Given $N>1$, and $\psi \in C^\infty(\mathbb{R}^d)$, $0 \leq \psi \leq 1$, which is radial and satisfies: $$ \begin{cases} \psi = 1 & \text{ on }\frac{3}{4}\leq |\xi| \leq \frac{5}4{} \\ \psi = 0 & \text{ on }|\xi| \leq \frac{1}{2} \text{ and on } |\xi| \geq 2 \end{cases} $$ We note that $\psi$ defined in this way is a smooth approximation of the characteristic function of the annulus $\frac{3}{4} \leq |\xi| \leq \frac{5}{4}$. For the $\psi$ defined above, we define an operator $P_N$ on $L^2(\mathbb{R}^d)$ by: $$(P_Nf)(\xi) = (K_N * f)(\xi)$$ where $$K_N(x) = \mathcal{F}^{-1}\Big[\psi\Big(\frac{x}{N}\Big)\Big] = \int_{\mathbb{R}^d} e^{2\pi i x \cdot \zeta}\psi\Big( \frac{\zeta}{N}\Big) d\zeta $$ I need to show that all $1 \leq p \leq q \leq \infty$, $$||P_N f||_{L^q(\mathbb{R}^d)} \leq CN^{\frac{d}{p}-\frac{d}{q}}||f||_{L^p(\mathbb{R}^d)}$$
So far, I have: Recall Young's convolution inequality which states that for $f \in L^p(\mathbb{R}^d)$ and $g \in L^q(\mathbb{R}^d)$ such that $\frac{1}{q} = \frac{1}{p} + \frac{1}{r} - 1$ with $1 \leq p,q,r\leq \infty$, we have $$||f*g||_q \leq ||f||_p ||g||_r$$ Now, we know that $P_Nf = K_N * f$. Then, we have that $$||P_N f||_q = ||K_N *f ||_q = ||f* K_N||_q \leq ||f||_p ||K_N||_r$$ Consider $||K_N||_r$. By definition, $$K_N(\xi) = \int_{\mathbb{R}^d}e^{2\pi i \xi \cdot x}\psi\Big( \frac{x}{N}\Big) dx = N^d \int_{\mathbb{R}^d}e^{2\pi i \xi \cdot (Nu)}\psi (u) du$$ $$\Rightarrow ||K_N||_r = N^d \cdot C(r,\xi)$$ where $$C(r,\xi) = \Big| \Big| \int_{\frac{1}{2}\leq |u|\leq \frac{3}{4}}e^{2\pi i \xi \cdot (Nu)} \psi( u ) du + \int_{\frac{3}{4} \leq |u|\leq \frac{5}{4}}e^{2\pi i \xi \cdot (Nu)} du + \int_{\frac{5}{4} \leq |u| \leq 2}e^{2\pi i \xi \cdot (Nu)}\psi (u) du\Big|\Big|_r$$
Where can I go from here?