I am reading the book Fourier Analysis by Javier Duandikoetxea and I am stuck in the proof of a lemma that it is the key part to prove the Marcinkiewicz multiplier theorem. In particular I am stuck in showing the following inequality:

$$\left(\int_{\mathbb{R}}\left|\int_\mathbb{R}{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd\xi\right)^{1/p}\leq C_p\|m'\|_{L^1}\|f\|_{L^p}$$ where $\mathcal{F}^{−1}$ means the inverse fourier transform, $f,m$ are two arbitrary functions in the Scwartz class and $C_p>0$ is a constant depending on $p$. I'm very confused, since the book claim it as obvious, but none of the usual tools for inequalities works here (Hölder, Minkowski, Cauchy-Schwarz, Plancherel, Parseval,...) so maybe there is some trick that solve this problem easily, but I can't see it.

Any help will be thanked.


1 Answer 1


It seems that the tool being used is the Minkowski integral inequality , with Proposition 3.6 of Duoandikoetxea.

Minkowski Integral inequality (Duoandikoetxea, Fourier Analysis,xviii (Preliminaries)) : Let $(X,\mu)$ and $(Y,\nu)$ be $\sigma$-finite measure spaces. For all appropriate $f(x,y) : X \times Y \to \mathbb R$ (that is ,for which the RHS of the inequality exists),$$ \left(\int_X \left|\int_{Y} f(x,y)d\nu(y)\right|^p d \mu(x)\right)^{\frac 1p} \leq \int_Y \left(\int_{X}|f(x,y)|^pd \mu(x)\right)^{\frac 1p} d \nu(y) $$ we write this as $$ \int_X \left|\int_{Y} f(x,y)d\nu(y)\right|^p d \mu(x) \leq \left(\int_Y \left(\int_{X}|f(x,y)|^pd \mu(x)\right)^{\frac 1p} d \nu(y)\right)^{p} $$ This can be thought of as an integral analogue of the triangle inequality for $L^p$ functions.

At this point, we cannot quite apply this exactly to our situation, because the measure given by $m'(t)dt$ is a signed measure (where some sets may have a negative measure) rather than a conventional measure. However, note that one of the assumptions on $m$ usually imposed is that it is of bounded variation (it's a consequence of the fact that $\|m'\|_{L^1}$ exists).

In that case, we have the functions $h_1(t)= m'(t)1_{m'(t) \geq 0}$ and $h_2(t) = m'(t)1_{m'(t) \leq 0}$. Call the measures $dm_1(t) = h_1(t)dt$ and $dm_2(t)= -h_2(t)dt$. It is clear that $\int_A m'(t)dt = m_1(A)-m_2(A)$ and $\int_A |m'(t)|dt = m_1(A) + m_2(A)$ for any measurable set $A$. This is a Jordan decomposition of $m'(t)dt$ that we've performed : getting two measures on which we'll execute the Minkowski integral inequality.

The way to start is to somehow relate the LHS with $d \nu(y) = m'(t)dt$ and separately with $dm_1$ and $dm_2$. For this, note that $$ \int_{A} m'(t)dt = m_1(A)-m_2(A) \implies \left|\int_{A} m'(t)dt\right| \leq |m_1(A)| + |m_2(A)| $$ and therefore $$ \left|\int_{A} m'(t)dt\right|^p \leq \left(|m_1(A)| + |m_2(A)|\right)^p \leq 2^p (|m_1(A)|^p + |m_2(A)|^p) $$

for any measurable set $A$. The last inequality follows from two inequalities $$ x +y \leq 2\max(x,y) \quad ; \quad \max(x^p,y^p) \leq (x^p+y^p) $$ for all $x,y$ positive. Once we see this, we can split any $L^p$-integrable function into a positive-negative part , approximate each part by indicators, and obtain for any $m_1(t)dt$ $L^p$-integrable $g$ that $$ \left|\int_{\mathbb R} g(t)m'(t)dt\right|^p \leq 2^p \left(\left|\int_{\Bbb R} g(t)dm_1(t)\right|^p + \left|\int_{\Bbb R} g(t)dm_2(t)\right|^p\right) $$

We get by using $g(t) = \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)$ for any fixed $\xi$ that $$ \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^p \leq 2^p \left(\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_1(t)\right|^p + \left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_2(t)\right|^p\right) $$ Integrating w.r.t $d\xi$, we have $$ \int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi\\ \leq 2^p\left(\int_{\mathbb R}\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_1(t)\right|^p d \xi + \int_{\mathbb R}\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_2(t)\right|^p d \xi\right) \tag{1} $$

