As far as I know, ket vectors are in the Hilbert space $L_2 (\mathbb{R})$ (the inner product defined as $$ \langle f, g \rangle = \int_{-\infty}^{\infty} f(x)\cdot\bar g(x) dx$$
However, bra vectors, which are in the dual space of the linear functionals from $L_2(\mathbb{R})$, but how can we find the bra vector of a given ket vector, i.e if $f \in L_2 (\mathbb{R})$, what is its corresponding bra vector ?
Plus, is the dual space isomorphic to the $L_2 (\mathbb{R})$ ?
On
For any $f \in L^2(\mathbb{R})$ I believe you are looking for the functional $\psi = \langle \cdot , f\rangle \in L^2(\mathbb{R})^*$, that is:
$$\psi(g) = \langle g, f\rangle = \int_\mathbb{R} g(x)\cdot \overline{f}(x)\,dx, \quad\forall g \in L^2(\mathbb{R})$$
Also, by the Riesz representation theorem, since $L^2(\mathbb{R})$ is a Hilbert space, every bounded functional $\psi \in L^2(\mathbb{R})^*$ can be uniquely represented by some $f \in L^2(\mathbb{R})$ such that $\psi = \langle \cdot , f\rangle$.
The mapping $f \mapsto \psi$ is in fact an isometrical isomorphism of $L^2(\mathbb{R})$ and $L^2(\mathbb{R})^*$, so in particular we get that $L^2(\mathbb{R})$ is reflexive: $L^2(\mathbb{R}) \cong L^2(\mathbb{R})^{**}$.
The dual space of a Hilbert space can be naturally identified with itself, so that every "ket" vector can change its hat and play a role of a "bra" vector. Other than this trivial identification, there is no other vector that will naturally be a "corresponding bra vector of a ket vector".