[proofplan] Both inclusions follow from a single tool: the **mixed-norm Minkowski integral inequality**, which compares the order in which one takes an $L^p$-norm in $x \in \mathbb{R}^n$ and an $\ell^r$-norm over the dyadic index $j$. The Triebel–Lizorkin norm is $\|f\|_{F^s_{p,q}} = \|(2^{js}\Delta_j f)_j\|_{L^p(\ell^q)}$ (inner $\ell^q$, outer $L^p$), while the Besov norm is $\|f\|_{B^s_{p,r}} = \|(2^{js}\Delta_j f)_j\|_{\ell^r(L^p)}$ (inner $L^p$, outer $\ell^r$). Minkowski's integral inequality, in the standard form $\|F\|_{L^B(\ell^A)} \le \|F\|_{\ell^A(L^B)}$ for $1 \le A \le B \le \infty$, says that **placing the larger exponent on the outside produces the smaller norm**. Applied with $(A, B) = (\min(p,q), p)$, this gives the lower embedding $B^s_{p,\min(p,q)} \hookrightarrow F^s_{p,q}$ after a pointwise $\ell^q \le \ell^{\min(p,q)}$ comparison; applied with $(A, B) = (p, \max(p,q))$, it gives the upper embedding $F^s_{p,q} \hookrightarrow B^s_{p,\max(p,q)}$ after a pointwise $\ell^{\max(p,q)} \le \ell^q$ comparison. [/proofplan]

[step:Recall the Besov and Triebel–Lizorkin norms] Fix the dyadic resolution of unity and Littlewood–Paley projectors $\Delta_j$ as in the previous proofs. The Besov norm is \begin{align*} \|f\|_{B^s_{p,r}(\mathbb{R}^n)} := \Bigl(\sum_{j=0}^\infty 2^{jsr}\,\|\Delta_j f\|_{L^p(\mathbb{R}^n)}^r\Bigr)^{1/r}, \qquad 1 \le r < \infty, \end{align*} with the obvious modification $\sup_j 2^{js}\|\Delta_j f\|_{L^p}$ for $r = \infty$. The Triebel–Lizorkin norm is \begin{align*} \|f\|_{F^s_{p,q}(\mathbb{R}^n)} := \Bigl\| \Bigl(\sum_{j=0}^\infty 2^{jsq}\,|\Delta_j f|^q\Bigr)^{1/q} \Bigr\|_{L^p(\mathbb{R}^n)}, \qquad 1 \le q < \infty, \end{align*} with the modification $\| \sup_j 2^{js}|\Delta_j f|\,\|_{L^p}$ for $q = \infty$. Define the sequence of functions $g_j : \mathbb{R}^n \to \mathbb{C}$ by $g_j(x) := 2^{js}\,\Delta_j f(x)$. Then \begin{align*} \|f\|_{F^s_{p,q}} &= \Bigl\| (g_j)_j \Bigr\|_{L^p(\mathbb{R}^n;\,\ell^q)} = \Bigl\| \|(g_j(x))_j\|_{\ell^q} \Bigr\|_{L^p(\mathbb{R}^n)}, \\ \|f\|_{B^s_{p,r}} &= \Bigl\| (g_j)_j \Bigr\|_{\ell^r(L^p(\mathbb{R}^n))} = \Bigl\| \|g_j\|_{L^p(\mathbb{R}^n)} \Bigr\|_{\ell^r}. \end{align*} The two norms differ only in the order of the $L^p$ (in $x \in \mathbb{R}^n$) and $\ell^r$ or $\ell^q$ (in $j \in \mathbb{N}_0$) operations. [/step]

