Is $K_t :mathbb{T}tomathbb{C}, smapstosum_{n∈mathbb{Z}}e^{-n^2 t}e^{ins}$ a summability kernel for $tto...












0












$begingroup$


In trying to solve with Fourier series the heat equation on the 1-torus, I stumbled into this family of functions:
$$forall t>0, K_t :mathbb{T}tomathbb{C}, smapstosum_{n∈mathbb{Z}}e^{-n^2 t}e^{ins},$$
i.e. a family of functions indexed by $tin(0,+infty)$ where the $t$-member is a function with Fourier coefficients given by $(e^{-n^2 t})_{ninmathbb{Z}}$.
It is clear that $forall t>0, K_tin C^infty(mathbb{T})$ and so, if $f$ is a distribution on $mathbb{T}$, the distribution $K_t*f$ is actually represented by an element of $C^infty(mathbb{T})$. Also, I've proved that the function:
$$u_f :mathbb{T}times(0,+infty)tomathbb{C}, (x,t)mapsto (K_t*f)(x)$$
is an element of $C^infty(mathbb{Ttimes(0,+infty)})$ and that:
$$partial_t u_f = partial_{xx}u_f.$$
So, I really want the family $(K_t)_{t>0}$ to be a summability kernel for $tto0^+$, because in this case $u_f$ would be a solution for the heat equation on the 1-torus with initial temperature $fin L^p(mathbb{T})$ with $1le p<infty$, in the sense that $$|u_f(cdot,t)-f|_p, tto 0^+.$$ So:




Is $(K_t)_{t>0}$ a summability kernel for $tto0^+$?




I noticed that the coefficients $e^{-t n^2}to 1, tto 0^+$ boundedly (by $1$), so $K_ttodelta_0, tto 0^+$ in distribution, and then at least there's some hope.
I tried with Poisson summation formula obtaining only a messier expression. Any ideas?






fourier-series holder-spaces







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$endgroup$

















    0












    $begingroup$


    In trying to solve with Fourier series the heat equation on the 1-torus, I stumbled into this family of functions:
    $$forall t>0, K_t :mathbb{T}tomathbb{C}, smapstosum_{n∈mathbb{Z}}e^{-n^2 t}e^{ins},$$
    i.e. a family of functions indexed by $tin(0,+infty)$ where the $t$-member is a function with Fourier coefficients given by $(e^{-n^2 t})_{ninmathbb{Z}}$.
    It is clear that $forall t>0, K_tin C^infty(mathbb{T})$ and so, if $f$ is a distribution on $mathbb{T}$, the distribution $K_t*f$ is actually represented by an element of $C^infty(mathbb{T})$. Also, I've proved that the function:
    $$u_f :mathbb{T}times(0,+infty)tomathbb{C}, (x,t)mapsto (K_t*f)(x)$$
    is an element of $C^infty(mathbb{Ttimes(0,+infty)})$ and that:
    $$partial_t u_f = partial_{xx}u_f.$$
    So, I really want the family $(K_t)_{t>0}$ to be a summability kernel for $tto0^+$, because in this case $u_f$ would be a solution for the heat equation on the 1-torus with initial temperature $fin L^p(mathbb{T})$ with $1le p<infty$, in the sense that $$|u_f(cdot,t)-f|_p, tto 0^+.$$ So:




    Is $(K_t)_{t>0}$ a summability kernel for $tto0^+$?




    I noticed that the coefficients $e^{-t n^2}to 1, tto 0^+$ boundedly (by $1$), so $K_ttodelta_0, tto 0^+$ in distribution, and then at least there's some hope.
    I tried with Poisson summation formula obtaining only a messier expression. Any ideas?






    fourier-series holder-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In trying to solve with Fourier series the heat equation on the 1-torus, I stumbled into this family of functions:
      $$forall t>0, K_t :mathbb{T}tomathbb{C}, smapstosum_{n∈mathbb{Z}}e^{-n^2 t}e^{ins},$$
      i.e. a family of functions indexed by $tin(0,+infty)$ where the $t$-member is a function with Fourier coefficients given by $(e^{-n^2 t})_{ninmathbb{Z}}$.
      It is clear that $forall t>0, K_tin C^infty(mathbb{T})$ and so, if $f$ is a distribution on $mathbb{T}$, the distribution $K_t*f$ is actually represented by an element of $C^infty(mathbb{T})$. Also, I've proved that the function:
      $$u_f :mathbb{T}times(0,+infty)tomathbb{C}, (x,t)mapsto (K_t*f)(x)$$
      is an element of $C^infty(mathbb{Ttimes(0,+infty)})$ and that:
      $$partial_t u_f = partial_{xx}u_f.$$
      So, I really want the family $(K_t)_{t>0}$ to be a summability kernel for $tto0^+$, because in this case $u_f$ would be a solution for the heat equation on the 1-torus with initial temperature $fin L^p(mathbb{T})$ with $1le p<infty$, in the sense that $$|u_f(cdot,t)-f|_p, tto 0^+.$$ So:




      Is $(K_t)_{t>0}$ a summability kernel for $tto0^+$?




