[proofplan]
The argument is the $TT^*$ method applied across two distinct admissible pairs. We define the Strichartz operator $T_{(q,r)}: L^2_x \to L^q_t L^r_x$ by $T_{(q,r)} f(t) := e^{it\Delta} f$, whose boundedness is the [homogeneous Strichartz Estimate](/theorems/???). Its adjoint $T_{(q,r)}^*: L^{q'}_t L^{r'}_x \to L^2_x$ is given by $T_{(q,r)}^* G = \int_\mathbb{R} e^{-is\Delta} G(s) \, d\mathcal{L}^1(s)$. The Duhamel operator $\Phi$ factors as $\Phi = T_{(q,r)} \circ T_{(\tilde q, \tilde r)}^*$, so its mixed-norm bound is the product of the operator norms of $T_{(q,r)}$ and $T_{(\tilde q, \tilde r)}^*$, both finite by the homogeneous estimate and its dual. The proof unfolds in four steps: (a) record the homogeneous estimate and its dual, (b) factor $\Phi$ as the composition $T_{(q,r)} T_{(\tilde q, \tilde r)}^*$, (c) compose the bounds, (d) extract the constant.
[/proofplan]
[step:Record the homogeneous Strichartz estimate and its dual for both admissible pairs]
For each admissible pair $(p, p_*) \in \{(q, r), (\tilde q, \tilde r)\}$, define the linear map
\begin{align*}
T_{(p, p_*)}: L^2(\mathbb{R}^n) &\to L^p_t L^{p_*}_x(\mathbb{R} \times \mathbb{R}^n) \\
f &\mapsto \big[(t, x) \mapsto (e^{it\Delta} f)(x)\big].
\end{align*}
By the [Homogeneous Strichartz Estimate](/theorems/???) applied to the admissible pair $(p, p_*)$ (the hypotheses $2 \le p, p_* \le \infty$, $2/p + n/p_* = n/2$, and the non-endpoint condition all hold by assumption), there exists $C_{p, p_*} > 0$ depending only on $n$ and $(p, p_*)$ such that for every $f \in L^2(\mathbb{R}^n)$,
\begin{align*}
\|T_{(p, p_*)} f\|_{L^p_t L^{p_*}_x} \le C_{p, p_*} \|f\|_{L^2_x}.
\end{align*}
By duality of bounded linear operators on Banach spaces, the adjoint map
\begin{align*}
T_{(p, p_*)}^*: L^{p'}_t L^{p_*'}_x &\to L^2(\mathbb{R}^n)
\end{align*}
has the same operator norm $\|T_{(p, p_*)}^*\| = \|T_{(p, p_*)}\| \le C_{p, p_*}$. To identify $T_{(p, p_*)}^*$ explicitly, observe that for $f \in L^2_x$ and $G \in L^{p'}_t L^{p_*'}_x$,
\begin{align*}
\langle T_{(p, p_*)} f, G \rangle_{L^p_t L^{p_*}_x \times L^{p'}_t L^{p_*'}_x}
= \int_\mathbb{R} \int_{\mathbb{R}^n} (e^{it\Delta} f)(x) \, \overline{G(t, x)} \, d\mathcal{L}^n(x) \, d\mathcal{L}^1(t).
\end{align*}
Since $e^{it\Delta}$ is unitary on $L^2_x$ (Theorem 3205), its $L^2$-adjoint is $e^{-it\Delta}$. Applying [Fubini's Theorem](/theorems/???) on the absolutely-integrable integrand (justified for $f \in \mathcal{S}_x$ and $G \in C_c(\mathbb{R}_t; \mathcal{S}_x)$, then extended by density),
\begin{align*}
\langle T_{(p, p_*)} f, G \rangle = \int_{\mathbb{R}^n} f(x) \, \overline{ \int_\mathbb{R} (e^{-it\Delta} G(t, \cdot))(x) \, d\mathcal{L}^1(t) } \, d\mathcal{L}^n(x).
\end{align*}
Therefore
\begin{align*}
T_{(p, p_*)}^* G = \int_\mathbb{R} e^{-is\Delta} G(s, \cdot) \, d\mathcal{L}^1(s) \in L^2(\mathbb{R}^n),
\end{align*}
with $\|T_{(p, p_*)}^* G\|_{L^2_x} \le C_{p, p_*} \|G\|_{L^{p'}_t L^{p_*'}_x}$.
