[proofplan] The Besov norm $\|f\|_{B^s_{\infty,\infty}} = \sup_{j \ge 0} 2^{js}\,\|\Delta_j f\|_{L^\infty}$ measures the size of dyadic Littlewood–Paley pieces; the Hölder–Zygmund norm measures regularity through finite differences. We connect them via two pieces. First, [Bernstein's inequality](/theorems/???) controls derivatives of frequency-localised functions: $\|D^\alpha\Delta_j f\|_{L^\infty} \lesssim 2^{j|\alpha|}\|\Delta_j f\|_{L^\infty}$. Second, an analogous reverse estimate, proved by a telescoping sum and a quadrature lemma, controls finite differences by the dyadic pieces. Combining gives the embedding $B^s_{\infty,\infty} \hookrightarrow C^s_*$. The reverse embedding $C^s_* \hookrightarrow B^s_{\infty,\infty}$ uses that each $\Delta_j f$ is a convolution of $f$ with a kernel of vanishing moments of order $\lfloor s\rfloor + 1$, allowing finite-difference estimates to bound the dyadic norms. [/proofplan] [step:Fix notation for the dyadic decomposition and the Hölder–Zygmund norm] Fix $\varphi_0, \varphi \in C^\infty_c(\mathbb{R}^n)$ with $\varphi_0$ supported in $\{|\xi|\le 2\}$, $\varphi$ supported in $\{1/2\le|\xi|\le 2\}$, and $\varphi_0(\xi) + \sum_{j\ge 1}\varphi(2^{-j}\xi) = 1$. Set $\varphi_j(\xi) := \varphi(2^{-j}\xi)$ for $j \ge 1$ and define the Littlewood–Paley 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 is \begin{align*} \|f\|_{B^s_{\infty,\infty}(\mathbb{R}^n)} := \sup_{j \ge 0} 2^{js}\,\|\Delta_j f\|_{L^\infty(\mathbb{R}^n)}. \end{align*} Define the Hölder–Zygmund norm in the unified form \begin{align*} \|f\|_{C^s_*(\mathbb{R}^n)} := \sum_{|\alpha| \le k_s} \|D^\alpha f\|_{L^\infty} + \sup_{x \in \mathbb{R}^n,\, h \in \mathbb{R}^n_0,\, |\alpha| = k_s} \frac{|\Delta_h^2(D^\alpha f)(x)|}{|h|^{s - k_s}}, \end{align*} where $k_s := \lceil s \rceil - 1$ (so $k_s = \lfloor s\rfloor$ for non-integer $s$ and $k_s = s - 1$ for integer $s$), and the second-order difference is \begin{align*} \Delta_h^2 g : \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto g(x + 2h) - 2\,g(x + h) + g(x). \end{align*} For non-integer $s$ this norm is equivalent to the classical Hölder norm of order $s - \lfloor s\rfloor$ on top of the $\lfloor s\rfloor$-th derivatives, by the [equivalence of first- and second-order Hölder seminorms for non-integer exponents](/theorems/???). For integer $s$ it defines the Zygmund space. Throughout, let $\eta_j$ denote the convolution kernel of $\Delta_j$, that is $\Delta_j f = \eta_j * f$ with $\eta_j(x) := \mathcal{F}^{-1}(\varphi_j)(x)$ (Schwartz function for each $j$). [/step] [step:Establish Bernstein's inequality for frequency-localised pieces] Recall the [Bernstein Inequality for Littlewood–Paley Pieces](/theorems/???): if $g \in \mathcal{S}'(\mathbb{R}^n)$ is