[step:State the mixed-norm Minkowski integral inequality] We use the following formulation of Minkowski's integral inequality, valid for non-negative measurable functions. The general form, applied to a Banach-space-valued integrand on a $\sigma$-finite measure space $(Y, \nu)$ taking values in a reflexive Banach space $(X, \|\cdot\|_X)$, states \begin{align*} \Bigl\| \int_Y F(y)\, d\nu(y) \Bigr\|_X \;\le\; \int_Y \|F(y)\|_X\, d\nu(y), \end{align*} which after specialising $X = L^B(\mathbb{R}^n)$ ($1 \le B < \infty$, reflexive for $B \in (1, \infty)$ and handled by approximation at the endpoints) and $\nu = $ counting measure on $\mathbb{N}_0$ yields the mixed-norm form below. **Mixed-norm Minkowski.** Let $1 \le A \le B \le \infty$. For any non-negative measurable $F : \mathbb{R}^n \times \mathbb{N}_0 \to [0, \infty)$, \begin{align*} \|F\|_{L^B_x(\ell^A_j)} \;\le\; \|F\|_{\ell^A_j(L^B_x)}, \end{align*} where the mixed-norm conventions are \begin{align*} \|F\|_{L^B_x(\ell^A_j)} &= \Bigl(\int_{\mathbb{R}^n} \Bigl(\sum_{j} F(x,j)^A\Bigr)^{B/A} d\mathcal{L}^n(x)\Bigr)^{1/B}, \\ \|F\|_{\ell^A_j(L^B_x)} &= \Bigl(\sum_j \Bigl(\int_{\mathbb{R}^n} F(x,j)^B\,d\mathcal{L}^n(x)\Bigr)^{A/B}\Bigr)^{1/A}. \end{align*} **This is the standard form of Minkowski's integral inequality**, applied with the two measures $\nu = $ counting measure on $\mathbb{N}_0$ and $\mu = \mathcal{L}^n$ on $\mathbb{R}^n$. It expresses the unconditional rule: *placing the larger exponent on the outside yields the smaller mixed-norm.* Equivalently, **smaller exponent outside dominates**. Verification reduction (used twice below): with $\rho := B/A \ge 1$ and $a_j(x) := F(x, j)^A \ge 0$, the inequality $\bigl\|\sum_j a_j\bigr\|_{L^\rho}^{1/A} \le \bigl(\sum_j \|a_j\|_{L^\rho}\bigr)^{1/A}$ is exactly the $L^\rho$-triangle inequality (valid for $\rho \ge 1$), which after raising to the $1/A$ power and unfolding is the displayed inequality above. For $A = B$ the inequality is the equality obtained from Fubini's theorem, $\ell^A_j(L^A_x) = L^A_x(\ell^A_j)$ with norm-preserving identification. [/step]

[step:Compare the $\ell^r$-norms pointwise via the nesting of $\ell^r$-spaces] For sequences of non-negative reals $(b_j)_{j \in \mathbb{N}_0}$, the $\ell^r$-norms are *nested in the reverse direction*: for $1 \le r_1 \le r_2 \le \infty$, \begin{align*} \|(b_j)\|_{\ell^{r_2}} \le \|(b_j)\|_{\ell^{r_1}}. \end{align*} This is the elementary inequality $(\sum b_j^{r_2})^{1/r_2} \le (\sum b_j^{r_1})^{1/r_1}$ valid for $r_1 \le r_2$ and non-negative $b_j$, e.g. by setting $c_j := b_j / \|(b_k)\|_{\ell^{r_1}}$ so $\sum c_j^{r_1} \le 1$, hence $c_j \le 1$ and $c_j^{r_2} \le c_j^{r_1}$. In particular, for any $r \le q$ and any $x \in \mathbb{R}^n$, \begin{align*} \|(g_j(x))_j\|_{\ell^q} \le \|(g_j(x))_j\|_{\ell^r}, \qquad r \le q, \end{align*} and for any $r \ge q$, \begin{align*} \|(g_j(x))_j\|_{\ell^r} \le \|(g_j(x))_j\|_{\ell^q}, \qquad r \ge q. \end{align*} [/step]

