Easy Filter with FFT and Convolution
Actually, yes. As Hilmar pointed out, it might be cheaper to use an ordinary filter, but that depends on the circumstances.

If you have two signals $A$ and $B$ with Fourier transform being $\mathcal{F}$, their convolution $A\ast B$ will be 

\begin{equation}
A\ast B = \tilde{\mathcal{F}}(\mathcal{F}(A)\cdot\mathcal{F}(B))
\end{equation}

With $\tilde{\mathcal{F}}$ being the inverse transform. This is the well known "convolution lemma". There is not much of a theory to see. In the Fourier domain, you just replace convolution by (point-wise) multiplication. Your question already contained the answer, to me it just appears that you were not sure about the details.