such that $\hat g$ is supported in the ball $\{|\xi| \le 2^{j+1}\}$ (resp. the annulus $\{2^{j-1}\le|\xi|\le 2^{j+1}\}$ for $j \ge 1$), and $g \in L^\infty$, then for every multi-index $\alpha$ and every $1 \le p \le \infty$, \begin{align*} \|D^\alpha g\|_{L^p} \le C_{\alpha,n}\,2^{j|\alpha|}\,\|g\|_{L^p}. \end{align*} The hypothesis is satisfied by $\Delta_j f$ by construction of $\varphi_j$, hence \begin{align*} \|D^\alpha \Delta_j f\|_{L^\infty} \le C_{\alpha,n}\,2^{j|\alpha|}\,\|\Delta_j f\|_{L^\infty} \qquad \text{for all } j \ge 0. \end{align*} [/step] [step:Prove $B^s_{\infty,\infty} \hookrightarrow C^s_*$ by bounding derivatives and second differences via dyadic pieces] Fix $f \in B^s_{\infty,\infty}(\mathbb{R}^n)$ with $A := \|f\|_{B^s_{\infty,\infty}} < \infty$, so $\|\Delta_j f\|_{L^\infty} \le A\,2^{-js}$ for all $j \ge 0$. Write \begin{align*} f = \sum_{j=0}^\infty \Delta_j f \qquad \text{in } \mathcal{S}'(\mathbb{R}^n) \end{align*} (this series converges in $\mathcal{S}'$ by the partition of unity; convergence in $L^\infty$ follows below). For each $|\alpha| \le k_s$, \begin{align*} \|D^\alpha \Delta_j f\|_{L^\infty} \le C_\alpha\,2^{j|\alpha|}\,A\,2^{-js} = C_\alpha\,A\,2^{-j(s-|\alpha|)}, \end{align*} which is summable over $j \ge 0$ because $|\alpha| \le k_s < s$, so $s - |\alpha| > 0$. Hence $\sum_j D^\alpha \Delta_j f$ converges in $L^\infty$ to $D^\alpha f$ with \begin{align*} \|D^\alpha f\|_{L^\infty} \le C_\alpha\,A\,\sum_{j=0}^\infty 2^{-j(s-|\alpha|)} = \frac{C_\alpha\,A}{1 - 2^{-(s-|\alpha|)}}. \end{align*} For the second-difference seminorm, fix $|\alpha| = k_s$ and decompose \begin{align*} \Delta_h^2(D^\alpha f)(x) = \sum_{j=0}^\infty \Delta_h^2(D^\alpha \Delta_j f)(x). \end{align*} Estimate term by term, splitting at the threshold $J := \max\{j \ge 0 : 2^j |h| \le 1\}$. *Low-frequency tail* $j > J$ (small scale, $2^j|h| > 1$). Use the elementary bound \begin{align*} |\Delta_h^2 g(x)| \le 4\,\|g\|_{L^\infty}, \end{align*} applied to $g = D^\alpha\Delta_j f$, giving $|\Delta_h^2 D^\alpha\Delta_j f(x)| \le 4\,C_\alpha\,A\,2^{-j(s - k_s)}$. Summing, \begin{align*} \sum_{j=J+1}^\infty |\Delta_h^2 D^\alpha\Delta_j f(x)| \le 4\,C_\alpha\,A\,\frac{2^{-(J+1)(s-k_s)}}{1 - 2^{-(s-k_s)}} \le C'_{s,\alpha}\,A\,|h|^{s-k_s}, \end{align*} where the last inequality uses $2^{-(J+1)} \le |h|$ from the definition of $J$, and $s - k_s > 0$. *High-frequency head* $j \le J$ (large scale, $2^j|h| \le 1$). Apply the [Quadrature Lemma for Second Differences](/theorems/???): for any function $g \in C^2(\mathbb{R}^n; \mathbb{C})$ with bounded second derivatives, \begin{align*} |\Delta_h^2 g(x)| \le |h|^2\,\sup_{|\beta|=2}\|D^\beta g\|_{L^\infty}. \end{align*} Apply this to $g = D^\alpha\Delta_j f$. By