What can we say about the Begaman transform of $fast g (t_2)- fast g(t_1)$?
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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
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up vote
0
down vote
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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 |
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