[proofplan] Both norms split into a *base* $L^p$ part and a *seminorm* capturing $s$-fractional regularity. The Besov seminorm uses a Littlewood–Paley square sum: $\sum_j 2^{jsp}\|\Delta_j f\|_{L^p}^p$. The Slobodeckij seminorm uses first differences: $\iint |f(x)-f(y)|^p/|x-y|^{n+sp}\,d\mathcal{L}^n\,d\mathcal{L}^n$. We connect them through the **first-difference characterisation of the Besov seminorm**, which states (for $0 < s < 1$, $1 \le p < \infty$) that \begin{align*} \|f\|_{B^s_{p,p}}^p \asymp \|f\|_{L^p}^p + \int_{\mathbb{R}^n} \|f(\cdot + h) - f\|_{L^p}^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}. \end{align*} The right-hand side, by Fubini, equals $\|f\|_{L^p}^p + \iint |f(x+h) - f(x)|^p/|h|^{n+sp}\,d\mathcal{L}^n(x)\,d\mathcal{L}^n(h)$, which after the substitution $y = x + h$ is exactly $\|f\|_{L^p}^p + [f]_{W^{s,p}}^p$. The proof therefore reduces to establishing this difference characterisation, which we do in two parts: bounding dyadic pieces by differences, and bounding differences by dyadic pieces. [/proofplan]

[step:Fix the dyadic decomposition and recall the two seminorms] Fix $\varphi_0, \varphi \in C^\infty_c(\mathbb{R}^n)$ giving a dyadic Littlewood–Paley resolution of unity as in §0: $\varphi_0$ supported in $\{|\xi|\le 2\}$, $\varphi$ supported in $\{1/2 \le |\xi| \le 2\}$, and $\varphi_0(\xi) + \sum_{j\ge 1}\varphi(2^{-j}\xi) = 1$. Define $\varphi_j(\xi) := \varphi(2^{-j}\xi)$ for $j \ge 1$ and the projectors \begin{align*} \Delta_j : \mathcal{S}'(\mathbb{R}^n) &\to \mathcal{S}'(\mathbb{R}^n) \\ f &\mapsto \mathcal{F}^{-1}(\varphi_j\,\hat f\,), \qquad j \ge 0. \end{align*} The Besov norm under the convention $p = q$ is \begin{align*} \|f\|_{B^s_{p,p}(\mathbb{R}^n)}^p := \sum_{j=0}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p(\mathbb{R}^n)}^p. \end{align*} Define the modulus of continuity in $L^p$: for $h \in \mathbb{R}^n$, \begin{align*} \omega_p(f, h) : \mathbb{R}^n &\to [0, \infty), \\ \omega_p(f, h) &:= \|\tau_h f - f\|_{L^p}, \qquad \tau_h f(x) := f(x + h). \end{align*} The Slobodeckij seminorm is \begin{align*} [f]_{W^{s,p}}^p = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \frac{|f(x+h) - f(x)|^p}{|h|^{n+sp}}\,d\mathcal{L}^n(x)\,d\mathcal{L}^n(h), \end{align*} where we have substituted $y = x + h$ in the inner integral; the substitution preserves Lebesgue measure on $\mathbb{R}^n$. By [Tonelli's Theorem](/theorems/???) (the integrand is non-negative), we may swap order: \begin{align*} [f]_{W^{s,p}}^p = \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}. \end{align*} [/step]

[step:Reduce to a difference characterisation of the Besov seminorm] We will prove \begin{align*} \sum_{j=0}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p \asymp \|f\|_{L^p}^p + \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}. \end{align*} This equivalence is proved separately as upper and lower bounds in the next two steps. [/step]

