Smoothness criterion on function: $L^2$-Bernstein inequality

By | September 24, 2015

Lemma. Suppose that \mu\in M\left(\mathbb{R}^{d}\right) and \mathrm{\mathrm{supp\,}\mu} is compact. Then \hat{\mu} is in C^{\infty} and

    \[ D^{\alpha}\hat{\mu}=\left(\left(-2\pi ix\right)^{\alpha}\mu\right)^{\wedge}. \]

Furthermore, if \mathrm{supp}\mu\subset D\left(0,R\right), then

    \[ \norm{D^{\alpha}\hat{\mu}}{\infty}\le\left(2\pi R\right)^{\left|\alpha\right|}\norm{\mu}{}. \]

Theorem. Assume f\in L^{2} and \hat{f} is supported on D\left(0,R\right). Then f is indefinitely differentiable and \left\Vert D^{\alpha}f\right\Vert _{2}\le\left(2\pi R\right)^{\left|\alpha\right|}\left\Vert f\right\Vert _{2}.

Proof. 우선 Hausdorff-Young의 부등식에 의하여

    \[ \norm{\hat{f}}2\le\norm f2 \]

가 성립한다. 그러므로 \hat{f}\in L^{2}이고 D\left(0,R\right)가 finite measure를 가지므로 \hat{f}\in L^{1}\cap L^{2}다. 그러므로 Fourier inversion formula에 의하여

    \[ f\left(x\right)=\int_{\mathbb{R}^{d}}\hat{f}\left(\xi\right)e^{2\pi ix\cdot\xi}dx \]

가 성립한다.

L^{1}\left(\mathbb{R}^{d}\right)\hookrightarrow M\left(\mathbb{R}^{d}\right)이므로 \hat{f}\left(x\right)dx는 finite Borel measure on \mathbb{R}^{d}이고
이는 supported on D\left(0,R\right)이므로 Lemma에 의하여 fC^{\infty}이다. 그리고

    \[ \norm{D^{\alpha}f}2=\norm{\widehat{D^{\alpha}f}}2=\norm{\left(2\pi i\xi\right)^{\alpha}\hat{f}}2\le\left(2\pi R\right)^{\left|\alpha\right|}\norm{\hat{f}}2=\left(2\pi R\right)^{\left|\alpha\right|}\norm f2 \]

를 얻는다.

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