[proofplan] The proof rests on a frequency-localisation argument: each $\Delta_j$ is supported in a fixed dyadic annulus $\{2^{j-1} \le |\xi| \le 2^{j+1}\}$, hence $\hat\psi_j \cdot \hat{\tilde\psi}_k = 0$ unless the supports overlap, which happens only for $|j - k| \le N$ where $N$ depends only on the resolutions. Writing $\Delta_j f = \Delta_j \sum_{k} \tilde\Delta_k f = \sum_{|k - j| \le N}\Delta_j\tilde\Delta_k f$, the reduction is to bound $\|\Delta_j \tilde\Delta_k f\|_{L^p}$ by $\|\tilde\Delta_k f\|_{L^p}$ uniformly. This last bound is a Young convolution inequality for the kernel of $\Delta_j$, which has $L^1$-norm bounded uniformly in $j$. Combining the finite-overlap structure with Minkowski's inequality on $\ell^q$ then yields the equivalence of norms. [/proofplan] [step:Establish that $\Delta_j$ has uniformly bounded convolution kernel and acts boundedly on $L^p$ uniformly in $j$] For $j \ge 1$, the projector $\Delta_j$ has Fourier symbol $\hat\psi_j(\xi) = \hat\psi(2^{-j}\xi)$. By the Fourier inversion formula on $\mathcal{S}(\mathbb{R}^n)$, the convolution kernel of $\Delta_j$ is \begin{align*} \psi_j : \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto (2\pi)^{-n}\int_{\mathbb{R}^n}\hat\psi(2^{-j}\xi)\,e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi) = 2^{jn}\,\psi(2^j x), \end{align*} where $\psi := \mathcal{F}^{-1}\hat\psi \in \mathcal{S}(\mathbb{R}^n)$ (since $\hat\psi \in C^\infty_c(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n)$). The substitution used the change of variables $\eta := 2^{-j}\xi$ with $d\mathcal{L}^n(\xi) = 2^{jn}\,d\mathcal{L}^n(\eta)$, giving the dyadic dilation factor $2^{jn}$ on the kernel. By the substitution $y := 2^j x$, $d\mathcal{L}^n(x) = 2^{-jn}\,d\mathcal{L}^n(y)$, \begin{align*} \|\psi_j\|_{L^1(\mathbb{R}^n)} = \int_{\mathbb{R}^n}|2^{jn}\,\psi(2^j x)|\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n}|\psi(y)|\,d\mathcal{L}^n(y) = \|\psi\|_{L^1(\mathbb{R}^n)} =: M_\psi. \end{align*} Since $\psi \in \mathcal{S}(\mathbb{R}^n)$, $M_\psi < \infty$, and $M_\psi$ depends only on $\hat\psi$. For $j = 0$, $\psi_0 := \mathcal{F}^{-1}\hat\psi_0 \in \mathcal{S}(\mathbb{R}^n)$ and $\|\psi_0\|_{L^1(\mathbb{R}^n)} =: M_{\psi_0} < \infty$ depends only on $\hat\psi_0$. By Young's convolution inequality, $\|\Delta_j f\|_{L^p(\mathbb{R}^n)} = \|\psi_j * f\|_{L^p(\mathbb{R}^n)} \le \|\psi_j\|_{L^1(\mathbb{R}^n)}\,\|f\|_{L^p(\mathbb{R}^n)} \le M\,\|f\|_{L^p(\mathbb{R}^n)}$ where $M := \max(M_\psi, M_{\psi_0})$, for every $1 \le p \le \infty$ and $j \ge 0$. The analogous statement holds for $\tilde\Delta_j$ with constant $\tilde M$ depending only on $\hat{\tilde\psi}, \hat{\tilde\psi}_0$. [/step] [step:Use the partition of unity to express $\Delta_j f$ as a finite sum of frequency-overlapping pieces] By construction $\sum_{k \ge 0}\hat{\tilde\psi}_k(\xi) = 1$ for every $\xi \in \mathbb{R}^n$. Multiplying $\hat f(\xi)$ by this and applying $\Delta_j$ yields \begin{align*} \widehat{\Delta_j f}(\xi) = \hat\psi_j(\xi)\,\hat f(\xi) = \hat\psi_j(\xi)\sum_{k \ge 0}\hat{\tilde\psi}_k(\xi)\,\hat f(\xi) = \sum_{k \ge 0}\hat\psi_j(\xi)\,\hat{\tilde\psi}_k(\xi)\,\hat f(\xi). \end{align*} By Fourier inversion (formally on $\mathcal{S}(\mathbb{R}^n)$ and extending to $\mathcal{S}'(\mathbb{R}^n)$ by the standard duality, using that the partial sums $\sum_{k \le K}\hat{\tilde\psi}_k$ converge in $\mathcal{S}$-multiplier sense to $1$), \begin{align*} \Delta_j f = \sum_{k \ge 0}\Delta_j\tilde\Delta_k f. \end{align*} [/step] [step:Bound $\|\Delta_j f\|_{L^p}$ by a finite linear combination of $\|\tilde\Delta_k f\|_{L^p}$ for nearby $k$] Apply the triangle inequality to the finite sum from Step 2 in $L^p(\mathbb{R}^n)$ and use the uniform $L^p$ bound on $\Delta_j$ from Step 1, \begin{align*} \|\Delta_j f\|_{L^p(\mathbb{R}^n)} = \Bigl\|\sum_{|k-j| \le