[proofplan] The two estimates rest on two complementary representations of the propagator. On the Fourier side, $e^{it\Delta}$ is the multiplier $e^{-it|\xi|^2}$, and since $|e^{-it|\xi|^2}| = 1$ pointwise, [Plancherel's Theorem](/theorems/???) immediately yields $L^2$ conservation. On the physical side, $e^{it\Delta}$ acts as convolution with the Schrödinger kernel $K_t(x) = (4\pi i t)^{-n/2} e^{i|x|^2/(4t)}$, whose pointwise modulus is $(4\pi |t|)^{-n/2}$; combining this with [Young's convolution inequality](/theorems/???) gives the dispersive bound. Throughout we use the symmetric normalisation $\hat{f}(\xi) = (2\pi)^{-n/2}\int f(x)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$. [/proofplan]

[step:Establish $L^2$ conservation via the Fourier multiplier representation] By definition of the propagator, for $f \in \mathcal{S}(\mathbb{R}^n)$ the Fourier transform satisfies \begin{align*} \widehat{e^{it\Delta} f}(\xi) = e^{-it|\xi|^2} \hat{f}(\xi), \qquad \xi \in \mathbb{R}^n. \end{align*} The function $\xi \mapsto e^{-it|\xi|^2}$ is measurable and satisfies $|e^{-it|\xi|^2}| = 1$ for every $\xi \in \mathbb{R}^n$. Therefore \begin{align*} \|\widehat{e^{it\Delta} f}\|_{L^2(\mathbb{R}^n)}^2 = \int_{\mathbb{R}^n} |e^{-it|\xi|^2}|^2 \, |\hat{f}(\xi)|^2 \, d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n} |\hat{f}(\xi)|^2 \, d\mathcal{L}^n(\xi) = \|\hat{f}\|_{L^2(\mathbb{R}^n)}^2. \end{align*} By [Plancherel's Theorem](/theorems/???), the symmetric normalisation gives $\|\hat{g}\|_{L^2} = \|g\|_{L^2}$ for all $g \in L^2(\mathbb{R}^n)$. Applying this to both $e^{it\Delta}f$ and $f$: \begin{align*} \|e^{it\Delta} f\|_{L^2(\mathbb{R}^n)} = \|\widehat{e^{it\Delta} f}\|_{L^2(\mathbb{R}^n)} = \|\hat{f}\|_{L^2(\mathbb{R}^n)} = \|f\|_{L^2(\mathbb{R}^n)}. \end{align*} For general $f \in L^2(\mathbb{R}^n)$, density of $\mathcal{S}(\mathbb{R}^n)$ in $L^2(\mathbb{R}^n)$ together with the bounded extension of the multiplier operator yields the same identity. [/step]

