi tcdot z- pi t^2- frac{pi}{2} z^2} dt, $$ where $ z= (z_1, z_2)in mathbb C times mathbb C = mathbb C^2$
Question: What can we say about the Bergaman transform of $H$?
Can we expect $| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} <infty$? Can we say that $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?
integration complex-analysis functional-analysis inequality intuition
asked Nov 17 at 19:58
Math Learner
3109
add a comment |
up vote
0
down vote
favorite
Let $f, gin mathcal{S}(mathbb R)$. Then the convolution of two Schwartz class function is again Schwartz class function, that is, $fast g in mathcal{S}(mathbb R).$
Now we define
$$ H(t)= H(t_1,t_2)= int_{t_1}^{t_2} frac{d}{dr}(fast g)(r) dr = fast g (t_2)- fast g(t_1), (t_1, t_2 in mathbb R)$$
Since $(fast g)'= f'ast g,$ we may notice that, by Holder inequality, $|H|_{L^{infty}(mathbb R^{2})} leq |f'|_{L^{p}} |g|_{L^{p'}} < infty$ ($p'$ is the Holder conjugate).
We consider the Bergaman transform of $H$ as follows:
$$ BH(z)= int_{mathbb R^{2}} H(t) e^{2pi tcdot z- pi t^2- frac{pi}{2} z^2} dt, $$ where $ z= (z_1, z_2)in mathbb C times mathbb C = mathbb C^2$
Question: What can we say about the Bergaman transform of $H$?
Can we expect $| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} <infty$? Can we say that $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?
integration complex-analysis functional-analysis inequality intuition
asked Nov 17 at 19:58
Math Learner
3109
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f, gin mathcal{S}(mathbb R)$. Then the convolution of two Schwartz class function is again Schwartz class function, that is, $fast g in mathcal{S}(mathbb R).$
Now we define
$$ H(t)= H(t_1,t_2)= int_{t_1}^{t_2} frac{d}{dr}(fast g)(r) dr = fast g (t_2)- fast g(t_1), (t_1, t_2 in mathbb R)$$
Since $(fast g)'= f'ast g,$ we may notice that, by Holder inequality, $|H|_{L^{infty}(mathbb R^{2})} leq |f'|_{L^{p}} |g|_{L^{p'}} < infty$ ($p'$ is the Holder conjugate).
We consider the Bergaman transform of $H$ as follows:
$$ BH(z)= int_{mathbb R^{2}} H(t) e^{2pi tcdot z- pi t^2- frac{pi}{2} z^2} dt, $$ where $ z= (z_1, z_2)in mathbb C times mathbb C = mathbb C^2$
Question: What can we say about the Bergaman transform of $H$?
Can we expect $| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} <infty$? Can we say that $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?
integration complex-analysis functional-analysis inequality intuition
asked Nov 17 at 19:58
Math Learner
3109
Let $f, gin mathcal{S}(mathbb R)$. Then the convolution of two Schwartz class function is again Schwartz class function, that is, $fast g in mathcal{S}(mathbb R).$
Now we define
$$ H(t)= H(t_1,t_2)= int_{t_1}^{t_2} frac{d}{dr}(fast g)(r) dr = fast g (t_2)- fast g(t_1), (t_1, t_2 in mathbb R)$$
Since $(fast g)'= f'ast g,$ we may notice that, by Holder inequality, $|H|_{L^{infty}(mathbb R^{2})} leq |f'|_{L^{p}} |g|_{L^{p'}} < infty$ ($p'$ is the Holder conjugate).
We consider the Bergaman transform of $H$ as follows:
$$ BH(z)= int_{mathbb R^{2}} H(t) e^{2pi tcdot z- pi t^2- frac{pi}{2} z^2} dt, $$ where $ z= (z_1, z_2)in mathbb C times mathbb C = mathbb C^2$
Question: What can we say about the Bergaman transform of $H$?
Can we expect $| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} <infty$? Can we say that $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?
integration complex-analysis functional-analysis inequality intuition
integration complex-analysis functional-analysis inequality intuition
asked Nov 17 at 19:58
Math Learner
3109
asked Nov 17 at 19:58
Math Learner
3109
edited Nov 17 at 20:43
asked Nov 17 at 19:58
Math Learner
3109
asked Nov 17 at 19:58
Math Learner
3109
asked Nov 17 at 19:58
Math Learner
3109
3109
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002747%2fwhat-can-we-say-about-the-begaman-transform-of-f-ast-g-t-2-f-ast-gt-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