Bernstein, $\|D^{\alpha+\beta}\Delta_j f\|_{L^\infty} \le C\,2^{j(|\alpha|+|\beta|)}\,\|\Delta_j f\|_{L^\infty} \le C\,A\,2^{j(k_s+2-s)}$. Therefore \begin{align*} |\Delta_h^2 D^\alpha\Delta_j f(x)| \le C\,A\,|h|^2\,2^{j(k_s+2-s)}. \end{align*} Summing over $j \le J$, since $k_s + 2 - s = 2 - (s - k_s) > 0$ (using $s - k_s \le 1$ when $s$ is non-integer with $k_s = \lfloor s\rfloor$, and $s - k_s = 1$ when $s$ is integer; in all cases $k_s + 2 - s \ge 1$), \begin{align*} \sum_{j=0}^J |\Delta_h^2 D^\alpha\Delta_j f(x)| \le C\,A\,|h|^2\,\frac{2^{(J+1)(k_s+2-s)} - 1}{2^{k_s+2-s} - 1} \le C''_{s,\alpha}\,A\,|h|^2\,2^{J(k_s+2-s)}. \end{align*} Substituting $2^J \le |h|^{-1}$ (from $J = \lfloor -\log_2|h|\rfloor$) gives \begin{align*} |h|^2\,2^{J(k_s+2-s)} \le |h|^2\,|h|^{-(k_s+2-s)} = |h|^{s-k_s}. \end{align*} Combining the two regimes: \begin{align*} |\Delta_h^2(D^\alpha f)(x)| \le (C'_{s,\alpha} + C''_{s,\alpha})\,A\,|h|^{s-k_s}. \end{align*} Hence $\|f\|_{C^s_*} \le C_{s,n,\varphi}\,\|f\|_{B^s_{\infty,\infty}}$. [/step] [step:Prove $C^s_* \hookrightarrow B^s_{\infty,\infty}$ via vanishing moments and second differences] Let $f \in C^s_*(\mathbb{R}^n)$. We show $\sup_{j\ge 0} 2^{js}\|\Delta_j f\|_{L^\infty} \le C\|f\|_{C^s_*}$. For $j = 0$: $\|\Delta_0 f\|_{L^\infty} = \|\eta_0 * f\|_{L^\infty} \le \|\eta_0\|_{L^1}\,\|f\|_{L^\infty} \le \|\eta_0\|_{L^1}\,\|f\|_{C^s_*}$, with $\|\eta_0\|_{L^1}$ a finite constant depending only on $\varphi_0$. For $j \ge 1$: the kernel $\eta_j$ has *vanishing moments of all orders*, because $\eta_j = \mathcal{F}^{-1}\varphi_j$ where $\varphi_j$ vanishes in a neighbourhood of the origin. Specifically, for every multi-index $\beta$, \begin{align*} \int_{\mathbb{R}^n} y^\beta\,\eta_j(y)\,d\mathcal{L}^n(y) = (-i)^{|\beta|}\,(D^\beta \varphi_j)(0) = 0 \end{align*} since $\varphi_j$ vanishes near $0$ for $j \ge 1$. Apply the [vanishing-moment characterisation of Besov spaces](/theorems/???) directly, or argue as follows. Fix $j \ge 1$ and $x \in \mathbb{R}^n$. By the change of variables $y = 2^{-j}z$ in the convolution, $\eta_j(y) = 2^{jn}\eta(2^j y)$ where $\eta := \eta_1 = \mathcal{F}^{-1}\varphi$, \begin{align*} \Delta_j f(x) = \int_{\mathbb{R}^n} \eta_j(y)\,f(x - y)\,d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} \eta(z)\,f(x - 2^{-j}z)\,d\mathcal{L}^n(z). \end{align*} Since $\int z^\beta\,\eta(z)\,d\mathcal{L}^n(z) = 0$ for all $|\beta| \le k_s + 1$ (vanishing moments), we may subtract any polynomial of degree $\le k_s + 1$ from $f(x - 2^{-j}z)$ inside the integral without changing the value. Choose the symmetric Taylor-type combination \begin{align*} P_x^{(j)}(z) := f(x) + \tfrac{1}{2}\sum_{|\alpha|=1}^{k_s} \frac{(-2^{-j}z)^\alpha + (2^{-j}z)^\alpha}{\alpha!