To apply the Minkowski integral inequality, we have for $i=1,2$, $$ \int_{\mathbb{R}}\left|\int_\mathbb{R}\underbrace{{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)}_{f(\xi,t)}\underbrace{dm_i(t)}_{d \nu(t)}\right|^p\underbrace{d\xi}_{d \mu(\xi)} $$

It's obvious that $d \xi$ is a sigma-finite measure on $\mathbb R$. We also have $\|m_i\|_{L^1}\leq \|m'\|_{L^1}$ so the $m_i$ are finite measures. Finally, assuming that the RHS of the integral inequality will be finite, we'll write the expression for it : $$ \int_{\mathbb{R}}\left|\int_\mathbb{R}\underbrace{{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)}_{f(\xi,t)}\underbrace{dm_i(t)}_{d \nu(t)}\right|^p\underbrace{d\xi}_{d \mu(\xi)} \leq \left(\int_{\Bbb R}\left(\int_{\Bbb R}|{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)|^p d\xi\right)^{\frac 1p} dm_i(t)\right)^{p} \tag{2} $$

We now use proposition $3.6$ of the book. (Page 59). Recall the operator $S_{a,b}$ defined in the book. It is fairly clear to see that $$ \mathcal F(S_{(t,\infty)}f) (\xi) = (\chi_{(t,\infty)}\hat{f})(\xi) $$

since $S_{a,b}$ is the operator associated with the multiplier $\chi_{a,b}$. Inverting the Fourier transform, $$ S_{(t,\infty)}f = \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f}) $$

However, this means that we can write the RHS of $(2)$ as $$ \left(\int_{\Bbb R}\left(\int_{\Bbb R}|{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)|^p d\xi\right)^{\frac 1p} dm_i(t) \right)^p = \left(\int_{\mathbb R} \|S_{(t,\infty)}\|_p dm_i(t)\right)^p $$

Using proposition $3.6$, there exists a constant $C_p$ independent of $t$ such that $$ \|S_{(t,\infty)}\|_p \leq C_p \|f\|_p\quad \forall -\infty \leq t \leq \infty $$

Substituting this above and taking the constant $C_p\|f\|_p$ out of the integral gives $$ \left(\int_{\mathbb R} \|S_{(t,\infty)}\|_p dm_i(t)\right)^p \leq \left(C_p \|f\|_p \int_{\mathbb R} dm_i(t)\right)^p \leq C_p^p \|f\|^p_p\|m'\|^p_1 $$

For $i=1,2$.Combining $2$ and $1$ now gives us $$ \int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi \leq 2^p\left(2C_p^p \|f\|^p_p\|m'\|^p_1\right) \leq 2^{p+1} C_p^p \|f\|_p^p \|m'\|_1^p $$

Taking the $p$th root gives $$ \left(\int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi \right)^{\frac 1p} \leq 2^{\frac{p+1}p}C_p \|f\|_p\|m'\|_1 $$

where the constant term depends only on $p$, as desired.

  • $\begingroup$ Wow, that is so nice... You took so much effort and it is extremely well detailed. I can't give you a bounty untill tomorrow, but you will get a well deserved bounty, because that is one of the best solutions I've ever gotten here. $\endgroup$
    – Marcos
    Mar 22, 2022 at 11:31
  • $\begingroup$ @Marcos Thank you so much for your appreciative comment. I agree I spent some time on this, and I leave it to you to decide if you'd like to bounty this. I hope others can serve you with similar solutions across your journey. The $2^p$ trick is something that I think everyone should know. $\endgroup$ Mar 22, 2022 at 11:33
  • $\begingroup$ You'll get a bounty for sure. I have to say that I am having some troubles reading Duandikoetxea 's book, and a solution as detailed as yours will help me a lot in my way through the book. $\endgroup$
    – Marcos
    Mar 22, 2022 at 11:38
  • $\begingroup$ @Marcos I think the book is somewhat terse : it doesn't explain details when it doesn't need to, and prefers to get on with the main narrative. It took me some time to see why this step went through, which the book flat-out claimed as obvious. I wish I could recommend a better book, but I'm not aware of one that contains the Marcinkiewicz multiplier theorem (actually, even the interpolation theorem). $\endgroup$ Mar 22, 2022 at 11:40
  • 1
    $\begingroup$ And yes, all the proof is farly standard (i've seen many times the $2^p$ trick) but the part of spliting the measure in two parts to be able to apply Minkwski was something I have never seen and which is very elegant. $\endgroup$
    – Marcos
    Mar 22, 2022 at 11:41

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