The proof combines Duhamel's formula with the [abstract Strichartz estimates](/theorems/639) applied to $U(t) = e^{it\Delta}$, which satisfies the unitarity and dispersive hypotheses with $\sigma = n/2$. **Step 1 (Duhamel decomposition).** The solution of $i\partial_t u + \Delta u = F$ with $u(0) = u_{\mathrm{in}}$ is given by Duhamel's formula: \begin{align*} u(t) = e^{it\Delta} u_{\mathrm{in}} + \int_0^t e^{i(t-s)\Delta} F(s)\, d\mathcal{L}^1(s). \end{align*} By the triangle inequality in $L^q_t L^r_x$: \begin{align*} \|u\|_{L^q_t L^r_x} \le \|e^{it\Delta} u_{\mathrm{in}}\|_{L^q_t L^r_x} + \left\|\int_0^t e^{i(t-s)\Delta} F(s)\, d\mathcal{L}^1(s)\right\|_{L^q_t L^r_x}. \end{align*} **Step 2 (Homogeneous term).** The first term is bounded by $C\|u_{\mathrm{in}}\|_{L^2}$ using part (1) of the [abstract Strichartz estimates](/theorems/639) with the $S$-admissible pair $(q, r)$ (the admissibility condition $2/q + n/r = n/2$ corresponds to $\sigma = n/2$). **Step 3 (Inhomogeneous term).** For the second term, we write $\int_0^t e^{i(t-s)\Delta} F(s)\, d\mathcal{L}^1(s) = \int_\mathbb{R} e^{i(t-s)\Delta} [\chi_{[0,t]}(s) F(s)]\, d\mathcal{L}^1(s)$. The restricted-time version of the inhomogeneous estimate follows from part (3) of the [abstract Strichartz estimates](/theorems/639): replacing $F$ by $\chi_{[0,t]} F$ does not increase the $L^{a'}_t L^{b'}_x$ norm, so \begin{align*} \left\|\int_0^t e^{i(t-s)\Delta} F(s)\, d\mathcal{L}^1(s)\right\|_{L^q_t L^r_x} \le C\, \|F\|_{L^{a'}_t L^{b'}_x}, \end{align*} where $(a, b)$ is any $S$-admissible pair. (A more careful argument using the Christ–Kiselev lemma or direct interpolation between endpoint cases confirms this for the retarded [integral](/page/Integral).) **Step 4 (Conclusion).** Combining Steps 2 and 3: \begin{align*} \|u\|_{L^q_t L^r_x} \le C\|u_{\mathrm{in}}\|_{L^2} + C\|F\|_{L^{a'}_t L^{b'}_x}. \end{align*} The statement that $u$ belongs to every $S$-admissible Strichartz space when $F = 0$ follows from applying (1) of the abstract estimates with each admissible pair in turn — the constant $C$ depends on the pair, but the bound holds for all of them simultaneously.