[step:Compute the Schrödinger kernel by a contour-shifted Gaussian integral] We claim that for $t \neq 0$ and for $f \in \mathcal{S}(\mathbb{R}^n)$, \begin{align*} (e^{it\Delta} f)(x) = \int_{\mathbb{R}^n} K_t(x - y) f(y) \, d\mathcal{L}^n(y), \qquad K_t(x) := (4\pi i t)^{-n/2} e^{i|x|^2/(4t)}, \end{align*} where $(4\pi i t)^{-n/2}$ is interpreted via the principal branch of $z^{-n/2}$ on $\mathbb{C} \setminus (-\infty, 0]$ with $\arg(it) = \frac{\pi}{2}\,\mathrm{sgn}(t)$, so $|(4\pi i t)^{-n/2}| = (4\pi |t|)^{-n/2}$. By the Fourier inversion formula and the definition of $e^{it\Delta}$, substituting $\hat{f}(\xi) = (2\pi)^{-n/2}\int f(y)\,e^{-iy\cdot\xi}\,d\mathcal{L}^n(y)$: \begin{align*} (e^{it\Delta} f)(x) &= (2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{i x \cdot \xi} e^{-it|\xi|^2} \hat{f}(\xi) \, d\mathcal{L}^n(\xi) \\ &= (2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{i x \cdot \xi} e^{-it|\xi|^2} \left[(2\pi)^{-n/2}\int_{\mathbb{R}^n} f(y)\,e^{-iy\cdot\xi}\,d\mathcal{L}^n(y)\right] d\mathcal{L}^n(\xi) \\ &= (2\pi)^{-n} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} e^{i(x - y) \cdot \xi} e^{-it|\xi|^2} f(y) \, d\mathcal{L}^n(y) \, d\mathcal{L}^n(\xi). \end{align*} We approximate $e^{-it|\xi|^2}$ by $e^{-(\varepsilon + it)|\xi|^2}$ with $\varepsilon > 0$, apply [Fubini's Theorem](/theorems/???) on the absolutely-integrable integrand $|f(y)| e^{-\varepsilon|\xi|^2}$, and let $\varepsilon \downarrow 0$ at the end. After Fubini: \begin{align*} (e^{it\Delta} f)(x) = \int_{\mathbb{R}^n} f(y) \left[ (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x-y) \cdot \xi - (\varepsilon + it)|\xi|^2} \, d\mathcal{L}^n(\xi) \right] d\mathcal{L}^n(y). \end{align*} Setting $z := \varepsilon + it$ (so $\operatorname{Re}(z) = \varepsilon > 0$), completing the square gives \begin{align*} i(x - y) \cdot \xi - z|\xi|^2 = -z \left| \xi - \frac{i(x-y)}{2z} \right|^2 - \frac{|x-y|^2}{4z}. \end{align*} By the standard Gaussian integral $\int_{\mathbb{R}^n} e^{-z|\eta|^2}\, d\mathcal{L}^n(\eta) = (\pi/z)^{n/2}$ (valid for $\operatorname{Re}(z) > 0$ with the principal branch) and the contour shift justified by Cauchy's theorem, \begin{align*} (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i(x-y)\cdot\xi - z|\xi|^2} \, d\mathcal{L}^n(\xi) = (2\pi)^{-n} \cdot \left( \frac{\pi}{z} \right)^{n/2} e^{-|x-y|^2/(4z)} = (4\pi z)^{-n/2} e^{-|x-y|^2/(4z)}. \end{align*} Sending $\varepsilon \downarrow 0$, so $z \to it$, gives the kernel $K_t(x-y) = (4\pi i t)^{-n/2} e^{i|x-y|^2/(4t)}$. To pass the limit through the outer $y$-integral, the bracketed quantity is bounded uniformly in $\varepsilon \in (0,1]$ by a constant times $(4\pi |t|/2)^{-n/2}$ for small $\varepsilon$; the product with $|f(y)|$ is dominated by an integrable function (since $f \in L^1$). [Dominated Convergence](/theorems/???) yields \begin{align*} (e^{it\Delta} f)(x) = \int_{\mathbb{R}^n} (4\pi i t)^{-n/2} e^{i|x-y|^2/(4t)} f(y) \, d\mathcal{L}^n(y), \end{align*} proving the kernel formula. [/step]

[step:Bound the $L^\infty$ norm by $\|K_t\|_\infty \|f\|_1$ via Young's convolution inequality] We apply the kernel formula from the previous step. For $f \in L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, density of $\mathcal{S}(\mathbb{R}^n)$ in $L^1(\mathbb{R}^n)$ extends the integral representation to \begin{align*} (e^{it\Delta} f)(x) = \int_{\mathbb{R}^n} K_t(x - y) f(y) \, d\mathcal{L}^n(y), \end{align*} valid almost everywhere in $x$. Taking absolute values and using $|K_t(x - y)| = (4\pi |t|)^{-n/2}$ uniformly in $x, y$: \begin{align*} |(e^{it\Delta} f)(x)| \le \int_{\mathbb{R}^n} |K_t(x - y)| \, |f(y)| \, d\mathcal{L}^n(y) = (4\pi |t|)^{-n/2} \|f\|_{L^1(\mathbb{R}^n)}. \end{align*} Taking essential suprema over $x \in \mathbb{R}^n$ gives \begin{align*} \|e^{it\Delta} f\|_{L^\infty(\mathbb{R}^n)} \le (4\pi |t|)^{-n/2} \|f\|_{L^1(\mathbb{R}^n)}, \end{align*} which is the dispersive estimate. This bound, combined with the $L^2$ conservation of the first step, completes the proof. [/step]