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It's easy to see that $\lVert f \rVert_{L^1} \geq \lVert \hat{f} \rVert_{L^\infty}$, but, when $f$ is positive, it's also true that $\lVert f \rVert_{L^1} \leq \lVert \hat{f} \rVert_{L^\infty}$. Can anyone tell me why is that? We assume that $f$ is a Schwartz function, so Fourier transform is well defined.

Thanks.

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Assume $f$ is positive, then $$\|f\|_{L^1} = \int_\mathbb{R} f(x)\mathrm{d}x =\hat{f}(0)$$

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