[/step]
[step:Factor the Duhamel operator as $T_{(q,r)} \circ T_{(\tilde q, \tilde r)}^*$]
We claim that for $F \in L^{\tilde q'}_t L^{\tilde r'}_x$,
\begin{align*}
\Phi[F](t, x) = \big( T_{(q, r)} \big[ T_{(\tilde q, \tilde r)}^* F \big] \big)(t, x), \qquad (t, x) \in \mathbb{R} \times \mathbb{R}^n.
\end{align*}
By the previous step, $T_{(\tilde q, \tilde r)}^* F = \int_\mathbb{R} e^{-is\Delta} F(s, \cdot) \, d\mathcal{L}^1(s) \in L^2(\mathbb{R}^n)$. Then
\begin{align*}
T_{(q,r)} \big[ T_{(\tilde q, \tilde r)}^* F \big](t, x) = \big( e^{it\Delta} T_{(\tilde q, \tilde r)}^* F \big)(x) = e^{it\Delta} \int_\mathbb{R} e^{-is\Delta} F(s, \cdot) \, d\mathcal{L}^1(s) \,(x).
\end{align*}
The propagator $e^{it\Delta}$ is a bounded linear operator on $L^2(\mathbb{R}^n)$ — in fact unitary by [Theorem 3205](/theorems/3205) — so it commutes with the Bochner integral $\int_\mathbb{R} \cdots d\mathcal{L}^1(s)$ (this is the standard fact that bounded linear operators commute with Bochner integration of strongly measurable, integrable functions; the Bochner integrability of $s \mapsto e^{-is\Delta} F(s, \cdot) \in L^2_x$ follows from the assumption $F \in L^{\tilde q'}_t L^{\tilde r'}_x$ together with the homogeneous estimate, which gives an $L^{\tilde q'}_t$-bound on $\|e^{-is\Delta} F(s, \cdot)\|_{L^2_x}$ when $\tilde q' < \infty$; for $\tilde q' = \infty$ a separate argument via the action on $\mathcal{S}$ followed by density applies, but in our setting $\tilde q' < \infty$ since the case $(\tilde q, \tilde r) = (\infty, 2)$ corresponds to $\tilde q' = 1$ and is admissible). Pulling $e^{it\Delta}$ inside,
\begin{align*}
T_{(q,r)} \big[ T_{(\tilde q, \tilde r)}^* F \big](t, x) = \int_\mathbb{R} e^{it\Delta} e^{-is\Delta} F(s, \cdot) \, d\mathcal{L}^1(s) \,(x) = \int_\mathbb{R} e^{i(t-s)\Delta} F(s, \cdot) \, d\mathcal{L}^1(s) \,(x) = \Phi[F](t, x),
\end{align*}
where we used the group property $e^{it\Delta} e^{-is\Delta} = e^{i(t-s)\Delta}$ (immediate from the multiplier representation, since the symbols multiply: $e^{-it|\xi|^2} \cdot e^{is|\xi|^2} = e^{-i(t-s)|\xi|^2}$).
[/step]
[step:Compose the homogeneous bounds to derive the inhomogeneous estimate]
By the factorisation $\Phi = T_{(q, r)} \circ T_{(\tilde q, \tilde r)}^*$ and submultiplicativity of operator norms,
\begin{align*}
\|\Phi\|_{L^{\tilde q'}_t L^{\tilde r'}_x \to L^q_t L^r_x} \le \|T_{(q, r)}\|_{L^2_x \to L^q_t L^r_x} \cdot \|T_{(\tilde q, \tilde r)}^*\|_{L^{\tilde q'}_t L^{\tilde r'}_x \to L^2_x} \le C_{q, r} \cdot C_{\tilde q, \tilde r}.
\end{align*}
Therefore for every $F \in L^{\tilde q'}_t L^{\tilde r'}_x$,
\begin{align*}
\|\Phi[F]\|_{L^q_t L^r_x} \le C_{q, r} \cdot C_{\tilde q, \tilde r} \cdot \|F\|_{L^{\tilde q'}_t L^{\tilde r'}_x}.
\end{align*}
Setting $C := C_{q, r} \cdot C_{\tilde q, \tilde r} = C(n, q, r, \tilde q, \tilde r) > 0$, this is the claimed estimate.
[/step]