[step:Establish $B^s_{p,\min(p,q)} \hookrightarrow F^s_{p,q}$ via Minkowski with $A = \min(p,q)$, $B = p$]Set $r := \min(p, q)$, so $r \le p$ and $r \le q$. We show $\|f\|_{F^s_{p,q}} \le \|f\|_{B^s_{p,r}}$. By the pointwise nesting from the previous step, $r \le q$ implies $\|(g_j(x))_j\|_{\ell^q} \le \|(g_j(x))_j\|_{\ell^r}$ for every $x \in \mathbb{R}^n$, so \begin{align*} \|f\|_{F^s_{p,q}} = \bigl\|\,\|(g_j)\|_{\ell^q}\,\bigr\|_{L^p(\mathbb{R}^n)} \le \bigl\|\,\|(g_j)\|_{\ell^r}\,\bigr\|_{L^p(\mathbb{R}^n)} = \|(g_j)\|_{L^p(\ell^r)}. \end{align*} Now apply the mixed-norm Minkowski inequality with $A := r$, $B := p$. The hypothesis $A \le B$ holds because $r = \min(p,q) \le p$. Hence \begin{align*} \|(g_j)\|_{L^p(\ell^r)} \le \|(g_j)\|_{\ell^r(L^p)} = \|f\|_{B^s_{p,r}} = \|f\|_{B^s_{p,\min(p,q)}}. \end{align*} Chaining the two estimates yields $\|f\|_{F^s_{p,q}} \le \|f\|_{B^s_{p,\min(p,q)}}$, the asserted embedding with constant $C_1 = 1$.[/step]

[guided]We must bound the $L^p(\ell^q)$-norm of $(g_j)$ by the $\ell^r(L^p)$-norm with $r = \min(p, q)$. The strategy is in two steps: (a) replace the inner $\ell^q$-norm by the larger $\ell^r$-norm, valid because $r \le q$ makes the $\ell^r$-norm dominate pointwise; (b) swap the order of $L^p$ (outside) and $\ell^r$ (inside) using Minkowski with $A = r \le p = B$. **Step (a): pointwise $\ell^q \le \ell^r$.** Since $r \le q$, the $\ell^r$-to-$\ell^q$ comparison from the previous step gives, for each $x \in \mathbb{R}^n$, \begin{align*} \|(g_j(x))_j\|_{\ell^q} \le \|(g_j(x))_j\|_{\ell^r}. \end{align*} Taking the $L^p$-norm in $x \in \mathbb{R}^n$ (a monotone operation on non-negative integrands), \begin{align*} \|f\|_{F^s_{p,q}} = \bigl\|\,\|(g_j)\|_{\ell^q}\,\bigr\|_{L^p} \le \bigl\|\,\|(g_j)\|_{\ell^r}\,\bigr\|_{L^p} = \|(g_j)\|_{L^p(\ell^r)}. \end{align*} **Step (b): swap $L^p$ and $\ell^r$ via Minkowski.** Apply the mixed-norm Minkowski inequality from the previous step with $A := r$ and $B := p$. The hypothesis $A \le B$ is exactly $r \le p$, which holds since $r = \min(p,q) \le p$. Thus \begin{align*} \|(g_j)\|_{L^p(\ell^r)} \;\le\; \|(g_j)\|_{\ell^r(L^p)} \;=\; \|f\|_{B^s_{p,r}} \;=\; \|f\|_{B^s_{p,\min(p,q)}}. \end{align*} This is the standard direction of Minkowski (larger exponent $B = p$ outside on the LHS yields the smaller norm). Combining (a) and (b), \begin{align*} \|f\|_{F^s_{p,q}} \le \|f\|_{B^s_{p,\min(p,q)}}, \end{align*} which is the asserted embedding $B^s_{p,\min(p,q)} \hookrightarrow F^s_{p,q}$ with constant $C_1 = 1$. Why this works: we needed to move *from* a configuration with $\ell^q$ inside *to* one with $\ell^r$ outside. The $\ell^q$ is too sharp ($q$ may be larger than $\min(p,q)$, giving too small a norm), so we first *enlarge* the inner norm by passing to $\ell^r$ (step a). Now the inner norm uses the small index $r$, and we can apply Minkowski to swap the order — but this swap is only "free" in one direction, namely when the index of the *outer* operation in the desired endpoint is the *larger* one. Here the desired endpoint has outer $\ell^r$ and inner $L^p$ with $p \ge r$, so $L^p$ is the larger exponent and Minkowski must be applied to bring $L^p$ from inside to outside. The starting point has $L^p$ outside, $\ell^r$ inside, which is what Minkowski's "larger-outside" inequality bounds. Both legs go in the right direction.[/guided]

[step:Establish $F^s_{p,q} \hookrightarrow B^s_{p,\max(p,q)}$ via Minkowski with $A = p$, $B = \max(p,q)$]Set $r := \max(p, q)$, so $r \ge p$ and $r \ge q$. We show $\|f\|_{B^s_{p,r}} \le \|f\|_{F^s_{p,q}}$. We split the argument into two monotone moves. **Move (a): Minkowski with exponents $(A, B) = (p, r)$ on $(g_j)$ to bound $\ell^r(L^p)$ by $L^p(\ell^r)$.** The desired inequality is \begin{align*} \|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)}, \qquad r \ge p. \end{align*} Set $\rho := r/p \ge 1$ and $a_j(x) := |g_j(x)|^p$. Then $\|(g_j)\|_{\ell^r(L^p)}^p = \bigl(\sum_j \|a_j\|_{L^1}^\rho\bigr)^{1/\rho}$. By the dual characterisation of the $\ell^\rho$-norm (for $\rho \ge 1$, conjugate exponent $\rho' = r/(r-p)$, with the convention $\rho' = \infty$ when $r = p$), \begin{align*} \Bigl(\sum_j \|a_j\|_{L^1}^\rho\Bigr)^{1/\rho} = \sup_{\|(c_j)\|_{\ell^{\rho'}} \le 1} \sum_j c_j\, \|a_j\|_{L^1} = \sup_{\|(c_j)\|_{\ell^{\rho'}} \le 1} \int_{\mathbb{R}^n} \sum_j c_j\, a_j(x)\,d\mathcal{L}^n(x). \end{align*} For each $x \in \mathbb{R}^n$, Hölder's inequality on $j$ with exponents $(\rho', \rho)$ gives $\sum_j c_j\, a_j(x) \le \|(c_j)\|_{\ell^{\rho'}}\bigl(\sum_j a_j(x)^\rho\bigr)^{1/\rho} = \|(c_j)\|_{\ell^{\rho'}}\bigl(\sum_j |g_j(x)|^r\bigr)^{p/r}$. Integrating and taking the supremum, \begin{align*} \|(g_j)\|_{\ell^r(L^p)}^p \le \int_{\mathbb{R}^n}\Bigl(\sum_j |g_j(x)|^r\Bigr)^{p/r}\,d\mathcal{L}^n(x) = \|(g_j)\|_{L^p(\ell^r)}^p. \end{align*} Raising to the $1/p$ power gives $\|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)}$. The proven inequality, in the mixed-norm notation of Step 2, is \begin{align*} \|F\|_{\ell^B(L^A)} \le \|F\|_{L^A(\ell^B)}, \qquad 1 \le A \le B \le \infty, \end{align*} applied with $A = p$, $B = r$ — exactly the **standard mixed-norm Minkowski inequality** of Step 2, with the two exponents now playing the roles $(A, B) = (p, r)$. The hypothesis $A \le B$ becomes $p \le r$, which holds since $r = \max(p,q) \ge p$. **Move (b): pointwise $\ell^r \le \ell^q$.** Since $r = \max(p, q) \ge q$, the pointwise $\ell$-norm nesting from Step 3 gives, for every $x \in \mathbb{R}^n$, \begin{align*} \|(g_j(x))_j\|_{\ell^r} \le \|(g_j(x))_j\|_{\ell^q}. \end{align*} Taking the $L^p$-norm in $x$, \begin{align*} \|(g_j)\|_{L^p(\ell^r)} \le \|(g_j)\|_{L^p(\ell^q)} = \|f\|_{F^s_{p,q}}. \end{align*} Combining moves (a) and (b), \begin{align*} \|f\|_{B^s_{p,\max(p,q)}} = \|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)} \le \|(g_j)\|_{L^p(\ell^q)} = \|f\|_{F^s_{p,q}}, \end{align*} which is the asserted embedding $F^s_{p,q} \hookrightarrow B^s_{p,\max(p,q)}$ with constant $C_2 = 1$.[/step]

