[proofplan] We compare the Lorentz quasinorms by exploiting the monotonicity of the decreasing rearrangement $f^*$. The key inequality is that for any non-increasing $g: (0, \infty) \to [0, \infty)$, the weighted $\ell^q$ norms with respect to the multiplicative measure $d\mathcal{L}^1(t)/t$ are decreasing in $q$. Applied to $g(t) = t^{1/p} f^*(t)$, this delivers the embedding $L^{p, q_1} \hookrightarrow L^{p, q_2}$ with explicit constant $C = (q_1/p)^{1/q_1}$ when $q_2 = \infty$, and a constant depending on $p, q_1, q_2$ in general. The case $q_2 = \infty$ requires a separate Hardy-type argument that does not follow from the abstract $\ell^q$ comparison. [/proofplan] [step:Recall the definitions of the Lorentz quasinorms] For a measurable function $f: (X, \mathcal{A}, \mu) \to \mathbb{C}$ on a $\sigma$-finite measure space, define the **distribution function** \begin{align*} d_f: [0, \infty) &\to [0, \infty] \\ \lambda &\mapsto \mu(\{x \in X : |f(x)| > \lambda\}) \end{align*} and the **decreasing rearrangement** \begin{align*} f^*: [0, \infty) &\to [0, \infty] \\ t &\mapsto \inf\{\lambda \ge 0 : d_f(\lambda) \le t\}. \end{align*} The map $f^*$ is non-increasing and right-continuous. The **Lorentz quasinorm** is \begin{align*} \|f\|_{L^{p,q}} = \begin{cases} \left( \int_0^\infty \left( t^{1/p} f^*(t) \right)^q \, \frac{d\mathcal{L}^1(t)}{t} \right)^{1/q}, & 1 \le q < \infty, \\ \sup_{t > 0} t^{1/p} f^*(t), & q = \infty. \end{cases} \end{align*} [/step] [step:Establish the pointwise bound $f^*(t) \le t^{-1/p} \|f\|_{L^{p,q_1}} (q_1/p)^{1/q_1}$ for $q_1 < \infty$] Fix $t > 0$ and $1 \le q_1 < \infty$. Since $f^*$ is non-increasing, for every $0 < s \le t$ we have $f^*(s) \ge f^*(t)$. Therefore \begin{align*} \|f\|_{L^{p,q_1}}^{q_1} &= \int_0^\infty \left( s^{1/p} f^*(s) \right)^{q_1} \frac{d\mathcal{L}^1(s)}{s} \\ &\ge \int_0^t \left( s^{1/p} f^*(s) \right)^{q_1} \frac{d\mathcal{L}^1(s)}{s} \\ &\ge f^*(t)^{q_1} \int_0^t s^{q_1/p - 1} \, d\mathcal{L}^1(s) \\ &= f^*(t)^{q_1} \cdot \frac{p}{q_1} \cdot t^{q_1/p}. \end{align*} The first inequality restricts the domain of integration from $(0, \infty)$ to $(0, t]$, which is valid because the integrand is non-negative. The second inequality replaces $f^*(s)$ by the smaller quantity $f^*(t)$ on the smaller domain. The final equality computes the elementary integral $\int_0^t s^{q_1/p - 1} \, d\mathcal{L}^1(s) = (p/q_1) \, t^{q_1/p}$, which is finite because $q_1/p > 0$. Rearranging: \begin{align*} t^{1/p} f^*(t) \le \left( \frac{q_1}{p} \right)^{1/q_1} \|f\|_{L^{p, q_1}}. \end{align*} [/step] [step:Conclude the embedding $L^{p, q_1} \hookrightarrow L^{p, \infty}$] Taking the supremum over $t > 0$ in the pointwise bound from the previous step: \begin{align*} \|f\|_{L^{p, \infty}} = \sup_{t > 0} t^{1/p} f^*(t) \le \left( \frac{q_1}{p} \right)^{1/q_1} \|f\|_{L^{p, q_1}}. \end{align*} This proves the embedding $L^{p, q_1} \hookrightarrow L^{p, \infty}$ for any $1 \le q_1 < \infty$, with embedding constant $C = (q_1/p)^{1/q_1}$. [/step] [step:Reduce the case $q_1 < q_2 < \infty$ to a weighted $\ell^q$ comparison] Now suppose $1 \le q_1 < q_2 < \infty$. Define $g: (0, \infty) \to [0, \infty)$ by $g(t) := t^{1/p} f^*(t)$. We aim to show \begin{align*} \|g\|_{L^{q_2}((0, \infty), d\mathcal{L}^1/t)} \le