Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $L_1(\mathbb{R})$.

share|improve this question
add comment

1 Answer

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)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.