N}\Delta_j\tilde\Delta_k f\Bigr\|_{L^p(\mathbb{R}^n)} \le \sum_{|k-j| \le N}\|\Delta_j\tilde\Delta_k f\|_{L^p(\mathbb{R}^n)} \le M\sum_{|k-j| \le N}\|\tilde\Delta_k f\|_{L^p(\mathbb{R}^n)}, \end{align*} where the last inequality is Young's convolution applied to $\Delta_j\tilde\Delta_k f = \psi_j * \tilde\Delta_k f$ as in Step 1. Note the indexing convention $k \ge 0$, so the sum on the right is over $\max(0, j - N) \le k \le j + N$, containing at most $2N + 1$ terms. [/step] [step:Convert the pointwise-in-$j$ bound into the Besov norm equivalence via Minkowski on $\ell^q$] Multiply both sides of the bound from Step 3 by $2^{js}$, \begin{align*} 2^{js}\,\|\Delta_j f\|_{L^p(\mathbb{R}^n)} \le M\,\sum_{|k - j| \le N} 2^{js}\,\|\tilde\Delta_k f\|_{L^p(\mathbb{R}^n)}. \end{align*} Substitute $j = k + r$ with $r \in \{-N, \ldots, N\}$. For $k \ge 0$ and $r$ in this range, $j \ge 0$ provided $k \ge -r$, automatically true if $r \ge 0$ and otherwise $k \ge |r|$. The factor $2^{js} = 2^{(k+r)s} = 2^{rs}\,2^{ks}$. Hence \begin{align*} 2^{js}\,\|\Delta_j f\|_{L^p} \le M\,\sum_{r = -N}^{N}\mathbb{1}_{\{k = j - r \ge 0\}}\,2^{rs}\,2^{ks}\,\|\tilde\Delta_k f\|_{L^p}. \end{align*} where $\mathbb{1}_{\{k = j - r \ge 0\}}$ denotes the indicator that the index $k = j - r$ is non-negative. Set $\Lambda := M\,\max_{|r| \le N} 2^{|r||s|} \cdot (2N+1)$, which depends only on $s$, $N$, and $M$ (i.e., only on $s$, $\hat\psi$, $\hat{\tilde\psi}$). For $1 \le q < \infty$, raise both sides to the $q$-th power, take $\ell^q$ over $j \ge 0$, and apply Minkowski's inequality (or a direct convolution-on-$\ell^q$ argument) to the finite sum on the right. The shifted-index sum is a discrete convolution of a finitely supported sequence (length $2N+1$) with the sequence $(2^{ks}\|\tilde\Delta_k f\|_{L^p})_{k \ge 0}$. By Young's inequality on $\ell^q$ ($\|a * b\|_{\ell^q} \le \|a\|_{\ell^1}\,\|b\|_{\ell^q}$ for $1 \le q \le \infty$), with $a$ the kernel $r \mapsto 2^{rs}\,M\,\mathbb{1}_{|r|\le N}$ (whose $\ell^1$-norm is $\le \Lambda$) and $b$ the sequence $(2^{ks}\|\tilde\Delta_k f\|_{L^p})_{k \ge 0}$, \begin{align*} \Bigl(\sum_{j \ge 0}\bigl(2^{js}\,\|\Delta_j f\|_{L^p}\bigr)^q\Bigr)^{1/q} \le \Lambda\,\Bigl(\sum_{k \ge 0}\bigl(2^{ks}\,\|\tilde\Delta_k f\|_{L^p}\bigr)^q\Bigr)^{1/q}, \end{align*} i.e., $\|f\|_{B^s_{p,q}(\{\Delta_j\})} \le \Lambda\,\|f\|_{B^s_{p,q}(\{\tilde\Delta_j\})}$. For $q = \infty$, the same argument with $\sup$ in place of the $\ell^q$ sum gives $\sup_{j \ge 0}2^{js}\,\|\Delta_j f\|_{L^p} \le \Lambda\,\sup_{k \ge 0}2^{ks}\,\|\tilde\Delta_k f\|_{L^p}$. Setting $C := \Lambda$, we have established \begin{align*} \|f\|_{B^s_{p,q}(\{\Delta_j\})} \le C\,\|f\|_{B^s_{p,q}(\{\tilde\Delta_j\})}. \end{align*} [/step] [step:Apply the same argument with the roles of $\{\Delta_j\}$ and $\{\tilde\Delta_j\}$ exchanged] The argument in Steps 1--4 used only the structural properties of the resolutions: that each is a partition of unity in frequency, supported on dyadic annuli of bounded ratio, with smooth bumps in $C^\infty_c$. Both $\{\Delta_j\}$ and $\{\tilde\Delta_j\}$ satisfy these properties by hypothesis, with the roles fully symmetric. Repeating Steps 1--4 with $\{\Delta_j\}$ and $\{\tilde\Delta_j\}$ exchanged yields the same conclusion in the opposite direction: \begin{align*} \|f\|_{B^s_{p,q}(\{\tilde\Delta_j\})} \le C'\,\|f\|_{B^s_{p,q}(\{\Delta_j\})}, \end{align*} where $C'$ depends only on $s$, $p$, $q$, $n$, $\hat\psi$, $\hat{\tilde\psi}$. Setting $C_{\mathrm{final}} := \max(C, C')$ gives the two-sided equivalence \begin{align*} C_{\mathrm{final}}^{-1}\,\|f\|_{B^s_{p,q}(\{\tilde\Delta_j\})} \le \|f\|_{B^s_{p,q}(\{\Delta_j\})} \le C_{\mathrm{final}}\,\|f\|_{B^s_{p,q}(\{\tilde\Delta_j\})}, \end{align*} valid for every $f \in \mathcal{S}'(\mathbb{R}^n)$. This completes the proof. [/step]