Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$ i$,$-1$, and$-i$. For$a$one of these eigenvalues, denote by$M_a$the corresponding eigenspace. It turns out then that$L^2(\mathbb{R})$is the direct sum of these$4$eigenspaces! In fact, this is easy linear algebra. Consider$f \in L^2(\mathbb{R})$. We want to find$f_a \in M_a$for each of the eigenvalues such that$f = f_1 + f_{-1} + f_{i} + f_{-i}$. Using the fact that$T^4 = 1$, we obtain the following 4 equations in 4 unknowns:$f = f_1 + f_{-1} + f_{i} + f_{-i}T(f) = f_1 - f_{-1} +i f_{i} -i f_{-i}T^2(f) = f_1 + f_{-1} - f_{i} - f_{-i}T^3(f) = f_1 - f_{-1} -i f_{i} +i f_{-i}$Solving these four equations yields the corresponding projection operators. As an example, for$f \in L^2(\mathbb{R})$, we get that$\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$is a fixed point for$T$. 1 The following is discussed in a little more detail on pages 337-339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well). Normalize the Fourier transform so that it is a unitary operator$T$on$L^2(\mathbb{R})$. One can then check that$T^4=1$. The eigenvalues are thus$1$,$i-1$, and$-i$. For$a$one of these eigenvalues, denote by$M_a$the corresponding eigenspace. It turns out then that$L^2(\mathbb{R})$is the direct sum of these$4$eigenspaces! In fact, this is easy linear algebra. Consider$f \in L^2(\mathbb{R})$. We want to find$f_a \in M_a$for each of the eigenvalues such that$f = f_1 + f_{-1} + f_{i} + f_{-i}$. Using the fact that$T^4 = 1$, we obtain the following 4 equations in 4 unknowns:$f = f_1 + f_{-1} + f_{i} + f_{-i}T(f) = f_1 - f_{-1} +i f_{i} -i f_{-i}T^2(f) = f_1 + f_{-1} - f_{i} - f_{-i}T^3(f) = f_1 - f_{-1} -i f_{i} +i f_{-i}$Solving these four equations yields the corresponding projection operators. As an example, for$f \in L^2(\mathbb{R})$, we get that$\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$is a fixed point for$T\$.