Last term we defined the (one-dimensional) Schwartz space as the following:
$$\mathscr{S}(\mathbb{R}) := \left\{ f \in C^{\infty}(\mathbb{R}) \mid \forall p,q \in \mathbb{N}_0 \,\exists c_{p,q} \in \mathbb{R} : \left| \, x^p f^{(q)}(x) \, \right| \leq c_{p,q} \right\}$$
where $f^{(q)}=\frac{d^q f}{dx^q}$. Afterwards we dealt with some common theorems considering the fourier transformation like:
$$\mathscr{F}\left[P\left(\frac{d}{dx}\right)f(x)\right](k) = P(ik) \cdot \mathscr{F}\left[ f \right] (k)$$
where
$$P\left(\frac{d}{dx}\right)=\sum_{k=0}^{n} a_k \cdot \frac{d^k}{dx^k}$$
, $i$ the imaginary unit and $P(ik)$ a polynomial in $ik$. I have a question concerning the following "proof" of the theorem (not the corollary, all from my notes):
$$\vdots$$
Corollary. Suppose $f \in \mathscr{S}(\mathbb{R}),$ then $\mathscr{F}[f](k) \in \mathscr{S}(\mathbb{R})$.
Proof. $P(k) \cdot R\left(\frac{d}{dk}\right) \mathscr{F}[f](k) = \mathscr{F} \left[ \underbrace{R(-ix) \cdot P \left(- i \frac{d}{dx}\right) f(x)}_{\in \mathscr{S}(\mathbb{R}), \, \text{because} \, f(x) \in \mathscr{S}(\mathbb{R})} \right] \implies \mathscr{F}[f]$ is defined and bounded $\implies \mathscr{F}[f] \in \mathscr{S}(\mathbb{R}) \hspace{3mm} \Box$
Theorem. $\mathscr{F}: \mathscr{S}(\mathbb{R}) \to \mathscr{S}(\mathbb{R})$ is an isomorphism.
Proof. The last corollary shows that $\mathscr{F}[f] \in \mathscr{S}(\mathbb{R})$, for $f \in \mathscr{S}(\mathbb{R})$. Isomorphism follows from $\mathscr{F}^2[f](x) \equiv \mathscr{F}\left[\mathscr{F}[f]\right](x)=f(-x). \hspace{3mm} \Box$
$$\vdots$$
I'm not sure if I understood it properly. I mean, the fourier transformation is linear (thus a homomorphism) and maps $\mathscr{S}(\mathbb{R}) \to \mathscr{S}(\mathbb{R})$. The bijectivity remains to be shown, right? I would say $\mathscr{F}^2[f](x) = f(-x) \implies \mathscr{F}^4[f](x) = \text{id}_{\mathscr{S}(\mathbb{R})} \implies \mathscr{F}^3[f] (x) = \mathscr{F}^{-1}[f](x)$. Thus we have an inverse Operator with
$$\mathscr{F}^{-1}[\mathscr{F}[f]](x)=\mathscr{F}^3[\mathscr{F}[f]](x)=\mathscr{F}^4[f](x)=\mathscr{F}^2[f(-x)](x)=f(x)=\text{id}_{\mathscr{S}(\mathbb{R})}$$
$$\mathscr{F}[\mathscr{F}^{-1}[f]](x)=\mathscr{F}[\mathscr{F}^3[f]](x)=\mathscr{F}^4[f](x)=\mathscr{F}^2[f(-x)](x)=f(x)=\text{id}_{\mathscr{S}(\mathbb{R})}$$
hence $\mathscr{F}$ and $\mathscr{F}^{-1}$ are bijective and inverse, which qualifies $\mathscr{F}$ for an automorphism on $\mathscr{S}(\mathbb{R})$. $\hspace{3mm} \Box$
Would that be sufficient?
Schwartz space is topological vector space so what is really usually meant by "isomorphism" is a bijection that not only preserves algebraic, but also topological structure. That means that you also need to know that your mapping is homeomorphism. In this case, three-fold composition of Fourier transform ($\mathcal F$) is its inverse so it's sufficient to show that single iteration of $\mathcal F$ is continuous mapping of Schwartz space into itself. That is most easily done by closed graph theorem if you know that. If you don't, think about getting acquainted with it - it's one of the most important tools in inspecting linear operators in infinite-dimensional topological vector spaces.