Show that $T(V)$ is finite-dimensional? Let $T: V\rightarrow V$ be the linear transformation defined as follows: If $f$ is in $T$, $g=T(f)$ is:
$$
g(x) = \int_{-\pi}^{\pi} (1+\cos(x-t))f(t) dt $$
Prove $T(V)$ is finite-dimensional. 
edit: Forgot to put the definition of $V$: the linear space of all real functions continuous on the interval $[-\pi,\pi]$.
edit 2 Dammit, thought the problem was written with a slash; re-edited to correct integrand.
 A: We have $$
T(f)(x)=\int_{-\pi}^\pi(1+\cos(x-t))f(t)dt=\int_{-\pi}^\pi(1+\cos x\cos t+\sin x\sin t)f(t)dt
$$
$$
=\int_{-\pi}^\pi f(t)dt+\left(\int_{-\pi}^\pi\cos tf(t)dt\right)\cos x+\left(\int_{-\pi}^\pi\sin tf(t)dt\right)\sin x
$$
$$
=L_1(f)1+L_2(f)\cos x+L_3(f)\sin x
$$
where $L_1, L_2, L_3$ are linear functionals. So $T(f)$ belongs to the span of the functions $1, \cos $ and $\sin $.
A: Another way to see this is to note that $g$, acting on $L^2[-\pi, \pi]$, is a sum of two linear operators $g = g_1 + g_2$ both of which have $1$-dimensional ranges. So the range of $g$ has dimension at most (equal to, in these case) $2$.
The operator $g_1$ is integration against $dx$; this is a linear functional that computes the $0$-th Fourier coefficient. $g_2$ is convolution with $\cos x$, which, by the Fourier transform, is a $1$-dimensional projection onto the subspace generated by $\cos x$.  
$g_1$ can also be viewed as a convolution(i.e. projection in Fourier basis). So your operator is an orthogonal projection onto the $2$-dimensional subspace generated by $\{1, \cos x\}$ (possibly up to some scaling constant that makes the Fourier transform unitary).