[guided]We must bound the $\ell^r(L^p)$-norm by the $L^p(\ell^q)$-norm with $r = \max(p, q)$. The strategy mirrors the previous step but with the role of the two pointwise $\ell$-comparisons reversed: (a) swap the order of $L^p$ (inside) and $\ell^r$ (outside) using Minkowski with $A = p \le r = B$; (b) replace the inner $\ell^r$-norm by the smaller $\ell^q$-norm, valid because $r \ge q$ makes the $\ell^q$-norm dominate pointwise. **Step (a): swap $\ell^r$ and $L^p$ via Minkowski.** The starting configuration $\ell^r(L^p)$ has the *larger* exponent $r$ on the outside. The mixed-norm Minkowski inequality with $A := p$, $B := r$ reads \begin{align*} \|(g_j)\|_{L^B_x(\ell^A_j)} = \|(g_j)\|_{L^r(\ell^p)} \le \|(g_j)\|_{\ell^p(L^r)} = \|(g_j)\|_{\ell^A_j(L^B_x)}. \end{align*} This is **the same Minkowski inequality** as in the previous step — the standard form $\|F\|_{L^B(\ell^A)} \le \|F\|_{\ell^A(L^B)}$ for $A \le B$ — applied with the two exponents now playing the roles $(A, B) = (p, r)$ instead of $(A, B) = (r, p)$. The hypothesis $A \le B$ becomes $p \le r$, which holds since $r = \max(p,q) \ge p$. But this gives the wrong endpoint: the output is $\|(g_j)\|_{\ell^p(L^r)}$, not the $\ell^r(L^p)$-norm we want. We need a different, finer argument: prove directly that $\|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)}$ for $r \ge p$. This is also a form of Minkowski, derived as follows. With the substitution $\rho := r/p \ge 1$ and $a_j(x) := |g_j(x)|^p$, the inequality becomes the $L^\rho$-triangle inequality applied to non-negative functions: \begin{align*} \|(g_j)\|_{\ell^r(L^p)}^p &= \Bigl(\sum_j \|g_j\|_{L^p}^r\Bigr)^{p/r} = \Bigl(\sum_j \|a_j\|_{L^1}^\rho\Bigr)^{1/\rho}. \end{align*} By duality of $\ell^\rho$ (for $\rho \ge 1$ with conjugate $\rho' = r/(r-p)$), \begin{align*} \Bigl(\sum_j \|a_j\|_{L^1}^\rho\Bigr)^{1/\rho} &= \sup_{(c_j) :\, \|(c_j)\|_{\ell^{\rho'}} \le 1} \sum_j c_j \|a_j\|_{L^1} = \sup_{(c_j) :\, \|(c_j)\|_{\ell^{\rho'}} \le 1} \int_{\mathbb{R}^n} \sum_j c_j\, a_j(x)\, d\mathcal{L}^n(x). \end{align*} For each $x$, Hölder's inequality on $j \in \mathbb{N}_0$ with exponents $(\rho', \rho)$ gives \begin{align*} \sum_j c_j\, a_j(x) \le \|(c_j)\|_{\ell^{\rho'}} \cdot \Bigl(\sum_j a_j(x)^\rho\Bigr)^{1/\rho} = \|(c_j)\|_{\ell^{\rho'}} \cdot \Bigl(\sum_j |g_j(x)|^r\Bigr)^{p/r}. \end{align*} Integrating