The proof uses the $TT^*$ method: (1) is reduced to (2) by duality, and (2) is proved by expanding the $L^2$ norm as a double [integral](/page/Integral) and applying the [Hardy–Littlewood–Sobolev inequality](/theorems/469) in the time variable. The estimate (3) follows by composing (1) and (2). **Step 1 (Duality reduction: (1) $\Leftrightarrow$ (2)).** By duality of $L^q_t L^r_x$, \begin{align*} \|U(t)f\|_{L^q_t L^r_x} = \sup_{\|F\|_{L^{q'}_t L^{r'}_x} = 1} \left|\int_\mathbb{R} \langle U(t)f, F(t)\rangle_{L^2}\, d\mathcal{L}^1(t)\right|. \end{align*} By unitarity, $\langle U(t)f, F(t)\rangle_{L^2} = \langle f, U^*(t)F(t)\rangle_{L^2}$, so the right side equals \begin{align*} \sup_{\|F\| = 1} \left|\left\langle f,\, \int_\mathbb{R} U^*(t)F(t)\, d\mathcal{L}^1(t)\right\rangle_{L^2}\right| \le \|f\|_{L^2}\, \sup_{\|F\| = 1} \left\|\int_\mathbb{R} U^*(t)F(t)\, d\mathcal{L}^1(t)\right\|_{L^2}. \end{align*} Thus (1) with constant $C$ follows from (2) with the same constant, and conversely. **Step 2 (Proof of (2): $L^2$ expansion).** We expand the squared $L^2$ norm: \begin{align*} \left\|\int_\mathbb{R} U^*(t)F(t)\, d\mathcal{L}^1(t)\right\|_{L^2}^2 &= \left\langle \int_\mathbb{R} U^*(s)F(s)\, d\mathcal{L}^1(s),\, \int_\mathbb{R} U^*(t)F(t)\, d\mathcal{L}^1(t)\right\rangle_{L^2} \\ &= \int_\mathbb{R}\!\int_\mathbb{R} \langle U(t)U^*(s)F(s),\, F(t)\rangle_{L^2}\, d\mathcal{L}^1(s)\, d\mathcal{L}^1(t). \end{align*} [claim:Dispersive Kernel Bound] For $2 \le r \le \infty$ and $(q,r)$ admissible, the integrand satisfies $|\langle U(t)U^*(s)F(s), F(t)\rangle_{L^2}| \le C_0\, |t-s|^{-\alpha}\, G(s)\, G(t)$, where $\alpha = \sigma(1 - 2/r)$ and $G(t) = \|F(t)\|_{L^{r'}_x}$. [/claim] [proof] By Hölder's inequality in $x$: $|\langle U(t)U^*(s)F(s), F(t)\rangle| \le \|U(t)U^*(s)F(s)\|_{L^r_x}\, \|F(t)\|_{L^{r'}_x}$. By interpolation between the dispersive estimate ($L^1 \to L^\infty$ with decay $|t-s|^{-\sigma}$) and unitarity ($L^2 \to L^2$ with no decay): $\|U(t)U^*(s)F(s)\|_{L^r_x} \le C_0\, |t-s|^{-\sigma(1-2/r)}\, \|F(s)\|_{L^{r'}_x}$. [/proof] **Step 3 (Hardy–Littlewood–Sobolev in time).** The double integral is now bounded by \begin{align*} C_0 \int_\mathbb{R}\!\int_\mathbb{R} |t-s|^{-\alpha}\, G(s)\, G(t)\, d\mathcal{L}^1(s)\, d\mathcal{L}^1(t) = C_0 \int_\mathbb{R} (|\cdot|^{-\alpha} * G)(t)\, G(t)\, d\mathcal{L}^1(t). \end{align*} By Hölder in $t$: $\int (|\cdot|^{-\alpha} * G)\, G\, d\mathcal{L}^1 \le \||\cdot|^{-\alpha} * G\|_{L^q_t}\, \|G\|_{L^{q'}_t}$. The [Hardy–Littlewood–Sobolev inequality](/theorems/469) (in one dimension) gives $\||\cdot|^{-\alpha} * G\|_{L^q_t} \le C\|G\|_{L^{q'}_t}$ provided $1 + 1/q = 1/q' + \alpha$, i.e. $\alpha = 1 - 2/q$. The admissibility condition $1/q + \sigma/r = \sigma/2$ is equivalent to $\sigma(1-2/r) = 1 - 2/q$, confirming $\alpha = 1 - 2/q$. Therefore \begin{align*} \left\|\int U^*(t)F(t)\, d\mathcal{L}^1(t)\right\|_{L^2}^2 \le C\, \|G\|_{L^{q'}_t}^2 = C\, \|F\|_{L^{q'}_t L^{r'}_x}^2. \end{align*} **Step 4 (Inhomogeneous estimate (3)).** Combining (1) applied to $g = \int U^*(s)F(s)\, d\mathcal{L}^1(s)$ with (2): \begin{align*} \left\|U(t)\int U^*(s)F(s)\, d\mathcal{L}^1(s)\right\|_{L^{q_1}_t L^{r_1}_x} &\le C\left\|\int U^*(s)F(s)\, d\mathcal{L}^1(s)\right\|_{L^2} \le C\|F\|_{L^{q'_2}_t L^{r'_2}_x}. \end{align*} The first inequality uses (1) with the admissible pair $(q_1, r_1)$, and the second uses (2) with $(q_2, r_2)$. Since $(q_1, r_1)$ and $(q_2, r_2)$ are independent, the inhomogeneous estimate mixes two different admissible pairs.