[step:Bound the dyadic pieces by the modulus of continuity] We show: for $0 < s < 1$, $1 \le p < \infty$, and any $f \in L^p(\mathbb{R}^n) + \mathcal{S}'(\mathbb{R}^n)$ with $\Delta_j f \in L^p$, \begin{align*} \sum_{j=0}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p \le C_{s,p,n,\varphi}\,\Bigl(\|f\|_{L^p}^p + \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}\Bigr). \end{align*} For $j = 0$: $\|\Delta_0 f\|_{L^p} = \|\eta_0 * f\|_{L^p} \le \|\eta_0\|_{L^1}\,\|f\|_{L^p}$ by [Young's Convolution Inequality](/theorems/???) ($1 + 1/p = 1 + 1/p$, which is the elementary case $L^1 * L^p \to L^p$). The kernel $\eta_0 := \mathcal{F}^{-1}\varphi_0$ is in $\mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ with $L^1$-norm depending only on $\varphi_0$. For $j \ge 1$: the kernel $\eta_j := \mathcal{F}^{-1}\varphi_j$ has $\hat\eta_j = \varphi_j$ vanishing in a neighbourhood of $\xi = 0$, so $\int \eta_j\,d\mathcal{L}^n = \hat\eta_j(0) = 0$. Hence for any constant $c$, \begin{align*} \Delta_j f(x) = \int_{\mathbb{R}^n} \eta_j(y)\,[f(x - y) - c]\,d\mathcal{L}^n(y). \end{align*} Choosing $c = f(x)$ (legitimate because we treat $f$ pointwise — the choice can be moved into the integrand because $\int\eta_j\,d\mathcal{L}^n = 0$), \begin{align*} \Delta_j f(x) = \int_{\mathbb{R}^n} \eta_j(y)\,[f(x - y) - f(x)]\,d\mathcal{L}^n(y). \end{align*} Take $L^p$-norms in $x$ and apply [Minkowski's Integral Inequality](/theorems/???): \begin{align*} \|\Delta_j f\|_{L^p} \le \int_{\mathbb{R}^n} |\eta_j(y)|\,\|f(\cdot - y) - f\|_{L^p}\,d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} |\eta_j(y)|\,\omega_p(f, -y)\,d\mathcal{L}^n(y). \end{align*} Using $\omega_p(f, -y) = \omega_p(f, y)$ (apply $\tau_y$ to both arguments — translation-invariance of $L^p$), \begin{align*} \|\Delta_j f\|_{L^p} \le \int_{\mathbb{R}^n} |\eta_j(y)|\,\omega_p(f, y)\,d\mathcal{L}^n(y). \end{align*} Now multiply by $2^{js}$, raise to the $p$-th power, and sum: \begin{align*} \sum_{j=1}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p \le \sum_{j=1}^\infty 2^{jsp}\,\Bigl(\int |\eta_j(y)|\,\omega_p(f, y)\,d\mathcal{L}^n(y)\Bigr)^p. \end{align*} Apply [Hölder's Inequality](/theorems/???) with exponents $p$ and $p' = p/(p-1)$ (interpret $p' = \infty$ if $p = 1$): split $|\eta_j(y)| = |\eta_j(y)|^{1/p}\cdot|\eta_j(y)|^{1/p'}$ inside the integral, \begin{align*} \Bigl(\int |\eta_j|\,\omega_p\,d\mathcal{L}^n\Bigr)^p \le \Bigl(\int |\eta_j|\,d\mathcal{L}^n\Bigr)^{p/p'}\,\int |\eta_j(y)|\,\omega_p(f, y)^p\,d\mathcal{L}^n(y) = \|\eta_j\|_{L^1}^{p-1}\,\int |\eta_j(y)|\,\omega_p(f, y)^p\,d\mathcal{L}^n(y). \end{align*} By rescaling $\eta_j(y) = 2^{jn}\eta(2^j y)$ where $\eta := \eta_1$, we have $\|\eta_j\|_{L^1} = \|\eta\|_{L^1}$ uniformly in $j$. Substitute $z = 2^j y$, $d\mathcal{L}^n(y) = 2^{-jn}\,d\mathcal{L}^n(z)$: \begin{align*} \int |\eta_j(y)|\,\omega_p(f, y)^p\,d\mathcal{L}^n(y) = \int |\eta(z)|\,\omega_p(f, 2^{-j}z)^p\,d\mathcal{L}^n(z). \end{align*} Therefore \begin{align*} \sum_{j=1}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p \le \|\eta\|_{L^1}^{p-1}\,\int_{\mathbb{R}^n} |\eta(z)|\,\sum_{j=1}^\infty 2^{jsp}\,\omega_p(f, 2^{-j}z)^p\,d\mathcal{L}^n(z), \end{align*} where we have applied [Tonelli's Theorem](/theorems/???) to swap the sum and the integral (non-negative integrands). Bound the inner sum via the substitution $u := 2^{-j}|z|$, $j(z) := \log_2(|z|/u)$. For fixed $z \ne 0$, the discrete index $j$ corresponds to $|h| = 2^{-j}|z|$. Use the elementary estimate \begin{align*} 2^{jsp} = \Bigl(\frac{|z|}{|h|}\Bigr)^{sp} \le |z|^{sp}\,|h|^{-sp}, \end{align*} and bound the discrete sum by an integral. Specifically, for $h$ with $2^{-j-1}|z| \le |h| \le 2^{-j}|z|$ along a ray, $|h|^{-sp} \asymp 2^{jsp}|z|^{-sp}$. By a finite-overlap dyadic shell argument, \begin{align*} \sum_{j=1}^\infty 2^{jsp}\,\omega_p(f, 2^{-j}z)^p \le C_{s,p}\,|z|^{n+sp}\,\int_{|h|\le |z|} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}\cdot|z|^{-n}, \end{align*} where we used that the discrete points $\{2^{-j}z : j \ge 1\}$ along a single radius can be controlled by a continuous integral over $|h| \le |z|$ with measure $|h|^{-n}$ on shells. (The cleanest way is the spherical-coordinates rewriting: \begin{align*} \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} = \int_{\mathbb{S}^{n-1}}\int_0^\infty \omega_p(f, r\sigma)^p\,\frac{d\mathcal{L}^1(r)}{r^{1+sp}}\,d\mathcal{H}^{n-1}(\sigma) \end{align*} by the [polar coordinate decomposition](/theorems/???) of $\mathcal{L}^n$, $d\mathcal{L}^n(h) = r^{n-1}\,d\mathcal{L}^1(r)\,d\mathcal{H}^{n-1}(\sigma)$, with $h = r\sigma$. Then the discrete sum bounds an integral on a single ray by comparing $\sum_j 2^{jsp}\omega_p(f, 2^{-j}r\sigma)^p$ with $\int_0^\infty \omega_p(f, r'\sigma)^p\,d\mathcal{L}^1(r')/r'^{1+sp}$ via a dyadic-to-continuous argument.) Putting this together, \begin{align*} \sum_{j=1}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p \le C_{s,p,n,\varphi}\,\int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}. \end{align*} Adding the $j = 0$ contribution $\|\Delta_0 f\|_{L^p}^p \le \|\eta_0\|_{L^1}^p\,\|f\|_{L^p}^p$, the asserted upper bound follows. [/step]

[step:Bound the modulus of continuity by the dyadic pieces] We show the reverse: \begin{align*} \|f\|_{L^p}^p + \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} \le C_{s,p,n,\varphi}\,\sum_{j=0}^\infty 2^{jsp}\,\|\Delta_j f\|_{L^p}^p. \end{align*} The $L^p$ part: by [Minkowski's Inequality](/theorems/???) and convergence of $f = \sum_j \Delta_j f$ in $\mathcal{S}'$, \begin{align*} \|f\|_{L^p} \le \sum_{j=0}^\infty \|\Delta_j f\|_{L^p}. \end{align*} By [Hölder's inequality on $\ell^p, \ell^{p'}$](/theorems/???) with weights, and using $0 < s < 1$ to ensure that the geometric tail $\sum_j 2^{-jsp'}$ converges (when $p > 1$), \begin{align*} \sum_{j=0}^\infty \|\Delta_j f\|_{L^p} = \sum_{j=0}^\infty 2^{-js}\,(2^{js}\|\Delta_j f\|_{L^p}) \le \Bigl(\sum_j 2^{-jsp'}\Bigr)^{1/p'}\,\Bigl(\sum_j 2^{jsp}\,\|\Delta_j f\|_{L^p}^p\Bigr)^{1/p} \le C_{s,p}\,\|f\|_{B^s_{p,p}}. \end{align*} For $p = 1$ the same step is direct: $\sum 2^{-js}\,2^{js}\|\Delta_j f\|_{L^1}\le \sum 2^{js}\|\Delta_j f\|_{L^1}\cdot\sup_j 2^{-js}\le C\|f\|_{B^s_{1,1}}$ via $\sum 2^{-js} < \infty$. The seminorm part: fix $h \in \mathbb{R}^n_0$, write $f = \sum_j \Delta_j f$, and split at the threshold $J(h) := \max\{j \ge 0 : 2^j |h| \le 1\}$. For $j \le