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Let $f$ be a function in the Schwartz space $\mathcal{S}(\mathbb{R})$ and let the Fourier transform of $f$ be defined by \begin{equation} \hat{f}(k) = \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} f(x) e^{-ikx} dx \end{equation} with inverse \begin{equation} f(x) = \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} \hat{f}(k) e^{ikx} dk \end{equation} Then obviously \begin{equation} |f(x)| \leq \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} |\hat{f}(k)| dk. \end{equation} Now I should be able to find the inequality \begin{equation} |f(x)|^2 \leq \frac{1}{2} \int_\mathbb{R} |\hat{f}(k)|^2(1+k^2) dk. \end{equation} I already tried partial integration, observed that $\int_\mathbb{R} \frac{1}{1+k^2}dk=\pi$ which may explain why the $\pi$ disappears, tried to find a way of applying Cauchy-Schwartz, but I always get stuck. Does anyone have an idea how to get this inequality?

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Cauchy-Schwarz does the job: $$\begin{align} \int_\mathbb{R} |\hat{f}(k)|\,dk&=\int_\mathbb{R} |\hat{f}(k)|\,(1+k^2)^{1/2}\,(1+k^2)^{-1/2}\,dk\\ &\le\Bigl(\int_\mathbb{R} |\hat{f}(k)|^2\,(1+k^2)\,dk\Bigr)^{1/2}\Bigl(\int_\mathbb{R} (1+k^2)^{-1}\,dk\Bigr)^{1/2}\\ &=\sqrt{\pi}\Bigl(\int_\mathbb{R} |\hat{f}(k)|^2\,(1+k^2)\,dk\Bigr)^{1/2}. \end{align}$$

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