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== Proof ==
'''Assumption.''' <math>f\in L^1 \cap L^2</math>, i.e. <math>\int |f(x)|dx, \int |f(x)|^2 dx<\infty</math>
'''Step 1. The equality holds if ''f'' is differentiable and ''f'<nowiki/>'' is bounded'''
Let <math>f^{\star}(y)=\bar{f}(-y), \phi(x)=(f\ast f^{\star})(x)=\int f(x-y) f^{\star}(y)dy=\int f(x-y)\bar{f}(-y)dy=\int f(x+t)\bar{f}(t)dt</math>, then <math>|\frac{\partial [f(x+t)\bar{f}(t)]}{\partial x}|=|f'(x+t)\bar{f}(t)|\leq C|f(t)|</math>, and the [[Dominated convergence theorem|Dominated Convergence Theorem]] implies the interchangibility of differentiation and integration, thus <math>\phi '(x)=\int f'(x+t)\bar{f}(t)dt</math>, <math>\phi</math> is differentiable, hence by [[Fourier inversion theorem]], <math>\int|f(x)|^2 dx=\phi (0)=\lim\limits_{L\rightarrow \infty}\int_{-L}^{L} \mathcal{F}(\phi)(\xi)exp(2\pi i\cdot 0\cdot \xi)d\xi=\lim\limits_{L\rightarrow \infty}\int_{-L}^{L} \mathcal{F}(\phi)(\xi)d\xi</math>
By [[convolution theorem]] of Fourier transform, <math>\mathcal{F}(\phi)=\mathcal{F}(f)\mathcal{F}(f^{\star})=|\mathcal{F}(f)|^2=|\hat{f}|^2</math>, <math>\lim\limits_{L\rightarrow \infty}\int_{-L}^{L} |\hat{f}(\xi)|^2 d\xi=\int |\hat{f}(\xi)|^2 d\xi</math> by [[Monotone convergence theorem|Monotone Convergence Theorem]], hence <math>\int |f(x)|^2 dx=\int |\hat{f}(\xi)|^2 d\xi</math>
'''Step 2. the General Case'''
Let <math>\rho _\epsilon</math> be a family of [[Mollifier|mollifiers]], <math>f_\epsilon=f \ast \rho_\epsilon</math>, then for each ε, <math>f_\epsilon'=f\ast \rho_\epsilon'</math>, <math>|f_\epsilon'|=|f\ast \rho_\epsilon'|\leq \|f\|_{L^2}\|\rho_\epsilon'\|_{L^2}</math> by [[Hölder's inequality]], hence <math>f_\epsilon</math> is differentiable and has a bounded derivative. By '''Step 1''', <math>\int |f_\epsilon(x)|^2 dx=\int |\hat{f_\epsilon }(\xi)|^2 d\xi</math>. By the property of mollification, the left hand side converges to <math>\|f\|_{L^2}^2</math> as <math>\epsilon\rightarrow 0</math>, and by [[convolution theorem]], <math>|\hat{f_\epsilon }|=|\hat{f}||\hat{\rho_\epsilon }|\rightarrow |\hat{f}| \text{ as }\epsilon\rightarrow 0 </math>, hence by [[Fatou's lemma|Fatou' lemma]], we have <math>\int |\hat{f}|^2 d\xi \leq \liminf \limits_{\epsilon\rightarrow 0}\int |\hat{f_\epsilon}|^2 d\xi = \liminf \limits_{\epsilon\rightarrow 0} \int |f_\epsilon|^2 dx =\int |f|^2 dx </math>, thus <math>|\hat{f}|^2 </math> is integrable. Thus the right hand side converges to <math>\|\hat{f}\|_{L^2}^2</math> as <math>\epsilon\rightarrow 0</math> by [[Dominated convergence theorem|Dominated Convergence Theorem]]. Q.E.D.
==See also==
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