in $x$ and taking the supremum over $(c_j)$ with $\|(c_j)\|_{\ell^{\rho'}} \le 1$, \begin{align*} \|(g_j)\|_{\ell^r(L^p)}^p &\le \int_{\mathbb{R}^n} \Bigl(\sum_j |g_j(x)|^r\Bigr)^{p/r}\,d\mathcal{L}^n(x) = \|(g_j)\|_{L^p(\ell^r)}^p. \end{align*} Taking $1/p$ powers gives $\|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)}$. (At $r = p$ this is the equality $\ell^p(L^p) = L^p(\ell^p)$ from Fubini's theorem.) This is a standard packaging of Minkowski's integral inequality: the triangle inequality in $L^\rho$ plus duality with Hölder gives the mixed-norm inequality $\|F\|_{\ell^B_j(L^A_x)} \le \|F\|_{L^A_x(\ell^B_j)}$ for $1 \le A \le B \le \infty$ — *the same standard direction* as in the previous step, with the variables relabeled (here $A = p$, $B = r$). **Step (b): pointwise $\ell^r \le \ell^q$.** Since $r \ge q$, the pointwise nesting from Step 3 gives \begin{align*} \|(g_j(x))_j\|_{\ell^r} \le \|(g_j(x))_j\|_{\ell^q}, \qquad x \in \mathbb{R}^n. \end{align*} Taking $L^p$-norm in $x$, \begin{align*} \|(g_j)\|_{L^p(\ell^r)} \le \|(g_j)\|_{L^p(\ell^q)} = \|f\|_{F^s_{p,q}}. \end{align*} Combining (a) and (b): \begin{align*} \|f\|_{B^s_{p,\max(p,q)}} = \|(g_j)\|_{\ell^r(L^p)} \le \|(g_j)\|_{L^p(\ell^r)} \le \|(g_j)\|_{L^p(\ell^q)} = \|f\|_{F^s_{p,q}}, \end{align*} which is the asserted embedding $F^s_{p,q} \hookrightarrow B^s_{p,\max(p,q)}$ with constant $C_2 = 1$. Why this works: in this direction we *start* with the desired endpoint $\ell^r(L^p)$ and want to bound it by $L^p(\ell^q)$. The first move (step a) uses Minkowski to convert the outer $\ell^r$-norm into an *inner* $\ell^r$-norm, and this is precisely the standard Minkowski form $\|F\|_{\ell^B(L^A)} \le \|F\|_{L^A(\ell^B)}$ for $A \le B$ (with $A = p$, $B = r$). The second move (step b) uses that with $r \ge q$ we have $\ell^r \subseteq \ell^q$ as norms, so we may *shrink* the inner $\ell^r$ to $\ell^q$ for free. Both moves are monotone in the right direction. The previous step's argument is symmetric: pointwise $\ell^q \le \ell^r$ enlarges the inner norm, then Minkowski moves $L^p$ outside; here, Minkowski moves $\ell^r$ inside, then pointwise $\ell^r \le \ell^q$ shrinks the inner norm.[/guided]

[step:Combine the two embeddings] Steps 4 and 5 give the asserted chain \begin{align*} B^s_{p,\min(p,q)}(\mathbb{R}^n) \hookrightarrow F^s_{p,q}(\mathbb{R}^n) \hookrightarrow B^s_{p,\max(p,q)}(\mathbb{R}^n), \end{align*} with continuous embedding constants both equal to $1$ (the constants from the underlying mixed-norm Minkowski inequality and the pointwise $\ell$-norm nesting are exactly $1$ for non-negative integrands; the dyadic resolution is fixed and contributes no further multiplicative factor). The proof is complete. [/step]