C(p, q_1, q_2) \, \|g\|_{L^{q_1}((0, \infty), d\mathcal{L}^1/t)}, \end{align*} where the constant $C(p, q_1, q_2)$ will be computed below. The function $g$ is **not** non-increasing in general (the factor $t^{1/p}$ is increasing while $f^*$ is non-increasing). However, for any $t > 0$, the previous step gives the pointwise bound \begin{align*} g(t) = t^{1/p} f^*(t) \le \left( \frac{q_1}{p} \right)^{1/q_1} \|f\|_{L^{p, q_1}}. \end{align*} This is a uniform bound on $g$ in terms of the $L^{p, q_1}$-norm of $f$. [/step] [step:Apply the elementary $\ell^{q_2} \le \ell^{q_1}$ inequality with the uniform bound on $g$] We compute $\|f\|_{L^{p, q_2}}^{q_2}$ by inserting the pointwise bound on $g$: \begin{align*} \|f\|_{L^{p, q_2}}^{q_2} &= \int_0^\infty g(t)^{q_2} \, \frac{d\mathcal{L}^1(t)}{t} \\ &= \int_0^\infty g(t)^{q_2 - q_1} \cdot g(t)^{q_1} \, \frac{d\mathcal{L}^1(t)}{t} \\ &\le \left[ \left( \frac{q_1}{p} \right)^{1/q_1} \|f\|_{L^{p, q_1}} \right]^{q_2 - q_1} \int_0^\infty g(t)^{q_1} \, \frac{d\mathcal{L}^1(t)}{t} \\ &= \left( \frac{q_1}{p} \right)^{(q_2 - q_1)/q_1} \|f\|_{L^{p, q_1}}^{q_2 - q_1} \cdot \|f\|_{L^{p, q_1}}^{q_1} \\ &= \left( \frac{q_1}{p} \right)^{(q_2 - q_1)/q_1} \|f\|_{L^{p, q_1}}^{q_2}. \end{align*} The decomposition $g^{q_2} = g^{q_2 - q_1} \cdot g^{q_1}$ is valid because $q_2 > q_1$, so $q_2 - q_1 > 0$ and $g \ge 0$. The inequality bounds $g^{q_2 - q_1}(t)$ by the uniform supremum bound from the previous step, raised to the positive power $q_2 - q_1$. The resulting integral $\int_0^\infty g(t)^{q_1} \, d\mathcal{L}^1(t)/t = \|f\|_{L^{p, q_1}}^{q_1}$ by definition. Taking $q_2$-th roots: \begin{align*} \|f\|_{L^{p, q_2}} \le \left( \frac{q_1}{p} \right)^{(q_2 - q_1)/(q_1 q_2)} \|f\|_{L^{p, q_1}}. \end{align*} [/step] [step:Verify the embedding for the special case $q_1 = q_2$ and assemble the conclusion] When $q_1 = q_2$, the embedding $L^{p, q_1} \hookrightarrow L^{p, q_2}$ is the identity map on the same space, with embedding constant $1$. (Setting $q_2 = q_1$ in the formula gives constant $(q_1/p)^0 = 1$, consistent with the identity embedding.) Combining the cases $q_2 = \infty$, $q_2 < \infty$ with $q_1 < q_2$, and $q_1 = q_2$, we obtain \begin{align*} \|f\|_{L^{p, q_2}} \le C(p, q_1, q_2) \|f\|_{L^{p, q_1}} \end{align*} for all $1 \le p < \infty$ and $1 \le q_1 \le q_2 \le \infty$. The constant \begin{align*} C(p, q_1, q_2) = \begin{cases} 1, & q_1 = q_2, \\ \left( \frac{q_1}{p} \right)^{(q_2 - q_1)/(q_1 q_2)}, & q_1 < q_2 < \infty, \\ \left( \frac{q_1}{p} \right)^{1/q_1}, & q_2 = \infty, \end{cases} \end{align*} depends only on $p$, $q_1$, $q_2$. The embedding map $L^{p, q_1} \to L^{p, q_2}$ is the identity on representatives, which is well-defined because elements of both spaces are equivalence classes of measurable functions modulo a.e. equality on the same measure space. The bound $\|f\|_{L^{p, q_2}} \le C \|f\|_{L^{p, q_1}}$ shows the identity map is bounded, hence continuous, completing the embedding. The special case $L^p = L^{p, p} \hookrightarrow L^{p, \infty}$ follows by taking $q_1 = q_2 = p$ in the first conclusion (giving $L^p = L^{p, p}$ with equal norms up to a constant, which is a standard consequence of the layer-cake representation) and then applying the embedding $L^{p, p} \hookrightarrow L^{p, \infty}$ with $q_1 = p$, yielding constant $C = 1$. [/step]