It's possible that I'm missing a constant. Is there any way to directly prove this without invoking the Fourier Inversion theorem?
Let $u\in C^\infty_0(\mathbb{R}^d)$, then show that $u(0)=\iint u(x)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\,\mathrm{d}\xi$.
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functional-analysis
pde
harmonic-analysis
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0The result is missing a factor of $1/(2\pi)$. Using distributions, note that $\int_{\mathbb{R}}e^{-i\xi x}\,d\xi=2\pi \delta(x)$. – 2017-02-27
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0Alternatively, one could write $u(0) = \int_{R^2} u(x) e^{-2\pi i x \cdot \xi} dx d\xi,$ if you do not wish to have the $2 \pi'$s out in front. – 2017-02-27
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0Why is that first statement true? – 2017-02-27
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2Any proof of this identity will be equivalent to proving the full Fourier inversion theorem (with very small modifications). I would focus on proving this theorem instead and get this particular results as a freebee. – 2017-02-27