$\left(\int_{\mathbb{R}}\left|\int_\mathbb{R}\chi_{(t,\infty)}(\xi)\hat{f}(\xi)m'(t)dt\right|^pd\xi\right)^{1/p}\leq C_p\|m'\|_{L^1}\|f\|_{L^1}$ 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:
$$\left(\int_{\mathbb{R}}\left|\int_\mathbb{R}\chi_{(t,\infty)}(\xi)\hat{f}(\xi)m'(t)dt\right|^pd\xi\right)^{1/p}\leq C_p\|m'\|_{L^1}\|f\|_{L^1}$$
Where $f$ and $m$ 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) so maybe there is some trick that solve this problem easily, but I can't see it.
Any help will be thanked.
Edit: It turns out that I missed something, the real inequality is:
$$\left(\int_{\mathbb{R}}\left|\int_\mathbb{R}{\mathcal{F}^{-1}}(\chi_{(t,\infty)}(\xi)\hat{f}(\xi))m'(t)dt\right|^pd\xi\right)^{1/p}\leq C_p\|m'\|_{L^1}\|f\|_{L^1}$$
where $\mathcal{F}^{−1}$ means inverse fourier transform (I don't know a better notation) In this case the inequality is correct, right?
 A: The inequality is not true because the LHS and RHS have different scaling behaviour ("units").
More precisely, define a dilation operator $(δ_α f)(x) = f(α^{-1} x)$. Then by change of variables $\lVert δ_α f \rVert_{L^1} = α \lVert f \rVert_{L^1}$ and $\widehat{δ_α f} = α\, δ_{α^{-1}} \hat{f}$.
Now consider replacing $m'$ by $δ_{α^{-1}} m'$ and $f$ by $δ_α f$. For the LHS we then get
\begin{align*}
\renewcommand{\d}{\mathop{}\!d}
 &\left( \int_\mathbb{R} \left| \int_{\mathbb{R}} χ_{(t, \infty)}(ξ) m'(α t) α\hat{f}(α ξ) \d t \right|^p \d ξ\right)^{1/p} \\
  &\quad= \left( \int_\mathbb{R} \left| \int_{\mathbb{R}} χ_{(α t, \infty)}(α ξ) m'(α t) \hat{f}(α ξ) \d (α t) \right|^p α^{-1} \d (α ξ)  \right)^{1/p} \\
  & \quad = α^{-1/p} \cdot \text{(original LHS)}
\end{align*}
Meanwhile the RHS of your inequality is unchanged by this replacement. So letting $α \to 0$ we can make the LHS arbitrarily large while keeping the RHS constant so no such $C_p$ can exist (except for $p = \infty$ I suppose).
