$lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$.












1












$begingroup$


How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?



The definition is,



$lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.



I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.






analysis fourier-analysis fourier-transform harmonic-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?



    The definition is,



    $lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.



    I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
    I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.






    analysis fourier-analysis fourier-transform harmonic-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?



      The definition is,



      $lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.



      I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
      I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.






      analysis fourier-analysis fourier-transform harmonic-analysis







      share|cite|improve this question









      $endgroup$




      How do I show that $lVert f rVert _{H^{s}(mathbb{R}^d )} leq C lVert f rVert_{L^{1} (mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?



      The definition is,



      $lVert f rVert_{H^{s}} = lVert langle nabla rangle^{s}f rVert_{L^2}$ and $langle nabla rangle^{s} = (1+|nabla|^2)^{frac{s}{2}}$.



      I tried there is a $f$ which is not $H^{s}$ but is $L^1$.
      I am gussing it is $langle D rangle ^{-s} frac{1}{|x|^{d/2}}$. But I can't.






      analysis fourier-analysis fourier-transform harmonic-analysis




      analysis fourier-analysis fourier-transform harmonic-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 3:22









      IdkwhatIdkwhat

      236




      236






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Idea: If you take $f(x) = delta(x)$, then
          you see that $hat f(xi) = text{const}$, say $1$. Hence we have
          begin{align}
          |f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
          end{align}
          However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.



          Spoiler:




          Let $d=1$. Take the sequence
          $f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
            $endgroup$
            – Idkwhat
            Dec 16 '18 at 9:13












          • $begingroup$
            Delta function is not a function.
            $endgroup$
            – Jacky Chong
            Dec 16 '18 at 18:10










          • $begingroup$
            Ooh, thank you.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 5:49










          • $begingroup$
            Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 8:37












          • $begingroup$
            Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
            $endgroup$
            – Jacky Chong
            Dec 17 '18 at 13:51











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Idea: If you take $f(x) = delta(x)$, then
          you see that $hat f(xi) = text{const}$, say $1$. Hence we have
          begin{align}
          |f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
          end{align}
          However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.



          Spoiler:




          Let $d=1$. Take the sequence
          $f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
            $endgroup$
            – Idkwhat
            Dec 16 '18 at 9:13












          • $begingroup$
            Delta function is not a function.
            $endgroup$
            – Jacky Chong
            Dec 16 '18 at 18:10










          • $begingroup$
            Ooh, thank you.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 5:49










          • $begingroup$
            Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 8:37












          • $begingroup$
            Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
            $endgroup$
            – Jacky Chong
            Dec 17 '18 at 13:51
















          1












          $begingroup$

          Idea: If you take $f(x) = delta(x)$, then
          you see that $hat f(xi) = text{const}$, say $1$. Hence we have
          begin{align}
          |f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
          end{align}
          However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.



          Spoiler:




          Let $d=1$. Take the sequence
          $f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
            $endgroup$
            – Idkwhat
            Dec 16 '18 at 9:13












          • $begingroup$
            Delta function is not a function.
            $endgroup$
            – Jacky Chong
            Dec 16 '18 at 18:10










          • $begingroup$
            Ooh, thank you.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 5:49










          • $begingroup$
            Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 8:37












          • $begingroup$
            Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
            $endgroup$
            – Jacky Chong
            Dec 17 '18 at 13:51














          1












          1








          1





          $begingroup$

          Idea: If you take $f(x) = delta(x)$, then
          you see that $hat f(xi) = text{const}$, say $1$. Hence we have
          begin{align}
          |f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
          end{align}
          However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.



          Spoiler:




          Let $d=1$. Take the sequence
          $f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$







          share|cite|improve this answer









          $endgroup$



          Idea: If you take $f(x) = delta(x)$, then
          you see that $hat f(xi) = text{const}$, say $1$. Hence we have
          begin{align}
          |f|_{H^{-d/2}(mathbb{R}^d)} = int_{mathbb{R}^d} frac{dxi}{(1+|xi|^2)^{d/2}} sim int^infty_0frac{r^{d-1}dr}{(1+r^d)} = infty.
          end{align}
          However, we see that $|f|_{L^1(mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.



          Spoiler:




          Let $d=1$. Take the sequence
          $f_N(x) = N(2pi)^{-1/2}exp(-N^2x^2/2)$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 5:24









          Jacky ChongJacky Chong

          18.8k21129




          18.8k21129












          • $begingroup$
            Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
            $endgroup$
            – Idkwhat
            Dec 16 '18 at 9:13












          • $begingroup$
            Delta function is not a function.
            $endgroup$
            – Jacky Chong
            Dec 16 '18 at 18:10










          • $begingroup$
            Ooh, thank you.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 5:49










          • $begingroup$
            Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 8:37












          • $begingroup$
            Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
            $endgroup$
            – Jacky Chong
            Dec 17 '18 at 13:51


















          • $begingroup$
            Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
            $endgroup$
            – Idkwhat
            Dec 16 '18 at 9:13












          • $begingroup$
            Delta function is not a function.
            $endgroup$
            – Jacky Chong
            Dec 16 '18 at 18:10










          • $begingroup$
            Ooh, thank you.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 5:49










          • $begingroup$
            Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
            $endgroup$
            – Idkwhat
            Dec 17 '18 at 8:37












          • $begingroup$
            Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
            $endgroup$
            – Jacky Chong
            Dec 17 '18 at 13:51
















          $begingroup$
          Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
          $endgroup$
          – Idkwhat
          Dec 16 '18 at 9:13






          $begingroup$
          Can I ask why 'Spoiler' required ? I think the 'Idea' suffices to show.
          $endgroup$
          – Idkwhat
          Dec 16 '18 at 9:13














          $begingroup$
          Delta function is not a function.
          $endgroup$
          – Jacky Chong
          Dec 16 '18 at 18:10




          $begingroup$
          Delta function is not a function.
          $endgroup$
          – Jacky Chong
          Dec 16 '18 at 18:10












          $begingroup$
          Ooh, thank you.
          $endgroup$
          – Idkwhat
          Dec 17 '18 at 5:49




          $begingroup$
          Ooh, thank you.
          $endgroup$
          – Idkwhat
          Dec 17 '18 at 5:49












          $begingroup$
          Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
          $endgroup$
          – Idkwhat
          Dec 17 '18 at 8:37






          $begingroup$
          Can I say $f:=lim_{N rightarrow infty} f_{N}$ is a function? I think that function is what I want but I can't be sure because it approaches to Dirac delta function as I know.
          $endgroup$
          – Idkwhat
          Dec 17 '18 at 8:37














          $begingroup$
          Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
          $endgroup$
          – Jacky Chong
          Dec 17 '18 at 13:51




          $begingroup$
          Your job is to prove there does not exist a constant such that the inequality holds not whether $f_N$ converge to the delta function or not.
          $endgroup$
          – Jacky Chong
          Dec 17 '18 at 13:51


















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