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 $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?











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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 $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?











    share|cite|improve this question


























      up vote
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      down vote

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      up vote
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      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 $left|| e^{-pi |z|^2} B(z_1, z_2)|_{L^{infty}_{z_1}} right|_{L^1_{z_2}} <infty$?











      share|cite|improve this question















      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






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      edited Nov 17 at 20:43

























      asked Nov 17 at 19:58









      Math Learner

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