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If $\int_0^1\lvert f\rvert^2=\sum\limits_{n\in\mathbf Z}\lvert\hat f_n\rvert^2$ then how can I derive $\int_0^1f\bar g =\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}$

$\hat f_n$ is the fourier transform of $f$ and I try to show the parseval identity, the problem is, one does not know if the function admits a fourier series, so I have to derive it only from the knowledge that LHS holds

I know normally the converse is true (Identitiy on the right implies the one on the left) but is there a trick to show the other direction If I take for example $f+g$ then

$\int_0^1\lvert f+g\rvert^2=\int_0^1\lvert f\rvert^2+\int_0^1\lvert g\rvert^2+\int_0^1f\bar g+\int_0^1g\bar f$

$\sum\limits_{n\in\mathbf Z}\lvert\hat f_n+\hat g_n\rvert^2=\sum\limits_{n\in\mathbf Z}\lvert\hat f_n\rvert^2+\sum\limits_{n\in\mathbf Z}\lvert\hat g_n\rvert^2+\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}+\sum\limits_{n\in\mathbf Z}\overline{\hat f_n}\hat g_n$

At least I have

$\int_0^1f\bar g+\int_0^1g\bar f=\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}+\sum\limits_{n\in\mathbf Z}\overline{\hat f_n}\hat g_n$

ketum
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  • Perhaps you could use the polarisation identity? – copper.hat Jun 06 '16 at 04:43
  • @copper.hat : so you interpret his question as : for every $h \in L^2([0,1])$, there are coefs $\hat{h}n$ such that $|h|{L^2}^2 = \sum_n |\hat{h}_n|^2$, but $\hat{h}_n$ doesn't have to be related to the Fourier series – reuns Jun 06 '16 at 04:49
  • and in that case you should look at wiki/Polarization_identity#For_vector_spaces_with_complex_scalars. otherwise if $\hat{f}_n$ are really the Fourier coefs, you should better write $f = \sum_n \hat{f}_n e_n$ where $(e_n) = (e^{2 i \pi n x})$ is an orthonormal basis of $L^2([0,1])$ – reuns Jun 06 '16 at 04:51
  • @user1952009: Well, I am assuming that there is an isometry $h \mapsto \hat{h}$. – copper.hat Jun 06 '16 at 04:53
  • @copper.hat : in that case there is also the isometry backward $l^2 \to L^2([0,1])$ i.e. $f = \sum_n \hat{f}_n \phi_n$ for some orthonormal basis $(\phi_n)$ – reuns Jun 06 '16 at 04:55
  • @user1952009 This is exactly the problem, the equality $f=\sum_n\hat f_n e_n$ holds for $f\in C^1$ (I proved it in a previous exercise) then I showed the left identity in the yellow box now I have to show this, If you understand german it is here ''Aufgabe 3 (iv) '' http://www.math.uzh.ch/index.php?file&key1=40906 – ketum Jun 06 '16 at 04:58
  • @ketum it holds also for any $f \in L^2([0,1])$ in the sense $|f - \sum_{|n| < N} \hat{f}n e_n |{L^2} \to 0$ as $N \to \infty$ (which is different to the pointwise convergence), this is how you make sense to the Fourier series of $f \in L^2([0,1])$ – reuns Jun 06 '16 at 05:02
  • @user1952009 Yes I proved also that $L^2$ convergence in part (i) does that imply that f admits a Fourier series ? – ketum Jun 06 '16 at 05:03
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    (this is why I didn't write $f(x) = \sum_n \hat{f}n e_n(x)$ which would mean the pointwise convergence) you have to understand how the pointwise convergence is very different to the $|.|{L^2}$ convergence – reuns Jun 06 '16 at 05:04
  • I'm not exactly sure what you are asking. If you have some isometry $f \mapsto \hat{f}$ then the right hand side follows immediately from the polarisation identity. – copper.hat Jun 06 '16 at 05:04
  • @user1952009: The point is, just the fact that the inner product can be written in terms of norms is sufficient. The result then follows from the isometry. – copper.hat Jun 06 '16 at 05:09
  • @user1952009: I'm not exactly sure what is being asked here, I am just pointing out that the isometry alone is sufficient to prove Parseval's identity. The OP might be asking about the structure of $L^2([0,1])$ or a dense subset, but that is not clear either... – copper.hat Jun 06 '16 at 05:16
  • @user1952009 how can I write in this case $\int_0^1 f(x)\bar {g(x)}$ ? maybe $||f\bar g-S_N(f\bar g)||_{L^2}\to 0$ ? – ketum Jun 06 '16 at 05:27
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    why do you think to $f \bar{g}$ as an element of $L^2([0,1])$ ?? it is not true (try with $f(x) = g(x)= x^{-1/3}$). no, it is just that $\int_0^1 f(x) \bar{g}(x) dx = \langle f,g \rangle =\langle \sum_n \hat{f}_n e_n, \sum_n \hat{g}_n e_n \rangle = \sum_n \langle \hat{f}_n e_n, \hat{g}_n e_n \rangle $ $= \sum_n \hat{f}_n \overline{\hat{g}_n} \langle e_n, e_n \rangle = \sum_n \hat{f}_n \overline{\hat{g}_n}$ (where $\langle \sum_n \hat{f}_n e_n, \sum_n \hat{g}_n e_n \rangle = \sum_n \langle \hat{f}_n e_n, \hat{g}_n e_n \rangle $ is because of the orthogalinty of $(e_n)$) – reuns Jun 06 '16 at 05:29
  • @user1952009 thanks and apologize but this is a bit new for me. So I 've only shown $f$ and its $S_N(f)$ agree in $L^2$ norm so don't I need to justify the middle (2nd) equality in your comment, with Cauchy-Schwarz or so ? – ketum Jun 06 '16 at 05:37
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    prove what you think you need to prove. Yes for proving that $\lim_{N \to \infty} \langle f - S_n(f),g \rangle \to 0$ you need some argument, Cauchy-Schwarz, or the definition $\int_0^1 ( f - S_n(f))(x) \bar{g}[x) dx$. Then, from $\lim_{N \to \infty} \langle f - S_n(f),g \rangle \to 0$ you know that $\langle f,g \rangle = \lim_{N \to \infty} \langle S_N(f),g \rangle = \lim_{N \to \infty} \langle S_N(f),S_N(g) \rangle$ – reuns Jun 06 '16 at 05:41

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