[proofplan]
The case $p = \infty$ is immediate from the definition of $\|\cdot\|_{0,0}$. For $1 \leq p < \infty$, we establish a pointwise decay bound $|f(x)| \leq C_N (1 + |x|)^{-N} \max_{|\alpha| \leq N} \|f\|_{\alpha,0}$ by exploiting the Schwartz semi-norms to control $|f(x)|$ in terms of a power of $|x|$. Choosing $N > n/p$ ensures $(1 + |x|)^{-Np}$ is integrable over $\mathbb{R}^n$, and raising the pointwise bound to the $p$-th power and integrating yields the $L^p$ estimate.
[/proofplan]
[step:Handle the case $p = \infty$ directly from the definition]
By definition, $\|f\|_{L^\infty} = \operatorname{ess\,sup}_x |f(x)| \leq \sup_x |f(x)| = \|f\|_{0,0}$.
[/step]
[step:Establish a pointwise decay bound using finitely many Schwartz semi-norms]
Choose any integer $N > n/p$.
[claim:Pointwise Decay Bound]
For every integer $N \geq 1$, there exists a constant $C_N > 0$ depending only on $n$ and $N$ such that
\begin{align*}
|f(x)| &\leq C_N \, (1 + |x|)^{-N} \max_{|\alpha| \leq N} \|f\|_{\alpha, 0}
\end{align*}
for all $f \in \mathcal{S}(\mathbb{R}^n)$ and all $x \in \mathbb{R}^n$.
[/claim]
[proof]
Write $M = \max_{|\alpha| \leq N} \|f\|_{\alpha, 0}$. For $|x| \leq 1$, we have $(1 + |x|)^{-N} \geq 2^{-N}$, so
\begin{align*}
|f(x)| &\leq \|f\|_{0,0} \leq M = M \cdot 2^N \cdot 2^{-N} \leq 2^N M (1+|x|)^{-N}.
\end{align*}
For $|x| > 1$, at least one coordinate satisfies $|x_j| \geq |x| / \sqrt{n}$. Taking $\alpha = N e_j$ (the multi-index with $N$ in position $j$ and $0$ elsewhere), the semi-norm bound gives $|x_j^N f(x)| \leq \|f\|_{Ne_j, 0} \leq M$, so
\begin{align*}
|f(x)| &\leq \frac{M}{|x_j|^N} \leq \frac{M \, n^{N/2}}{|x|^N} \leq 2^N n^{N/2} \, M \, (1 + |x|)^{-N},
\end{align*}
where the last inequality uses $(1 + |x|)^{-N} \geq 2^{-N} |x|^{-N}$ for $|x| > 1$. Setting $C_N = 2^N n^{N/2}$ covers both cases.
[/proof]
[/step]
[step:Integrate the $p$-th power of the decay bound to obtain the $L^p$ estimate]
Applying the pointwise bound with the chosen $N > n/p$ and writing $M = \max_{|\alpha| \leq N} \|f\|_{\alpha, 0}$:
\begin{align*}
\|f\|_{L^p}^p &= \int_{\mathbb{R}^n} |f(x)|^p \, d\mathcal{L}^n(x) \leq C_N^p \, M^p \int_{\mathbb{R}^n} (1 + |x|)^{-Np} \, d\mathcal{L}^n(x).
\end{align*}
Passing to polar coordinates via the decomposition $d\mathcal{L}^n(y) = r^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(r)$ for $y = rw$ with $r > 0$ and $w \in \mathbb{S}^{n-1}$, the [integral](/page/Integral) becomes
\begin{align*}
\int_{\mathbb{R}^n} (1 + |x|)^{-Np} \, d\mathcal{L}^n(x) &= c_n \int_0^\infty (1+r)^{-Np} r^{n-1} \, d\mathcal{L}^1(r),
\end{align*}
where $c_n = \mathcal{H}^{n-1}(\mathbb{S}^{n-1})$ is the surface area of the unit sphere. The integrand decays like $r^{n-1-Np}$ as $r \to \infty$, and $n - 1 - Np < -1$ since $Np > n$, so the integral converges to a finite constant $I_{n,p}$. Therefore
\begin{align*}
\|f\|_{L^p} &\leq C_N \, I_{n,p}^{1/p} \max_{|\alpha| \leq N} \|f\|_{\alpha, 0},
\end{align*}
which is the claimed estimate with $C_{n,p} = C_N \, I_{n,p}^{1/p}$ and $N = \lfloor n/p \rfloor + 1$.
[guided]
We now integrate the pointwise decay bound from the previous step. With $N > n/p$ and $M = \max_{|\alpha| \leq N} \|f\|_{\alpha, 0}$, the bound $|f(x)| \leq C_N M (1 + |x|)^{-N}$ gives
\begin{align*}
\|f\|_{L^p}^p &= \int_{\mathbb{R}^n} |f(x)|^p \, d\mathcal{L}^n(x) \leq C_N^p \, M^p \int_{\mathbb{R}^n} (1 + |x|)^{-Np} \, d\mathcal{L}^n(x).
\end{align*}
The question reduces to whether the integral $\int_{\mathbb{R}^n} (1 + |x|)^{-Np} \, d\mathcal{L}^n(x)$ converges. To evaluate it, pass to polar coordinates: write $x = rw$ with $r = |x| > 0$ and $w \in \mathbb{S}^{n-1}$, so $d\mathcal{L}^n(x) = r^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(r)$ where $d\sigma$ is the surface measure on $\mathbb{S}^{n-1}$. Integrating over the angular variable gives the factor $c_n = \mathcal{H}^{n-1}(\mathbb{S}^{n-1})$:
\begin{align*}
\int_{\mathbb{R}^n} (1 + |x|)^{-Np} \, d\mathcal{L}^n(x) &= c_n \int_0^\infty (1+r)^{-Np} r^{n-1} \, d\mathcal{L}^1(r).
\end{align*}
For large $r$, the integrand behaves like $r^{n-1-Np}$. This is integrable near $\infty$ if and only if $n - 1 - Np < -1$, i.e., $Np > n$. Since we chose $N > n/p$, this condition holds, and the integral converges to a finite constant $I_{n,p}$. Taking $p$-th roots:
\begin{align*}
\|f\|_{L^p} &\leq C_N \, I_{n,p}^{1/p} \, M = C_{n,p} \max_{|\alpha| \leq N} \|f\|_{\alpha, 0}.
\end{align*}
The right-hand side is a finite maximum of Schwartz semi-norms times a constant, which is a [continuous](/page/Continuity) semi-norm on $\mathcal{S}(\mathbb{R}^n)$. This establishes that the inclusion $\iota: \mathcal{S}(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n)$ is continuous.
[/guided]
[/step]