      I noticed that the coefficients $e^{-t n^2}to 1, tto 0^+$ boundedly (by $1$), so $K_ttodelta_0, tto 0^+$ in distribution, and then at least there's some hope.
      I tried with Poisson summation formula obtaining only a messier expression. Any ideas?






      fourier-series holder-spaces







      share|cite|improve this question











      $endgroup$




      In trying to solve with Fourier series the heat equation on the 1-torus, I stumbled into this family of functions:
      $$forall t>0, K_t :mathbb{T}tomathbb{C}, smapstosum_{n∈mathbb{Z}}e^{-n^2 t}e^{ins},$$
      i.e. a family of functions indexed by $tin(0,+infty)$ where the $t$-member is a function with Fourier coefficients given by $(e^{-n^2 t})_{ninmathbb{Z}}$.
      It is clear that $forall t>0, K_tin C^infty(mathbb{T})$ and so, if $f$ is a distribution on $mathbb{T}$, the distribution $K_t*f$ is actually represented by an element of $C^infty(mathbb{T})$. Also, I've proved that the function:
      $$u_f :mathbb{T}times(0,+infty)tomathbb{C}, (x,t)mapsto (K_t*f)(x)$$
      is an element of $C^infty(mathbb{Ttimes(0,+infty)})$ and that:
      $$partial_t u_f = partial_{xx}u_f.$$
      So, I really want the family $(K_t)_{t>0}$ to be a summability kernel for $tto0^+$, because in this case $u_f$ would be a solution for the heat equation on the 1-torus with initial temperature $fin L^p(mathbb{T})$ with $1le p<infty$, in the sense that $$|u_f(cdot,t)-f|_p, tto 0^+.$$ So:




      Is $(K_t)_{t>0}$ a summability kernel for $tto0^+$?




      I noticed that the coefficients $e^{-t n^2}to 1, tto 0^+$ boundedly (by $1$), so $K_ttodelta_0, tto 0^+$ in distribution, and then at least there's some hope.
      I tried with Poisson summation formula obtaining only a messier expression. Any ideas?






      fourier-series holder-spaces




      fourier-series holder-spaces






      share|cite|improve this question















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      edited Feb 7 at 12:02







      Bob

















      asked Jan 30 at 16:09









      BobBob

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          $begingroup$

          By Poisson summation formula, we have
          $$
          frac1 {2pi}K_t(s)=frac{1}{2pi}sum_{nin Bbb Z} e^{-n^2 t}e^{ins} = frac 1 {2sqrt {pi t}}sum_{jinBbb Z}e^{-frac {(s-2pi j)^2}{4t}}.
          $$           Extend $f$ $2pi$-periodically to $Bbb R$. We obtain for $xinBbb T=[-pi,pi]$
          $$begin{eqnarray}
          (K_t*f)(x) &=&frac1{2pi}int_{-pi}^pi K_t(s)f(x-s)mathrm{d}s\
          &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi}^pi e^{-frac {(s-2pi j)^2}{4t}}f(x-s)mathrm ds\&=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s-2jpi)mathrm ds\
          &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s)mathrm ds\&=&
          frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}f(x-s)mathrm ds.
          end{eqnarray}$$           Note that the heat kernel $p_t(x) = frac 1 {2sqrt {pi t}} e^{-frac {x^2}{4t}}$ is a summability kernel as $trightarrow 0^+$. It follows
          $$begin{eqnarray}
          |K_t*f -f|_{L^p(Bbb T)}&le& frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}|f(cdot-s)-f(cdot)|_{L^p(Bbb T)}mathrm ds\&=& int_{-infty}^{infty} p_t(s)|tau_s f-f|_{L^p(Bbb T)}mathrm d sxrightarrow{tto 0^+} 0.
          end{eqnarray}$$






