Real positive index Sobolev spaces are Hilbert spaces












0












$begingroup$


I'm trying to prove that, for $kgeq 0$, Sobolev spaces defined in this way:



$H^k(mathbb{T})={fin L^2(mathbb{T}): sum_{n=-infty}^{+infty}(1+n^2)^k|hat{f}(n)|^2 < +infty}$



are Hilbert spaces over $mathbb{C}$ with respect to the inner product:



$(f,g)=sum_{n=-infty}^{+infty}(1+n^2)^khat{f}(n)overline{hat{g}(n)}$,



where $hat{f}$ is the Fourier transform of $f$ in $mathbb{T}=[-pi, pi)$.



I proved that $H^k(mathbb{T})$ is a vector space over $mathbb{C}$ $forall kgeq 0$ and that is an inner product space. Now I need to prove that $H^k(mathbb{T})$ is complete with respect to the distance induced by the norm $||cdot||=(cdot,cdot)^{1/2}$.



So I considered a Cauchy sequence ${f_m}_{minmathbb{N}}subseteq H^k(mathbb{T})$. This means in particular that, $forall ninmathbb{Z}$, the sequence ${hat{f_m}(n)}_{minmathbb{N}}$ is a Cauchy sequence in $mathbb{C}$, therefore it converges to some $g(n)inmathbb{C}$, because $mathbb{C}$ is complete. So I defined



$f(x):=sum_{n=-infty}^{+infty} g(n)e^{inx}$, $forall xinmathbb{T}$.



I managed to prove that $hat{f}(n)=g(n)$, $forall ninmathbb{Z}$, but now I'm finding troubles in showing that $fin H^k(mathbb{T})$ and that $f_mrightarrow f$ with respect to the norm in $H^k(mathbb{T})$.



Is my idea correct? How could I proceed?






real-analysis fourier-analysis hilbert-spaces fourier-transform







share|cite|improve this question









$endgroup$












  • $begingroup$
    how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
    $endgroup$
    – mathworker21
    Jan 7 at 19:14
















0












$begingroup$


I'm trying to prove that, for $kgeq 0$, Sobolev spaces defined in this way:



$H^k(mathbb{T})={fin L^2(mathbb{T}): sum_{n=-infty}^{+infty}(1+n^2)^k|hat{f}(n)|^2 < +infty}$



are Hilbert spaces over $mathbb{C}$ with respect to the inner product:



$(f,g)=sum_{n=-infty}^{+infty}(1+n^2)^khat{f}(n)overline{hat{g}(n)}$,



where $hat{f}$ is the Fourier transform of $f$ in $mathbb{T}=[-pi, pi)$.



I proved that $H^k(mathbb{T})$ is a vector space over $mathbb{C}$ $forall kgeq 0$ and that is an inner product space. Now I need to prove that $H^k(mathbb{T})$ is complete with respect to the distance induced by the norm $||cdot||=(cdot,cdot)^{1/2}$.



So I considered a Cauchy sequence ${f_m}_{minmathbb{N}}subseteq H^k(mathbb{T})$. This means in particular that, $forall ninmathbb{Z}$, the sequence ${hat{f_m}(n)}_{minmathbb{N}}$ is a Cauchy sequence in $mathbb{C}$, therefore it converges to some $g(n)inmathbb{C}$, because $mathbb{C}$ is complete. So I defined



$f(x):=sum_{n=-infty}^{+infty} g(n)e^{inx}$, $forall xinmathbb{T}$.



I managed to prove that $hat{f}(n)=g(n)$, $forall ninmathbb{Z}$, but now I'm finding troubles in showing that $fin H^k(mathbb{T})$ and that $f_mrightarrow f$ with respect to the norm in $H^k(mathbb{T})$.



Is my idea correct? How could I proceed?






real-analysis fourier-analysis hilbert-spaces fourier-transform







share|cite|improve this question









$endgroup$












  • $begingroup$
    how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
    $endgroup$
    – mathworker21
    Jan 7 at 19:14














0












0








0





$begingroup$


I'm trying to prove that, for $kgeq 0$, Sobolev spaces defined in this way:



$H^k(mathbb{T})={fin L^2(mathbb{T}): sum_{n=-infty}^{+infty}(1+n^2)^k|hat{f}(n)|^2 < +infty}$



are Hilbert spaces over $mathbb{C}$ with respect to the inner product:



$(f,g)=sum_{n=-infty}^{+infty}(1+n^2)^khat{f}(n)overline{hat{g}(n)}$,



where $hat{f}$ is the Fourier transform of $f$ in $mathbb{T}=[-pi, pi)$.



