# Proving Plancherel's identity.

I want to prove that $$\Arrowvert \hat{f} \Arrowvert_{l^2(\mathbb{Z})} = \Arrowvert f \Arrowvert _{L^2([-L,L])},$$

Could anyone give me a hint please?

• Show that the Fourier series a continuously differentiable periodic function $f$ converges uniformly to $f$. Use this to show what you want for such $f$. Then approximate. May 14, 2017 at 20:45
• @DisintegratingByParts I have seen this math.stackexchange.com/questions/1353893/… but here he uses Fourier transform How can I prove it without using Fourier transform? May 19, 2017 at 23:37

The theorem you want to prove is $$\int_{-L}^{L}|f(x)|^2dx = \frac{1}{2L}\sum_{n=-\infty}^{\infty}\left|\int_{-L}^{L}f(t)e^{-in\pi t/L}dt\right|^2$$ This is cast into the $L^2$ framework by considering the orthonormal set $$e_n(x) = \frac{1}{\sqrt{2L}}e^{in\pi x/L}.$$ In this framework, the theorem is $\|f\|^2 = \sum_{n=-\infty}^{\infty}|(f,e_n)|^2$ where $(f,g) = \int_{-L}^{L}f(t)\overline{g(t)}dt$ is the inner product on the complex space $L^2[-L,L]$. There are several ways to prove this theorem because of the following:

Theorem: Let $H$ be a Hilbert space, and let $\{ e_n \}_{n=-\infty}^{\infty}$ be an orthonormal subset of $H$. Then the following are equivalent:

1. $\|f\|^2 = \sum_{n}|(f,e_n)|^2$ for all $f \in H$.

2. $\|f\|^2 = \sum_{n}|(f,e_n)|^2$ for all $f$ in a dense subspace $M$ of $H$.

3. $\sum_{n}(f,e_n)(e_n,g)=(f,g)$ for all $f,g\in H$.

4. $\sum_{n=-N}^{N}(f,e_n)e_n$ converges to $f$ in the norm of $H$ as $N\rightarrow\infty$.

5. $\sum_{n=-N}^{N}(f,e_n)e_n$ converges to $f$ in the norm of $H$ as $N\rightarrow\infty$ for all $f$ in a dense subspace of $H$.

6. The only $f\in H$ for which $(f,e_n)=0$ holds for all $n$ is $f=0$.

Proving (5) is one of the eaiser ways in this particular case because it can be shown that the space $C^1_p[-L,L]$ of periodic continuous differentiable functions is dense in $H$, and classical uniform pointwise convergence results for the Fourier series then imply $L^2$ norm convergence. Indeed, for any such $f$, it follows that \begin{align} (f,e_n) & = \frac{1}{\sqrt{2L}}\int_{-\pi}^{\pi}f(t)e^{-in\pi x/L}f(t)dt \\ & = \left.\frac{1}{\sqrt{2L}}f(t)\frac{e^{-in\pi t/L}}{-in\pi/L}\right|_{-L}^{L}+\frac{1}{\sqrt{2L}in\pi/ L}\int_{-L}^{L}f'(t)e^{-int/L}dt \\ & = \frac{1}{\sqrt{2L}in\pi/ L}\int_{-L}^{L}f'(t)e^{-int/L}dt. \end{align} So the following series converges pointwise absolutely and uniformly on $[-L,L]$ $$g(x)=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}(f,e_n)e_n(x)$$ because $\sum_{n=-N}^{N}|(f,e_n)|^2 \le \|f\|^2$ always holds for any $f$ and for any orthonormal set $\{ e_n \}$, and because $$\sum_{n=-N}^{N}|(f,e_n)e_n(x)| \le \sum_{n=-N}^{N}|(f,e_n)|\frac{1}{\sqrt{2/L}n\pi} \\ \le \frac{1}{\sqrt{2/L}}|(f,e_0)|+\frac{1}{\sqrt{2/L}\pi}\sum_{n=-N,n\ne 0}^{N}\frac{1}{n}|(f',e_n)| \\ \le \frac{1}{\sqrt{2/L}}|(f,e_0)|+\frac{1}{\sqrt{2/L}\pi}\left(\sum_{n=-N,n\ne 0}^{N}\frac{1}{n^2}\right)^{1/2}\left(\sum_{n=-N,n\ne 0}^{N}|(f',e_n)|^2\right)^{1/2}.$$ So the series $\sum_{n=-N}^{N}(f,e_n)e_n(x)$ converges absolutely and uniformly to some continuous functions $g$. Actually it must converge everywhere to $f$ because of classical convergence theorems for the Fourier series. Because the series converges absolutely and uniformly to $g=f$, then (5) holds for all $f$ in the dense subspace $C^1_p[-L,L]$ because the series converges uniformly to $f$ for all $f\in C_p^1[-L,L]$, which certainly implies $L^2$ norm convergence.

