Applying Fourier transform to $f(x)-\int_\limits{-\infty}^{\infty}K(x-t)f(t)dt=g(x)$

Question

Notes:

$f(x)-\int_\limits{-\infty}^{\infty}K(x-t)f(t)dt=g(x)$ where $f,g,k\in L_1{\mathbb{R}}$ therefore the equation is solvable with a unique solution iff $1-\hat{K(\xi)}\neq 0\:\:,\forall\xi\in\mathbb{R}$.

Applying the Fourier transform to both sides of the equation:

$\hat{f}(\xi)-\hat{K}(\xi)\hat{f}(\xi)=\hat{g}(\xi)\implies \hat{f}(\xi)=\frac{\hat{g}(\xi)}{1-\hat{k}(\xi)}$

Fourier transform:$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{it\xi}dt\:\:,\:\: \xi\in\mathbb{R}$

Question:

Considering $\int_\limits{-\infty}^{\infty}K(x-t)f(t)$. Why did the integral disappeared when applying the Fourier transform?

Thanks in advance!

Answer 1

\begin{align*} \int_{-\infty}^{\infty}K(x-t)f(t)dt=(K\ast f)(x), \end{align*} and \begin{align*} (K\ast f)^{\wedge}(\xi)=\widehat{K}(\xi)\widehat{f}(\xi) \end{align*} for $K,f\in L^{1}({\bf{R}})$.