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Let $$ g(x) = v.p.\int_{-1/2}^{1/2}{e^{-itx}\over t\ln{|t|}}dt. $$ Please help me prove, that this function is not the Fourier transform of any function of $L_1(\mathbb{R})$.

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up vote 3 down vote accepted

Let $h$ be the distribution (with compact support) given by $$\langle h,\phi\rangle=v.p.\int_{-1/2}^{1/2}\frac{\phi(t)}{t\log|t|}dt$$ (for any smooth test function $\phi$). As $h$ has compact support, its Fourier transform is the (analytic) function $g(x)=\langle h, \exp(-ix\cdot)\rangle$. If $h$ is the Fourier transform of a $L^1$-function $f$ then $f=h$ in the sense of distributions. However, as $h|_{\mathbb{R}-\{0\}}=\frac{1}{t\log|t|}$ (that is, if $\phi$ vanishes in a neighbourhood of $0$, then the $v.p.$ integral becomes the ordinary integral), it would mean $f=\frac{1}{t\log|t|}$ a.e., but that is not an $L^1$ function (the integral $\int_0^{1/2} 1/(t\log|t|)dt$ diverges, as the substitution $s=\log t$ shows)

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Excuse me, can you explain to me why " If h is the Fourier transform of a L1-function f then f=h in the sense of distributions". – user31497 May 21 '12 at 15:32
Maybe, it's g(x) is Fourier transform of a $L_1$-function f? And $\exp(-ix\cdot)$ it's not a finit function, so how we can find $\langle h, \exp(-ix\cdot)\rangle$. – user31497 May 23 '12 at 17:17

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