It indeed lies in the Schwartz space, which is the space $$
\mathcal{S} := \left\{f: \mathbb{R} \to \mathbb{R} \,:\,
\forall \alpha,\beta \in \mathbb{N}\,
\sup_{x\in\mathbb{R}} \left|x^\alpha f^{(\beta)}(x)\right| < \infty \right\}
$$
where $f^{(n)}$ denotes the $n$-th derivative of $f$. This definition looks a bit intimidating at first, but it's easy to understand. A function $f$ lies in $\mathcal{S}$ if it decays rapidly, i.e. faster than any polynomial grows (thus $x^nf(x)$ is bounded for every n), and if the same is true for every derivative of $f$.
This space is closed under fourier transform, i.e. if $f \in S$, then so is the fourier transform $\mathcal{F}(f)$. The same is true from the reverse fourier transform. The space of compactly supported $C^\infty$ functions is quite obviously a subspace of $\mathcal{S}$, hence it's image under $\mathcal{F}$ is thus also a subspace of $\mathcal{S}$.
The problem with your reasoing is that you're thinking about pointwise limits. Under those, $\mathcal{S}$ isn't complete, so knowing something about the limit doesn't tell you much about $\mathcal{S}$. There are topologies on $\mathcal{S}$ which make it a complete space, and under those you indeed cannot approximate the step function by functions in $\mathcal{S}$. The "trick" of those topologies is basically to force convergence of all the derivatives, which e.g. prevents you from approximating any discontinous functions.
This won't be the most rigorous answer, but just a representation my intuition with regards to the Fourier transform.
In nonrelativistic quantum mechanics there are two important operators which act on $L^2(\mathbb{R})$.
The position operator,
$$ (\hat{x} \psi)(x) = x\psi(x),$$
and the momentum operator,
$$ (\hat{p} \psi)(x) = -i \frac{\partial \psi}{\partial x} (x).$$
Now the "eigenvectors" of $\hat{x}$ are $a_{x'}(x) = \delta(x-x')$ and the "eigenvectors" of $\hat{p}$ are $b_p(x) = e^{ipx}$.
Now if you study linear algebra at all you know that the choice of basis is arbitrary. Also you would know that the eigenvectors of a Hermitian operator always form a complete basis. Although our operators our Hermitian, their eigen-functions certainly don't live in $L^2(\mathbb{R})$. I'm going to completely ignore that fact and pretend that I can use them as basis.
For instance the expansion of $\psi(x)$ in terms of $a_x$'s will be,
$$ \psi(x) = \int \mathrm dx'\ c(x') a_{x'}(x),$$
we can use what we know about delta functions to conclude that the appropiate expansion coefficients, $c(x')$, are in fact $c(x')=\psi(x')$.
If we want to expand $\psi$ in the momentum basis (the $b_p$'s) then we write the superposition and try to determine the coefficients.
$$\boxed{ \psi(x) = \int \mathrm dp \ c(p) b_{p}(x)},$$
the coefficient function, $c(p)$, is a vector in $L^2(\mathbb{R})$ as well. In fact what we have writtend down here is an isomorphism between the x-basis and the p-basis. However the bahavior of the operators on functions expressed in this new bases is different.
$$ \hat{x} \psi(x) = \int \mathrm dp \ c(p) x e^{ipx} = \int \mathrm dp \ c(p) \left(-i\frac{\partial}{\partial p} \right)e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p} e^{ipx} $$
$$ \boxed{ \hat{x} = i \frac{\partial}{\partial p}} $$
$$ \hat{p} \psi(x) = \int \mathrm dp \ c(p) \left(-i \frac{\partial}{\partial x} \right) e^{ipx} = \int \mathrm dp \ c(p) p e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p} e^{ipx} $$
$$ \boxed{ \hat{p} = p} $$
This means that in the p-basis the eigenvalues of $\hat{x}$ are $a_x(p) = e^{-ipx}$. I can write $c(p)$ in terms of the x-basis then by the integral,
$$ c(p) = \int \mathrm dx \ \phi(x) e^{-ipx} $$
I've used $\phi(x)$ to represent the coefficients in this expansion, but in practice we actually know the vector in the x-basis which is isomorphic to $c$, that would be $\psi(x)$.
$$ \boxed{ c(p) = \int \mathrm dx \ \psi(x) e^{-ipx} }$$
One issue is the normalization of the transforms are off. This is because the don't have finite norms. For that you need to actually prove the Fourier inversion theorem.
Best Answer
The exponential function $e^{zt}$ is the eigenfunction of the differentiation operator, with eigenvalue $z$. The Fourier Transform essentially expresses a function as a linear combination of $e^{i\omega t}$ basis functions, so differentiating this linear combination is simply multiplication by the eigenvalue $i\omega$. This is analogous to applying a linear operator to a linear combination of basis vectors - the result is entirely determined by the operator's mapping of the basis vectors.