          share|cite|improve this answer











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            $begingroup$

            By Poisson summation formula, we have
            $$
            frac1 {2pi}K_t(s)=frac{1}{2pi}sum_{nin Bbb Z} e^{-n^2 t}e^{ins} = frac 1 {2sqrt {pi t}}sum_{jinBbb Z}e^{-frac {(s-2pi j)^2}{4t}}.
            $$             Extend $f$ $2pi$-periodically to $Bbb R$. We obtain for $xinBbb T=[-pi,pi]$
            $$begin{eqnarray}
            (K_t*f)(x) &=&frac1{2pi}int_{-pi}^pi K_t(s)f(x-s)mathrm{d}s\
            &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi}^pi e^{-frac {(s-2pi j)^2}{4t}}f(x-s)mathrm ds\&=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s-2jpi)mathrm ds\
            &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s)mathrm ds\&=&
            frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}f(x-s)mathrm ds.
            end{eqnarray}$$             Note that the heat kernel $p_t(x) = frac 1 {2sqrt {pi t}} e^{-frac {x^2}{4t}}$ is a summability kernel as $trightarrow 0^+$. It follows
            $$begin{eqnarray}
            |K_t*f -f|_{L^p(Bbb T)}&le& frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}|f(cdot-s)-f(cdot)|_{L^p(Bbb T)}mathrm ds\&=& int_{-infty}^{infty} p_t(s)|tau_s f-f|_{L^p(Bbb T)}mathrm d sxrightarrow{tto 0^+} 0.
            end{eqnarray}$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              By Poisson summation formula, we have
              $$
              frac1 {2pi}K_t(s)=frac{1}{2pi}sum_{nin Bbb Z} e^{-n^2 t}e^{ins} = frac 1 {2sqrt {pi t}}sum_{jinBbb Z}e^{-frac {(s-2pi j)^2}{4t}}.
              $$               Extend $f$ $2pi$-periodically to $Bbb R$. We obtain for $xinBbb T=[-pi,pi]$
              $$begin{eqnarray}
              (K_t*f)(x) &=&frac1{2pi}int_{-pi}^pi K_t(s)f(x-s)mathrm{d}s\
              &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi}^pi e^{-frac {(s-2pi j)^2}{4t}}f(x-s)mathrm ds\&=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s-2jpi)mathrm ds\
              &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s)mathrm ds\&=&
              frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}f(x-s)mathrm ds.
              end{eqnarray}$$               Note that the heat kernel $p_t(x) = frac 1 {2sqrt {pi t}} e^{-frac {x^2}{4t}}$ is a summability kernel as $trightarrow 0^+$. It follows
              $$begin{eqnarray}
              |K_t*f -f|_{L^p(Bbb T)}&le& frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}|f(cdot-s)-f(cdot)|_{L^p(Bbb T)}mathrm ds\&=& int_{-infty}^{infty} p_t(s)|tau_s f-f|_{L^p(Bbb T)}mathrm d sxrightarrow{tto 0^+} 0.
              end{eqnarray}$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                By Poisson summation formula, we have
                $$
                frac1 {2pi}K_t(s)=frac{1}{2pi}sum_{nin Bbb Z} e^{-n^2 t}e^{ins} = frac 1 {2sqrt {pi t}}sum_{jinBbb Z}e^{-frac {(s-2pi j)^2}{4t}}.
                $$                 Extend $f$ $2pi$-periodically to $Bbb R$. We obtain for $xinBbb T=[-pi,pi]$
                $$begin{eqnarray}
                (K_t*f)(x) &=&frac1{2pi}int_{-pi}^pi K_t(s)f(x-s)mathrm{d}s\
                &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi}^pi e^{-frac {(s-2pi j)^2}{4t}}f(x-s)mathrm ds\&=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s-2jpi)mathrm ds\
                &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s)mathrm ds\&=&
                frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}f(x-s)mathrm ds.
                end{eqnarray}$$                 Note that the heat kernel $p_t(x) = frac 1 {2sqrt {pi t}} e^{-frac {x^2}{4t}}$ is a summability kernel as $trightarrow 0^+$. It follows
                $$begin{eqnarray}
                |K_t*f -f|_{L^p(Bbb T)}&le& frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}|f(cdot-s)-f(cdot)|_{L^p(Bbb T)}mathrm ds\&=& int_{-infty}^{infty} p_t(s)|tau_s f-f|_{L^p(Bbb T)}mathrm d sxrightarrow{tto 0^+} 0.
                end{eqnarray}$$






                share|cite|improve this answer











                $endgroup$



                By Poisson summation formula, we have
                $$
                frac1 {2pi}K_t(s)=frac{1}{2pi}sum_{nin Bbb Z} e^{-n^2 t}e^{ins} = frac 1 {2sqrt {pi t}}sum_{jinBbb Z}e^{-frac {(s-2pi j)^2}{4t}}.
                $$                 Extend $f$ $2pi$-periodically to $Bbb R$. We obtain for $xinBbb T=[-pi,pi]$
                $$begin{eqnarray}
                (K_t*f)(x) &=&frac1{2pi}int_{-pi}^pi K_t(s)f(x-s)mathrm{d}s\
                &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi}^pi e^{-frac {(s-2pi j)^2}{4t}}f(x-s)mathrm ds\&=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s-2jpi)mathrm ds\
                &=&frac 1 {2sqrt {pi t}}sum_{jinBbb Z}int_{-pi-2jpi}^{pi-2jpi} e^{-frac {s^2}{4t}}f(x-s)mathrm ds\&=&
                frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}f(x-s)mathrm ds.
                end{eqnarray}$$                 Note that the heat kernel $p_t(x) = frac 1 {2sqrt {pi t}} e^{-frac {x^2}{4t}}$ is a summability kernel as $trightarrow 0^+$. It follows
                $$begin{eqnarray}
                |K_t*f -f|_{L^p(Bbb T)}&le& frac 1 {2sqrt {pi t}}int_{-infty}^{infty} e^{-frac {s^2}{4t}}|f(cdot-s)-f(cdot)|_{L^p(Bbb T)}mathrm ds\&=& int_{-infty}^{infty} p_t(s)|tau_s f-f|_{L^p(Bbb T)}mathrm d sxrightarrow{tto 0^+} 0.
                end{eqnarray}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 30 at 17:46

























                answered Jan 30 at 17:41









                SongSong

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                18.5k21651






























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