Given a filter with input $x(n)$ and output $y(n)$, it is always possible to compute the Fourier transform of he input and of the output, say $X(f)$ and $Y(f)$. The ratio of these two quantities is called the transfer function. For now, let us denote it by $T(f)$. Interestingly, we will see that the transfer function do not depend on $x$, and thus is a global characteristic of the system. More than that, we will see that the transfer function is intimately linked to the impulse response of the system.
Convolution enables to express the output of a filter characterized by its impulse response. Consider a system with impulse response $h(n)$ and an input $$ x(n) = X_0 e^{j2\pi f_0n}. $$ Its output is given by \begin{eqnarray*} y(n) & = & \sum_m h(m) X_0~ e^{j2\pi f_0(n-m)} \\ & = & X_0 e^{j2\pi f_0n} \sum_m h(m) e^{-j2\pi f_0m}. \end{eqnarray*} We recognize above the expression of the Fourier transform of $h(m)$ at the frequency $f_0$: $$ \boxed{ \displaystyle{H(f_0) = \sum_m h(m) e^{-j2\pi f_0m}}}. $$ \index{Filtres!Fonction de transfert|fin} Hence, the output can be simply written as $$ y(n) = X_0 e^{j2\pi f_0n} H(f_0). $$ For a linear system excited by a complex exponential at frequency $f_0$ , we obtain that output is the same signal, up to a complex factor $H(f_0)$. This gives us another insight on the interest of the decomposition on complex exponentials: they are eigen-functions of filters, and $H(f_0)$ plays the role of the associated eigenvalue. \index{Filters!eigenfunctions|end}
Consider now an arbitrary signal $x(n)$. It is possible to express $x(n)$ as an infinite sum of complex exponentials (this is nothing but the inverse Fourier transform); $$ x(n) = \int_{[1]} X(f) e^{j2\pi f n} \dr{f}. $$ To each component $X(f) e^{j2\pi fn}$ corresponds an output $X(f)H(f) e^{j2\pi fn}$, and, by superposition, $$ y(n) = \int_{[1]} X(f)H(f) e^{j2\pi f n} \dr{f}. $$ Thefore, we see that the Fourier transform of the output, $Y(f)$, is simply $$ \eqboxc{Y(f) = X(f)H(f)}. $$ The time domain description, in terms of convolution product, becomes a simple product in the Fourier domain. $$ \eqboxd{[x\mystar h] (n) \flecheTF ~X(f)H(f)}. $$ It is easy to check that reciprocally, $$ \eqboxd{x(n)h(n) \flecheTF [X\mystar H] (f)}. $$ Try to check it as an exercise. You will need to introduce a convolution for function of a continuous variable, following the model of the convolution for discrete signals.
Begin with the Fourier transform of $x(n)y(n)$, and replace the signals by their expression as the inverse Fourier transform:
\begin{align} \mathrm{FT}[x(n)y(n)] & = \sum_n x(n)y(n) e^{-j2\pi f n} \\ & = \sum_n \int X(f_1)e^{j\pi f_1 n} \dr{f_1} \int Y(f_2)e^{j\pi f_2 n} \dr{f_2} e^{-j2\pi f n} \\ & = \int\!\!\!\int X(f_1) Y(f_2) \sum_n e^{j\pi f_1 n}e^{j\pi f_2 n} e^{-j2\pi f n} \dr{f_1} \dr{f_2} \end{align}It remains to note that the sum of exponentials is nothing but the Fourier transform of the complex exponential $e^{j\pi (f_1+f_2) n}$, and thus that $$ \sum_n e^{j\pi f_1 n}e^{j\pi f_2 n} e^{-j2\pi f n} = \delta(f-f_1-f_2). $$ Therefore, the double integral above reduces to a simple one, since $f_2=f-f_1$, and we obtain $$ \mathrm{FT}[x(n)y(n)] = \int X(f_1) Y(f-f_1) \dr{f_1} =[X*Y](f). $$
(Another proof is possible, beginning with the inverse Fourier transform of the convolution $[X*Y](f)$, and decomposing the exponential so as to exhibit the inverse Fourier transform of $x(n)$ and $y(n)$). Try it.
The transform of a convolution into a simple product, and reciprocally, constitutes the Plancherel theorem: \index{Convolution!Plancherel theorem|end} $$ \boxed{ \begin{array}{lcl} [x*y](t) & \flecheTF & X(f)Y(f), \\ x(t)y(t) & \flecheTF & [X*Y](f). \end{array} } $$
This theorem has several important consequences.
The Fourier transform of $x(n)y(n)^{*}$ is $$ \displaystyle{x(n)y(n)^* \flecheTF \int_{[1]} X(u)Y(f-u)^* \dr{u}}, $$
since $\mathrm{FT}{y^*(n)}=Y^*(-f)$. Therefore, $$ \mathrm{FT}{x(n)y(n)^{*}} = \int_{[1]} X(u)Y(u-f)^* \dr{u}, $$ that is, for $f=0$, $\def\mystar{*}$ $$ \eqboxb{ \displaystyle{\sum_{-\infty}^{+\infty} x(n)y^\mystar (n) = \int_{[1]} X(u)Y^\mystar(u) \dr{u}}}. $$ \index{Convolution!Plancherel-Parseval theorem|end} This relation shows that \textit{the scalar product is conserved} in the different basis for signals. This property is called the Plancherel-Parseval theorem. Using this relation with $y(n)=x(n)$, we have $$ \eqboxb{ \displaystyle{\sum_{-\infty}^{+\infty} |x(n)|^2 = \int_{[1]} |X(f)|^2 \dr{f}}}, $$ which is a relation indicating \textit{energy conservation}. It is the Parseval relation. \index{Convolution!Parseval's relation|end}