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In Muscalu, Schlag "Classical and multilinear harmonic analysis, Volume 1" (2013), Excercise 11.1 is to prove, basically, that there exists a function $f\in L^p \quad \forall\ p>1$ such, that the Fourier-transform $\mathcal{F}f$ of $f$ is infinite on a (previously chosen) hyperplane $$H(\theta, s) := \{ x\in \mathbb{R}^n | x\cdot \theta = s\} \qquad (\theta \in S^{n-1}, s\in \mathbb{R})$$ I have arrived at the function $$f_{\theta, s}(x) =\frac{1}{1+|x\cdot \theta -s|}$$ (the $\cdot$ denotes the standard inner product) I could show, that $f\in L^p(\mathbb{R}^n) \quad \forall\ p>1$, but I have failed to show, that it's Fourier-transform is identical to $\infty$ on $H(\theta,s)$. How can I show this?

The hint which lead me to this choice of $f_{\theta,s}$ is from this book: Fourier Analysis And Convexity, p. 218

EDIT:
Thanks to @paulgarret I was able to prove the following
Let $U$ be a unitary matrix with $U\theta = e_n$ and let $$f(x) = \frac{e^{-\sum_{j=1}^{n-1} x_j^2}}{1+|x_n|}$$ Then by $f_{\theta, s} := f(Ux-se_n)$ we have $$\widehat{f_{\theta,s}}(\xi) =\infty \quad \forall \ \xi \in H(\theta,s)$$ and $$\Vert f_{\theta,s} \Vert_p^p = \Vert f \Vert_p^p = \frac{2}{p-1} \left ( \frac{\pi}{p} \right ) ^{\frac{n-1}{2}} < \infty \qquad \forall \ p>1$$ The only remaining Question is, how can we construct such a $U$ to give a concrete example?

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I doubt that $f_{\theta, s}$ is in $L^p(\mathbb{R}^n)$. Take for example $\theta=(0\ldots 1)$ and $s=0$. Then $$f_{\theta, s}(x_1\ldots x_n)=\frac{1}{1+\lvert x_n\rvert}$$ and $$\int_{\mathbb{R}^n}\left(\frac{1}{1+\lvert x_n\rvert}\right)^p dx_1\ldots dx_n=\int dx_2\ldots dx_n \int \left(\frac{1}{1+|x_n|}\right)^p dx_1=\infty.$$ –  Giuseppe Negro Aug 21 '13 at 18:20
    
Please take a closer look. $p>1$ works out just fine; I arrive at $$\Vert f\Vert_p^p = \frac{2}{p-1}$$ –  AlexR Aug 21 '13 at 18:26
    
@GiuseppeNegro Here's my calculation: $$\Vert f \Vert_p^p = \int_{\mathbb{R}^n} |f(x)|^p dx = \int_{\mathbb{R}} \int_{H(\theta,t)} |f(x)|^p dS(x) dt = \int_{\mathbb{R}} \int_{H(\theta,t)} \frac{1}{(1+|t-s|)^p} dS(x) dt = \int_{\mathbb{R}} \frac{1}{(1+|t-s|)^p} dt = 2\int_s^{\infty} \frac{1}{(1+t-s)^p} dt = 2\int_1^\infty \frac{1}{u^p} du = \frac{2}{p-1} \qquad \forall\ p>1$$ –  AlexR Aug 21 '13 at 19:00
    
@GiueseppeNegro's point is that your proposed function is constant in $n-1$ out of $n$ directions, so its integral (or any power) diverges, much as $\int_{\mathbb R^2}{dx\,dy\over 1+x^2}$ diverges. Thus, you need to do something in those $n-1$ other directions to arrange convergence. –  paul garrett Aug 21 '13 at 19:03
    
@paulgarrett I cannot see the problem here, plus the "authority argument" from the reference book. –  AlexR Aug 21 '13 at 19:14

2 Answers 2

up vote 2 down vote accepted

To write a preliminary answer as expanded comment, to explain why the tentative answer has a certain problem, although it does make progress: first, a simpler analogue where a function is constant in one direction: take $f(x,y)={1\over 1+x^2}$. Then $$ \int_{\mathbb R^2} f(x,y)\;dx\,dy\;=\; \int_{\mathbb R^2}{dx\,dy\over 1+x^2} \;=\;\int_{\mathbb R} {dx\over 1+x^2}\cdot \int_{\mathbb R}1\;dy $$ The integral over $x$ is fine, but the integral in $y$ is the integral of $1$ over the whole real line, giving $+\infty$. Thus, as in your attempt, as noted by @GiuseppeNegro, this glitch has to be addressed.

