Suppose $f\in C\cap\mathcal S'(\mathbb R^d)$ is a (may not be necessary bounded) continuous function such that $\hat f \in\mathcal S'(\mathbb R^d)$ is a continuous linear function on $C_b^\infty(\mathbb R^d)$ (smooth functions with bounded derivatives).
It's true that $f(x)=\langle \hat f,e^{2\pi i x(\cdot)}\rangle$ holds for all $x\in\mathbb R^d$?
Indeed $\hat f$ is not a measure in general
The answer is yes because Fourier operator $\mathcal{F} :\mathcal{S}(\mathbb{R}^d) \longrightarrow \mathcal{S}(\mathbb{R}^d)$ is continuous isomorphism and it can be considered as the restriction of Fourier operators $\mathcal{F} :\mathcal{S}'(\mathbb{R}^d) \longrightarrow \mathcal{S}'(\mathbb{R}^d)$ (weak* continuos isomorphism), furthermore $\widehat{f} \in \mathcal{S}(\mathbb{R}^d) \subset C_{b}^{\infty} (\mathbb{R}^d)$.