Solving for values of a Fourier transform

Question

Let $f(x)$ be an integrable function on $\bf R$. Given constants $A,B>0$, does there always exist an $x$ satisfying the following equation? $$A=Bx+\hat{f}(x)=Bx+\int^\infty_{-\infty}f(t)e^{ixt}dt$$ Perhaps one may need to impose further conditions on $f(x)$ for this to be true?

Answer 1

$\hat f$ is continuous and $\lim_{x\to\pm\infty}\hat f(x)=0$. Then $B\,x+\hat f(x)$ is continuous and $\lim_{x\to\pm\infty}\hat f(x)=\pm\infty$. The mean value theorem shows that the equation $B\,x+\hat f(x)=A$ has at least one solution.