By rotation, we might as well suppose the hyperplane is the $x_n=0$ hyperplane. Thus, you want a function to blow up as $x_n\to 0$, but/and be integrable on the whole $\mathbb R^n$. Ok, as a function of $x_n$, something like ${1\over \sqrt{|x_n|}(1+x_n^2)}$. In $x_1,\ldots,x_{n-1}$, you want reasonable behavior, so a Gaussian (handy for FT, also) like $e^{-\pi(x_1^2+\ldots+x_{n-1}^2)}$ would do. The product has the right sort of behavior. Take FT, ...

(Although one should generally tend to believe "authorities", there is the auxiliary issue of being alert to disconnects between one's understanding of what is being asserted/asked, what is really asked/asserted, and, still, what one has actually done. Partial but incomplete progress should not be confused with "failure", nor with "success", for example.)

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Okay, thanks a bunch; I'll go for $$f(x) = \frac{e^{-(x_1^2 + \ldots + x_{n-1}^2)}}{1+|x_n|}$$ for the moment and see where it takes me. –  AlexR Aug 21 '13 at 19:29

Here is an interesting idea, for two dimensions, and I'm sure it can be generalized. Let $f(x,y) = \frac{1}{\log |x| (1 + |yx|)}$. This function is in $L^p$: $\int |f(x,y)|^p dy dx = \int \frac{1}{(\log |x|)^p} \int \frac{du}{|x|(1+|u|)^p} dx < \infty$, and just barely (needs $p>1$ for the logarithm)!

Now take the Fourier transform. Integrating in $y$ first is already divergent (though I'm not sure about being $\infty$. The F.T. seems to exist as a principal value). It's not very clear here. If $\xi = 0$ then we definitely have $\infty$, but in other places the function is not Lebesgue integrable, so I don't know what to make of it.

To generalize this specific example, maybe $f(x_1,\ldots,x_{n-1},x_n) = \frac{1}{\log r (1+|x_n| r_{}^{n-1})}$ where $r= \sqrt{ |x_1|^2 + \ldots + |x_{n-1}|^2}$

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Thank you, but the function specified above seems alot handier for these purposes, as the gaussian $$\sqrt{\pi} = \int_\mathbb{R} e^{-x^2} dx$$ Is also very nice for FT. That $\mathcal{F}f \in L^q$ need not be checked, as $\mathcal{F} : L^p \to L^q$ is known to be continuous. –  AlexR Aug 22 '13 at 14:11
    
Ohh the FT there is already not convergent I see now. Also to construct the U take first column to be theta then arbitrarily extend to a basis and use orthogonalization (keep theta as one of the directions) –  Evan Aug 22 '13 at 14:31
    
Actually figured that out myself^^ though, because $U\theta = e_n$, we need to pick the last row / column to be $\theta$ ;-) –  AlexR Aug 22 '13 at 14:33
    
so does it make a difference that the value is not infinity (just undefined)? seems only infinite at the origin –  Evan Aug 22 '13 at 14:43
    
What I need is, that $\mathcal{F}f$ cannot be restricted to a bounded subset $S$ of a Hyperplane. This means, that we need $\mathcal{F}f$ to be such a function, that the restriction $$\left.\mathcal{F}f\right |_S$$ is not well-defined (or $\Vert \left . \mathcal{F}f \right |_S \Vert_1 = \infty$, i.e. $\mathcal{F}f \notin L^1(S)$) –  AlexR Aug 22 '13 at 15:39

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