= Solution
Use the unitary normalization of the <Fourier transform>,
$$
\mathcal Ff(\xi)=(2\pi)^{-n/2}\int_{\mathbb R^n}e^{-ix\cdot\xi}f(x)\,dx,\qquad
\mathcal Gg(x)=(2\pi)^{-n/2}\int_{\mathbb R^n}e^{ix\cdot\xi}g(\xi)\,d\xi.
$$
The <Schwartz space> $\mathcal S(\mathbb R^n)$ consists of <smooth function>s for which every <Schwartz seminorm> $\sup_x|x^\alpha\partial^\beta f(x)|$ is finite. Such functions and all their <derivatives> are integrable. Differentiating under the integral and integrating by parts give
$$
\partial_\xi^\alpha\mathcal Ff=(-i)^{|\alpha|}\mathcal F(x^\alpha f),\qquad
\xi^\beta\mathcal Ff=(-i)^{|\beta|}\mathcal F(\partial^\beta f).
$$
Consequently
$$
\sup_\xi|\xi^\beta\partial_\xi^\alpha\mathcal Ff(\xi)|
\leq(2\pi)^{-n/2}\|\partial^\beta(x^\alpha f)\|_1.
$$
Expand the <derivative> by the product rule and insert the integrable weight $(1+|x|)^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal F:\mathcal S\to\mathcal S$. The identical argument with the opposite sign proves continuity of $\mathcal G$.
For <surjectivity> and <injectivity> an inversion proof is needed. If $f\in\mathcal S$, multiply $\mathcal Ff(\xi)$ by $e^{-\varepsilon|\xi|^2/2}$. <Fubini's theorem> and the elementary <Gaussian integral> give
$$
\mathcal G\bigl(e^{-\varepsilon|\xi|^2/2}\mathcal Ff\bigr)(x)
=\int_{\mathbb R^n}f(y)\rho_\varepsilon(x-y)\,dy,
\qquad
\rho_\varepsilon(x)=(2\pi\varepsilon)^{-n/2}e^{-|x|^2/(2\varepsilon)}.
$$
The Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal Ff\in\mathcal S\subset L^1$, so dominated convergence gives $\mathcal G\mathcal Ff(x)$. This proves $\mathcal G\mathcal F=I$. Reversing the signs in the same calculation proves $\mathcal F\mathcal G=I$. Thus
$$
\boxed{\mathcal F:\mathcal S(\mathbb R^n)\longrightarrow\mathcal S(\mathbb R^n)\text{ is a continuous bijection, with inverse }\mathcal G.}
$$
Its inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming <Fourier inversion> as an unexplained theorem.
Inversion gives $\mathcal F^2f(x)=f(-x)$ and the <Parseval identity>. Indeed, inserting the integral for $\mathcal Ff$ into $\int\mathcal Ff\,\overline{\mathcal Fg}$ and using $f,\mathcal Fg\in L^1$ gives $\int f\overline g$. Density of <Schwartz functions> in $L^2$ extends $\mathcal F$ to a <unitary operator> on $L^2$, the <Plancherel theorem>. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, and
$$
\mathcal F(f*g)=(2\pi)^{n/2}(\mathcal Ff)(\mathcal Fg),\qquad
\mathcal F(\partial_jf)=i\xi_j\mathcal Ff.
$$
These formulas explain why the <Fourier transform> diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the <tempered distributions>: with the complex-linear distributional pairing, $\langle\mathcal FT,\phi\rangle=\langle T,\mathcal F\phi\rangle$, since the Fourier <kernel> is symmetric in its variables.
The connection with <creation and annihilation operators> is especially transparent for the <harmonic oscillator>. On the common invariant dense domain $\mathcal S$, set
$$
a_j=\frac{x_j+\partial_j}{\sqrt2},\qquad
a_j^\dagger=\frac{x_j-\partial_j}{\sqrt2}.
$$
<Integration by parts> shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; <commutators> below are computed on $\mathcal S$, avoiding undefined products on arbitrary $L^2$ <vectors>. The identity $[\partial_j,x_k]=\delta_{jk}I$ gives the <canonical commutation relations>
$$
[a_j,a_k^\dagger]=\delta_{jk}I,\qquad[a_j,a_k]=[a_j^\dagger,a_k^\dagger]=0.
$$
It also gives
$$
H=-\Delta+|x|^2=2\sum_j a_j^\dagger a_j+nI.
$$
The normalized vacuum $h_0(x)=\pi^{-n/4}e^{-|x|^2/2}$ satisfies $a_jh_0=0$. Define the <Hermite functions>
$$
h_\alpha=\frac{(a_1^\dagger)^{\alpha_1}\cdots(a_n^\dagger)^{\alpha_n}}{\sqrt{\alpha_1!\cdots\alpha_n!}}h_0.
$$
Moving <annihilation operators> past <creation operators> with the commutation relations proves
$$
a_jh_\alpha=\sqrt{\alpha_j}\,h_{\alpha-e_j},\qquad
a_j^\dagger h_\alpha=\sqrt{\alpha_j+1}\,h_{\alpha+e_j},\qquad
Hh_\alpha=(2|\alpha|+n)h_\alpha.
$$
These same relations show that the <Hermite functions> are <orthonormal>: unequal occupations are orthogonal <eigenvectors> of one of the commuting symmetric <number operators> $a_j^\dagger a_j$, and the ladder relations recursively give <norm> one.
For completeness, let $g\in L^2$ be orthogonal to every $h_\alpha$. The Hermite <polynomials> have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all <polynomial> moments of $h(x)=g(x)e^{-|x|^2/2}\in L^1$ vanish. Its Fourier integral extends to an <entire function> of $\zeta\in\mathbb C^n$:
$$
F(\zeta)=\int g(x)e^{-|x|^2/2}e^{-ix\cdot\zeta}\,dx.
$$
The <Cauchy-Schwarz inequality>, with the Gaussian absorbing every exponential linear weight and every <polynomial>, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the <entire function> is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero <Fourier transform>, convolving it with $\rho_\varepsilon$ gives zero by the <Gaussian integral> and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The <Hermite functions> form an <orthonormal basis> of $L^2$.
Finally, the elementary <Gaussian integral> gives $\mathcal Fh_0=h_0$. The <derivative> and multiplication identities above imply
$$
\mathcal F a_j=i a_j\mathcal F,\qquad
\mathcal F a_j^\dagger=-i a_j^\dagger\mathcal F,
$$
where the operators on the right act on the frequency variable. Hence
$$
\boxed{\mathcal Fh_\alpha=(-i)^{|\alpha|}h_\alpha.}
$$
Thus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier <eigenvalues>. On <Schwartz functions> the oscillator powers control weighted <derivatives>, and Hermite coefficients decay faster than every power of $1+|\alpha|$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of <Schwartz space>.
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