• why did you put $1/\sqrt(2L)$ in the orthonormal set? May 23, 2017 at 4:38
• @weakmathematician : $\int_{-L}^{L}\left|\frac{e^{in\pi x/L}}{\sqrt{2L}}\right|^2dx = \int_{-L}^{L}\frac{1}{2L}dx = \frac{2L}{2L}=1$. May 23, 2017 at 5:04
• Do this theorem have a name ? May 23, 2017 at 8:54
• @weakmathematician : If a function is continuously differentiable and periodic, then its Fourier series converges uniformly to the function; and uniform convergence implies $L^2$ convergence. So the proof reduces to classical convergence results of Fourier series. May 23, 2017 at 10:58
• @weakmathematician : If you know $C(T)$ is dense in $L^2$, then you can show that $C^1(T)$ is dense in $C(T)$ to achieve the result. May 29, 2017 at 10:31

If you have a quantum mechanical heart, you can consider an infinite dimensional Hilbert space $$H$$ over $$[-L, L]$$, and its dual space $$H^{*}$$. You then can construct projective spaces of those $$P(H)$$ and $$P(H^{*})$$ and define the vectors on them $$|f\rangle$$ and $$\langle f|$$ respectively. The inner-product may be cast as $$\langle f|f\rangle$$. To write the $$L^2$$ norm you essentially need to resolve the identity as $$I=\int_{-L}^{L}|x\rangle\!\langle x|dx$$. Inserting the identity in the inner product you get $$\langle f|f\rangle=\langle f|I|f\rangle=\int_{-L}^{L}\langle f|x\rangle\!\langle x|f\rangle dx=\int_{-L}^{L}|f(x)|^{2}dx$$ On the other hand, you can pick up an orthonormal countable basis for the projective space $$\{|n\rangle\}_{n\in\mathbb{Z}}$$, so that any vector on it (including $$|f\rangle$$) may be expanded in the basis due to the completeness $$|f\rangle=\sum_{n\in\mathbb{Z}}c_{n}|n\rangle$$ The coefficients of the expansion due to the orthonormality are clearly $$c_{n}=\langle n|f\rangle=\langle n|I|f\rangle=\int_{-L}^{L}\langle n|x\rangle\!\langle x|f\rangle dx=\int_{-L}^{L}n^{*}(x)f(x)dx$$ The inner product is then $$\langle f|f\rangle=\sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}c^{*}_{m}c_{n}\langle m|n\rangle=\sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}c^{*}_{m}c_{n}\delta_{mn}=\sum_{n\in\mathbb{Z}}|c_{n}|^{2}=\sum_{n\in\mathbb{Z}}\left|\int_{-L}^{L}n^{*}(x)f(x)dx\right |^{2}$$ Hence $$\int_{-L}^{L}|f(x)|^{2}dx=\sum_{n\in\mathbb{Z}}\left|\int_{-L}^{L}n^{*}(x)f(x)dx\right|^{2}$$ or in your notation $$||f||_{L^{2}[-L, L]}=||f||_{l^{2}(\mathbb{Z})}$$

• And if you don't see why identity is $I=\int_{[-L, L]}|x><x|dx$, consider expanding the vector on the projective space $|f>$ in the uncountable basis $\{|x>\}$ (which you can do by completeness) $|f>=\int_{[-L, L]}f(x)|x>dx$, on the other hand $<y|f>=\int_{[-L, L]}f(x)<y|x>dx=\int_{[-L, L]}f(x)\delta(x-y)dx=f(y)$, so $|f>=\int_{[-L, L]}f(x)|x>dx=\int_{[-L, L]}<x|f>|x>dx=\int_{[-L, L]}|x><x|f>dx$, this must hold for any $|f>$, so $I=\int_{[-L, L]}|x><x|dx$. Similarly you can show that in the countable basis $I=\sum_{n\in\mathbb{Z}}|n><n|$. Jun 13, 2017 at 14:04