I proved that $H^k(mathbb{T})$ is a vector space over $mathbb{C}$ $forall kgeq 0$ and that is an inner product space. Now I need to prove that $H^k(mathbb{T})$ is complete with respect to the distance induced by the norm $||cdot||=(cdot,cdot)^{1/2}$.



So I considered a Cauchy sequence ${f_m}_{minmathbb{N}}subseteq H^k(mathbb{T})$. This means in particular that, $forall ninmathbb{Z}$, the sequence ${hat{f_m}(n)}_{minmathbb{N}}$ is a Cauchy sequence in $mathbb{C}$, therefore it converges to some $g(n)inmathbb{C}$, because $mathbb{C}$ is complete. So I defined



$f(x):=sum_{n=-infty}^{+infty} g(n)e^{inx}$, $forall xinmathbb{T}$.



I managed to prove that $hat{f}(n)=g(n)$, $forall ninmathbb{Z}$, but now I'm finding troubles in showing that $fin H^k(mathbb{T})$ and that $f_mrightarrow f$ with respect to the norm in $H^k(mathbb{T})$.



Is my idea correct? How could I proceed?






real-analysis fourier-analysis hilbert-spaces fourier-transform







share|cite|improve this question









$endgroup$




I'm trying to prove that, for $kgeq 0$, Sobolev spaces defined in this way:



$H^k(mathbb{T})={fin L^2(mathbb{T}): sum_{n=-infty}^{+infty}(1+n^2)^k|hat{f}(n)|^2 < +infty}$



are Hilbert spaces over $mathbb{C}$ with respect to the inner product:



$(f,g)=sum_{n=-infty}^{+infty}(1+n^2)^khat{f}(n)overline{hat{g}(n)}$,



where $hat{f}$ is the Fourier transform of $f$ in $mathbb{T}=[-pi, pi)$.



I proved that $H^k(mathbb{T})$ is a vector space over $mathbb{C}$ $forall kgeq 0$ and that is an inner product space. Now I need to prove that $H^k(mathbb{T})$ is complete with respect to the distance induced by the norm $||cdot||=(cdot,cdot)^{1/2}$.



So I considered a Cauchy sequence ${f_m}_{minmathbb{N}}subseteq H^k(mathbb{T})$. This means in particular that, $forall ninmathbb{Z}$, the sequence ${hat{f_m}(n)}_{minmathbb{N}}$ is a Cauchy sequence in $mathbb{C}$, therefore it converges to some $g(n)inmathbb{C}$, because $mathbb{C}$ is complete. So I defined



$f(x):=sum_{n=-infty}^{+infty} g(n)e^{inx}$, $forall xinmathbb{T}$.



I managed to prove that $hat{f}(n)=g(n)$, $forall ninmathbb{Z}$, but now I'm finding troubles in showing that $fin H^k(mathbb{T})$ and that $f_mrightarrow f$ with respect to the norm in $H^k(mathbb{T})$.



Is my idea correct? How could I proceed?






real-analysis fourier-analysis hilbert-spaces fourier-transform




real-analysis fourier-analysis hilbert-spaces fourier-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 18:55









LukathLukath

575




575












  • $begingroup$
    how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
    $endgroup$
    – mathworker21
    Jan 7 at 19:14


















  • $begingroup$
    how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
    $endgroup$
    – mathworker21
    Jan 7 at 19:14
















$begingroup$
how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
$endgroup$
– mathworker21
Jan 7 at 19:14




$begingroup$
how do you know the definition of $f$ is valid? That is, how do you know the infinite sum converges? Moreover, how did you show that $hat{f}(n) = g(n)$ for all $n in mathbb{Z}$?
$endgroup$
– mathworker21
Jan 7 at 19:14










