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