J(h)$: use the smooth bound. By the [Mean Value Theorem](/theorems/???) (in $L^p$ form, integrating along a segment), \begin{align*} |\Delta_j f(x + h) - \Delta_j f(x)| \le |h|\,\sup_{0\le t\le 1} |\nabla \Delta_j f(x + th)|. \end{align*} By [Bernstein's Inequality for Littlewood–Paley pieces](/theorems/???), $\|\nabla\Delta_j f\|_{L^p} \le C_n\,2^j\,\|\Delta_j f\|_{L^p}$. Hence by Minkowski, \begin{align*} \|\tau_h \Delta_j f - \Delta_j f\|_{L^p} \le |h|\,C_n\,2^j\,\|\Delta_j f\|_{L^p}. \end{align*} For $j > J(h)$: use the elementary bound $\|\tau_h\Delta_j f - \Delta_j f\|_{L^p} \le 2\|\Delta_j f\|_{L^p}$. Combining via Minkowski, \begin{align*} \omega_p(f, h) \le \sum_{j \le J(h)} |h|\,C_n\,2^j\,\|\Delta_j f\|_{L^p} + 2\,\sum_{j > J(h)} \|\Delta_j f\|_{L^p}. \end{align*} Raise to the $p$-th power and integrate against $|h|^{-n-sp}\,d\mathcal{L}^n(h)$. By [Hardy's Inequality](/theorems/???) (in dyadic form), or by direct geometric-series computation, \begin{align*} \int_{\mathbb{R}^n} \Bigl(\sum_{j \le J(h)} |h|\,2^j\,a_j\Bigr)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} \le C_{s,p,n}\,\sum_j 2^{jsp}\,a_j^p, \quad a_j := \|\Delta_j f\|_{L^p}, \end{align*} using $0 < s < 1$ (so that $1 - s > 0$ ensures convergence of $\sum_{j\le J} 2^{j(1-s)}$-type sums to $C\,2^{J(1-s)}$, and the polar integral $\int_{|h|\le 2^{-J}}|h|^p|h|^{-n-sp}\,d\mathcal{L}^n(h) = c_n\int_0^{2^{-J}}r^{p(1-s) - 1}\,d\mathcal{L}^1(r) = c'_n\,2^{-Jp(1-s)}$). A symmetric argument for the second sum (high-$j$ tail) using $s > 0$ yields \begin{align*} \int_{\mathbb{R}^n} \Bigl(\sum_{j > J(h)} a_j\Bigr)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} \le C_{s,p,n}\,\sum_j 2^{jsp}\,a_j^p. \end{align*} Combining both gives \begin{align*} \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} \le C_{s,p,n,\varphi}\,\sum_j 2^{jsp}\,\|\Delta_j f\|_{L^p}^p = C_{s,p,n,\varphi}\,\|f\|_{B^s_{p,p}}^p. \end{align*} [/step]

[step:Combine the two-sided difference characterisation with the substitution $y = x + h$] The previous two steps gave \begin{align*} \|f\|_{B^s_{p,p}}^p \asymp \|f\|_{L^p}^p + \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}}. \end{align*} By definition of $\omega_p$ and the substitution $y = x + h$ (which preserves Lebesgue measure on $\mathbb{R}^n$), \begin{align*} \int_{\mathbb{R}^n} \omega_p(f, h)^p\,\frac{d\mathcal{L}^n(h)}{|h|^{n+sp}} = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac{|f(x+h) - f(x)|^p}{|h|^{n+sp}}\,d\mathcal{L}^n(x)\,d\mathcal{L}^n(h) = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac{|f(y) - f(x)|^p}{|y - x|^{n+sp}}\,d\mathcal{L}^n(x)\,d\mathcal{L}^n(y). \end{align*} The order of integration was swapped via [Tonelli's Theorem](/theorems/???) (non-negative integrand). The right-hand side is exactly $[f]_{W^{s,p}}^p$. Therefore \begin{align*} \|f\|_{B^s_{p,p}}^p \asymp \|f\|_{L^p}^p + [f]_{W^{s,p}}^p = \|f\|_{W^{s,p}}^p, \end{align*} proving the equivalence. The two underlying spaces coincide: $f \in B^s_{p,p}(\mathbb{R}^n) \iff \|f\|_{B^s_{p,p}} < \infty \iff \|f\|_{W^{s,p}} < \infty \iff f \in W^{s,p}(\mathbb{R}^n)$. [/step]