It is well-known that, for $f,g \in L^{1}(\mathbb R).$ Then, by Fubini's theorem, one can derive, $\widehat{f\ast g} = \hat{f} \cdot \hat{g},$ (that is, Fourier transform takes, convolution to point wise multiplication).
Also, we note that the fact, that, for $f, g\in L^{2}(\mathbb R)$ one has, $\widehat{f\ast g} =\hat{f} \cdot \hat{g}.$
Suppose $f, \hat{f} \in L^{p}(\mathbb R)\cap C(\mathbb R) (1<p<\infty, p\neq 2),$ and $g\in \mathcal{S}(\mathbb R),$ (Schwartz space)
My Question is: Can we expect, $\widehat{f\ast g}= \hat{f} \cdot \hat{g}$ ? If yes, how to prove it ?
Edited :(with hope may be this is helpful).
For $f\in L^{1}(\mathbb R), $ we define $f^{\vee}(x):=\hat{f}(-x)=\int_{\mathbb R} f(\xi) e^{2\pi i \xi\cdot x} d\xi, (x\in \mathbb R).$
Let $f, g\in L^{2}(\mathbb R).$ Then $\hat{f}\hat{g}\in L^{1},$ by Plancherel's theorem and H\"olders inequality, so $(\hat{f}\hat{g})^{\vee}$ make sense. Given $x\in \mathbb R,$ let $h(y)=\overline{g(x-y)}.$ It is easy to see that, $\hat{h}(\xi)=\overline{\hat{g}(\xi)}e^{-2\pi i\xi \cdot x},$ so since $\mathcal{F}$ is unitary on $L^{2},$ $$f\ast g (x)= \int f\bar{h}= \int \hat{f}\bar{\hat{h}}=\int \hat{f}(\xi)\hat{g}(\xi)e^{2\pi i \xi \cdot x} d\xi =(\hat{f}\hat{g})^{\vee}(x).$$
Thanks,
If $1<p<2$, it holds.
Note that $f \in C(R)$ implies $f \in L^{\infty}$. Since $f$ is in $L^p \cap L^\infty$, we get $f \in L^2$. Thus we get the desired result.
For $f \in L^p$ with $ 2<p<\infty$, Fourier transform of $f$ is not defined unless $f \in L^q$ for some $1 \leq q \leq 2$.