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From the Hilbert transform theory it follows that $$\forall p\in(1,2], \forall f\in L^p(\mathbb{R}), \left\|f-\left(x\mapsto\int_{-M}^M\hat{f}(\xi)e^{2\pi i\xi x}\operatorname{d}\xi \right)\right\|_{L^p(\mathbb{R})}\rightarrow0, M\rightarrow+\infty,\ \ \ \ (*)$$ where $\hat f$ is the Fourier transform of $f$.

What about if $p\in(2,+\infty)$?

Obviously, there's a problem in defining what should be the meaning of $\int_{-M}^M\hat{f}(\xi)e^{2\pi i\xi x}\operatorname{d}\xi$, but using the relation between convolution and Fourier transform valid at least in $L^2(\mathbb{R})$, we can easily see that we can replace it with $\varphi_M*f$ where: $$\varphi_M(x):= \frac{\sin (2\pi Mx)}{\pi x}.$$ Thanks to Young inequality for convolutions, we have that $\varphi_M*f \in L^\infty(\mathbb{R})$ if $f\in L^p(\mathbb{R})$ with $2<p<\infty$, but I don't see a way to guarantee that $\varphi_M*f \in L^p(\mathbb{R})$ in order to obtain something meaningful in $(*)$.

Is it true in general that $\varphi_M*f\in L^p(\mathbb{R})$? If it is, does $(*)$ hold? If not, is there a way to recover something analogous?

Bob
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  • If $p>2$ the Fourier transform of an $L^p$ function needs not be a function, it can be a distribution with positive order. – Giuseppe Negro Sep 24 '18 at 18:50
  • @GiuseppeNegro as you can read in the question, I replaced the term that involves Fourier transform with a convolution – Bob Sep 24 '18 at 18:52
  • You are right that it is not so obvious as I was trying to put it. However, I still think that the convolution is not going to be an element of Lp. That's only a guess, motivated by the distributional fact I cited above – Giuseppe Negro Sep 24 '18 at 19:03

1 Answers1

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Fix $2<p<\infty$.

It is known from the Hilbert transform theory that the characteristic function of an interval $I$ is a Fourier multiplier of $L^p(\mathbb{R})$, i.e. the operator initially defined on the dense subspace $L^2(\mathbb{R})\cap L^p(\mathbb{R})$ by $$T_I(f)=\mathcal{F}^{-1}\left(\chi_I\mathcal{F}(f)\right),$$ where $\mathcal{F}$ is the Fourier transform, extends uniquely by continuity to a continuous operator from $L^p(\mathbb{R})$ into itself.

Now, if $f\in L^p(\mathbb{R})$ choose a sequence of simple functions $(f_n)_{n\in\mathbb{N}}$ that converges in $L^p(\mathbb{R})$ to $f$ and get $M>0$. Then $\varphi_M*f_n\in L^p(\mathbb{R})\cap L^2(\mathbb{R})$ and so: $$\|\varphi_M*f_n-T_{[-M,M]}(f)\|_p =\|T_{[-M,M]}(f_n)-T_{[-M,M]}(f)\|_p\rightarrow0, n\rightarrow\infty.$$ On the other hand, for Young inequality we have that $$\|\varphi_M*f_n -\varphi_M*f\|_\infty = \|\varphi_M*(f_n-f)\|_\infty\le \|\varphi_M\|_{p'}\|f_n-f\|_p\rightarrow0, n\rightarrow\infty$$ where $p'\in(1,2)$ is such that $\frac{1}{p}+\frac{1}{p'}=1$.

Then, taking if necessary a subsequence, we can guarantee that $\varphi_M*f_n\rightarrow T_{[-M,M]}(f), n\rightarrow\infty$ pointwise a.e., so we get for a.e. $x\in\mathbb{R}$ that: $$(\varphi_M*f)(x)=T_{[-M,M]}(f)(x)$$ and so $\varphi_M*f$ is a.e. equal to a member of $L^p(\mathbb{R})$ and so it is in $L^p(\mathbb{R})$.

This argument also shows that $$\varphi_M* = T_{[-M,M]}.$$ Now, from the fact that the set of Fourier transform of simple functions are dense in $L^p(\mathbb{R})$ and from the fact that if $g$ is the Fourier transform of a simple function then: $$\|\varphi_M*g-g\|_p\rightarrow0, M\rightarrow\infty$$ we get that: $$\forall f\in L^p(\mathbb{R}), \|\varphi_M*f-f\|_p\rightarrow0, M\rightarrow\infty$$ is equivalent to: $$\sup_{M>0}\|\varphi_M*\|_{p\rightarrow p}<\infty.$$ But we have shown that: $$\varphi_M*=T_{[-M,M]}$$ and from the Hilbert transform theory (see e.g. Javier Duoandikoetxea - Fourier Analysis, chapter 3 on Hilbert transform, paragraph 5 on multipliers, proposition 3.6) we know that: $$\sup_{M>0}\|T_{[-M,M]}\|_{p\rightarrow p}<\infty$$ and so we have obtained that: $$\forall f\in L^p(\mathbb{R}), \|\varphi_M*f-f\|_p\rightarrow0, M\rightarrow\infty.$$

Bob
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  • This surprised me, I must confess. I had bet that the answer was negative. So, forgive my ignorance on Fourier multipliers, but is it true that $T_{[-M,M]} f\to f$ in $L^p$ for $p>2$? I see that you essentially show the equivalence of this and $\phi_M\ast f\to f$ in $L^p$. – Giuseppe Negro Sep 28 '18 at 09:46
  • The point is not to prove directly that $\forall f\in L^p(\mathbb{R}), |T_{[-M,M]}-f|p\rightarrow0, M\rightarrow\infty$. Knowing that if $f$ is the Fourier transform of a simple function then $|\varphi_Mf-f|_p\rightarrow0, M\rightarrow\infty$, by density, the claim $\forall f\in L^p(\mathbb{R}), |\varphi_Mf-f|_p\rightarrow0, M\rightarrow\infty$ is equivalent to $\sup{M>0} |\varphi_M*|{p\rightarrow p}<\infty$, which is equivalent to $\sup{M>0} |T_{[-M,M]}|_{p\rightarrow p}<\infty$, and this last fact holds true thanks to the Hilbert transform theory – Bob Sep 28 '18 at 09:59
  • Yes, ok, I got that. I rephrase my question. I have a problem, I don't know what you mean by "Hilbert transform theory". So, to convince myself that your proof is correct, I need to convince myself that there are no hidden pitfalls. Now, I understand that, if the "Hilbert transform theory" can prove that $T_{[-M, M]}f\to f$ in $L^p$, even with $p>2$, then your proof will be correct. So, can you please explain me why this convergence property is true? Or point me to a reference. Thank you. – Giuseppe Negro Sep 28 '18 at 10:06
  • Do you agree with me that the point is to prove that $\sup_{M>0}|T_{[-M,M]}|_{p\rightarrow p}<\infty$? If this is the case, I can show you how – Bob Sep 28 '18 at 10:11
  • I agree with you on that point. – Giuseppe Negro Sep 28 '18 at 10:30
  • You can find a reference into: Javier Duoandikoetxea - Fourier Analysis, chapter 3 on Hilbert transform, paragraph 5 on multipliers, proposition 3.6 and corollary 3.7 – Bob Sep 28 '18 at 10:35
  • It would be great if you added that to the main text, in the relevant position. – Giuseppe Negro Sep 28 '18 at 10:40
  • Thank you for the reference! (and for this answer) – Giuseppe Negro Sep 28 '18 at 13:46