1 Answer
1






active

oldest

votes


















0












$begingroup$

I would personally take the $L^2$ route, to avoid issues I brought up in my above comment. Note that $(f_m)_m$ is Cauchy in $L^2$: for any $epsilon > 0$ and for $m,l$ large enough, we have $$||f_m-f_l||_2^2 = sum_{n=-infty}^infty |widehat{f_m}-widehat{f_l}|^2 le sum_{n=-infty}^infty (1+n^2)^k |widehat{f_m}-widehat{f_l}|^2 le epsilon.$$ Since we know $L^2$ is complete, let $f$ be the $L^2$ limit of the $f_m$'s. It remains to show $f in H^k(mathbb{T})$ and $f_m to f$ in $H^k$. But these should just be some easy triangle inequality or Cauchy-Schwarz arguments.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
    $endgroup$
    – Lukath
    Jan 8 at 12:47











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0












$begingroup$

I would personally take the $L^2$ route, to avoid issues I brought up in my above comment. Note that $(f_m)_m$ is Cauchy in $L^2$: for any $epsilon > 0$ and for $m,l$ large enough, we have $$||f_m-f_l||_2^2 = sum_{n=-infty}^infty |widehat{f_m}-widehat{f_l}|^2 le sum_{n=-infty}^infty (1+n^2)^k |widehat{f_m}-widehat{f_l}|^2 le epsilon.$$ Since we know $L^2$ is complete, let $f$ be the $L^2$ limit of the $f_m$'s. It remains to show $f in H^k(mathbb{T})$ and $f_m to f$ in $H^k$. But these should just be some easy triangle inequality or Cauchy-Schwarz arguments.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
    $endgroup$
    – Lukath
    Jan 8 at 12:47
















0












$begingroup$

I would personally take the $L^2$ route, to avoid issues I brought up in my above comment. Note that $(f_m)_m$ is Cauchy in $L^2$: for any $epsilon > 0$ and for $m,l$ large enough, we have $$||f_m-f_l||_2^2 = sum_{n=-infty}^infty |widehat{f_m}-widehat{f_l}|^2 le sum_{n=-infty}^infty (1+n^2)^k |widehat{f_m}-widehat{f_l}|^2 le epsilon.$$ Since we know $L^2$ is complete, let $f$ be the $L^2$ limit of the $f_m$'s. It remains to show $f in H^k(mathbb{T})$ and $f_m to f$ in $H^k$. But these should just be some easy triangle inequality or Cauchy-Schwarz arguments.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
    $endgroup$
    – Lukath
    Jan 8 at 12:47














0












0








0





$begingroup$

I would personally take the $L^2$ route, to avoid issues I brought up in my above comment. Note that $(f_m)_m$ is Cauchy in $L^2$: for any $epsilon > 0$ and for $m,l$ large enough, we have $$||f_m-f_l||_2^2 = sum_{n=-infty}^infty |widehat{f_m}-widehat{f_l}|^2 le sum_{n=-infty}^infty (1+n^2)^k |widehat{f_m}-widehat{f_l}|^2 le epsilon.$$ Since we know $L^2$ is complete, let $f$ be the $L^2$ limit of the $f_m$'s. It remains to show $f in H^k(mathbb{T})$ and $f_m to f$ in $H^k$. But these should just be some easy triangle inequality or Cauchy-Schwarz arguments.






share|cite|improve this answer









$endgroup$



I would personally take the $L^2$ route, to avoid issues I brought up in my above comment. Note that $(f_m)_m$ is Cauchy in $L^2$: for any $epsilon > 0$ and for $m,l$ large enough, we have $$||f_m-f_l||_2^2 = sum_{n=-infty}^infty |widehat{f_m}-widehat{f_l}|^2 le sum_{n=-infty}^infty (1+n^2)^k |widehat{f_m}-widehat{f_l}|^2 le epsilon.$$ Since we know $L^2$ is complete, let $f$ be the $L^2$ limit of the $f_m$'s. It remains to show $f in H^k(mathbb{T})$ and $f_m to f$ in $H^k$. But these should just be some easy triangle inequality or Cauchy-Schwarz arguments.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 19:19









mathworker21mathworker21

8,8871928




8,8871928












  • $begingroup$
    You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
    $endgroup$
    – Lukath
    Jan 8 at 12:47


















  • $begingroup$
    You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
    $endgroup$
    – Lukath
    Jan 8 at 12:47
















$begingroup$
You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
$endgroup$
– Lukath
Jan 8 at 12:47




$begingroup$
You're right, I did a huge mistake in one inequality. Still I cannot show the rest... how can I be sure that the Fourier coefficients of $f$ decrease sufficiently fast in order to have $||f||$ finite?
$endgroup$
– Lukath
Jan 8 at 12:47


















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