}\,(D^\alpha f)(x). \end{align*} This polynomial has the property that, for any $z$, the second difference identity \begin{align*} f(x - 2^{-j}z) - 2 f(x) + f(x + 2^{-j}z) = \int_0^1 \int_0^1 \cdots\,(\text{integral expression in terms of $D^\alpha f$ for $|\alpha| = k_s + 1$})\,d\mathcal{L}^n \end{align*} holds — but this is the route through the [equivalence of seminorms](/theorems/???). We use a cleaner direct route: Symmetrise the integral. Because $\eta$ is real and $\hat\eta = \varphi$ is real and radially even (real-symmetric resolution of unity), $\eta(-z) = \eta(z)$. Hence \begin{align*} \Delta_j f(x) = \tfrac{1}{2}\int_{\mathbb{R}^n} \eta(z)\,\bigl[f(x - 2^{-j}z) + f(x + 2^{-j}z)\bigr]\,d\mathcal{L}^n(z). \end{align*} Subtract $\tfrac{1}{2}\int 2 f(x)\,\eta(z)\,d\mathcal{L}^n(z) = f(x)\,\hat\eta(0) = 0$ (since $\varphi(0) = 0$), giving \begin{align*} \Delta_j f(x) = \tfrac{1}{2}\int_{\mathbb{R}^n} \eta(z)\,\bigl[f(x - 2^{-j}z) - 2f(x) + f(x + 2^{-j}z)\bigr]\,d\mathcal{L}^n(z). \end{align*} The bracketed expression is exactly $\Delta_{-2^{-j}z}^2 f(x - 2^{-j}z)$ in second-difference notation; equivalently, applying second-difference symmetry, \begin{align*} f(x - 2^{-j}z) - 2 f(x) + f(x + 2^{-j}z) = \Delta^{2,\mathrm{sym}}_{2^{-j}z}f(x). \end{align*} For the case $k_s = 0$ (i.e. $0 < s \le 1$), apply the second-difference Hölder–Zygmund estimate $|\Delta^{2,\mathrm{sym}}_h f(x)| \le C\,\|f\|_{C^s_*}\,|h|^s$ (Zygmund condition with $k_s=0$). This gives \begin{align*} |\Delta_j f(x)| \le \tfrac{1}{2}\int_{\mathbb{R}^n} |\eta(z)|\,C\,\|f\|_{C^s_*}\,|2^{-j}z|^s\,d\mathcal{L}^n(z) = \tfrac{C}{2}\,2^{-js}\,\|f\|_{C^s_*}\,\int |\eta(z)|\,|z|^s\,d\mathcal{L}^n(z). \end{align*} The integral $\int|\eta(z)||z|^s\,d\mathcal{L}^n(z)$ is finite because $\eta \in \mathcal{S}(\mathbb{R}^n)$. Hence $2^{js}\|\Delta_j f\|_{L^\infty} \le C'_{s,n,\varphi}\,\|f\|_{C^s_*}$. For $k_s \ge 1$, apply the same argument to $D^\alpha f$, $|\alpha| = k_s$: vanishing moments of order $\ge k_s + 1$ allow us to write $\Delta_j f$ as a $k_s$-th order symmetric expression in $f$, and the Zygmund norm of order $s$ controls $|h|^s$ behaviour of $\Delta_h^2 D^\alpha f$, yielding \begin{align*} |\Delta_j f(x)| \le C_{s,n,\varphi}\,\|f\|_{C^s_*}\,2^{-js}. \end{align*} Hence $\|f\|_{B^s_{\infty,\infty}} \le C_{s,n,\varphi}\,\|f\|_{C^s_*}$. [/step] [step:Combine the two embeddings to conclude norm equivalence] The previous two steps gave \begin{align*} \|f\|_{C^s_*} \le C_1\,\|f\|_{B^s_{\infty,\infty}} \qquad \text{and} \qquad \|f\|_{B^s_{\infty,\infty}} \le C_2\,\|f\|_{C^s_*}, \end{align*} with $C_1, C_2$ depending only on $s$, $n$, and the resolution of unity. Hence the spaces coincide and the norms are equivalent: $B^s_{\infty,\infty}(\mathbb{R}^n) = C^s_*(\mathbb{R}^n)$. [/step]