Find all functions $f(x) = c[T(f)](x)$ for a linear transformation $T$ I'm working through Apostol's calculus vol 2, and I'm not sure how to finish the following question (2.4, #29, part f):

Let $V$ be the set of all real functions continuous on the interval of $[-\pi, \pi]$ and let $T: V \to V$ be the linear transformation defined as:
$$
g(x) = \int_{-\pi}^{\pi} (1 + cos(x-t))f(t) dt
$$
Find all real $c \neq 0$ and all nonzero $f$ in $V$ such that $T(f) = cf$. 

So far, I'm sure that the constant functions $C_k(x) =k$ satisfy the relationship with $c = 2\pi k$, since:
$$
\begin{align}
[T(f)](x) &= \int_{-\pi}^{\pi} f(t) dt  \\
&+ \cos(x)\int_{-\pi}^{\pi} \cos(t) f(t) dt \\ &-\sin(x)\int_{-\pi}^{\pi} \sin(t) f(t) dt
\end{align}
$$
and for $C_k$, the second two terms are $0$.
But I'm not sure how to prove that these are the only functions that satisfy the relationship. I think the second two terms are always going to be $0$ -- since if there was a $\sin(x)$ or $\cos(x)$ term in $f(x)$, it those two integrals would be $0$. 
 A: Assume $f$ and $c\in \mathbb{R}$, $c\ne0$ satisfies  $T(f) = cf$. Then we can write $$ f(x) = \frac{1}{c} T(f)(x) = \frac{1}{c} \int_{-\pi}^\pi \left( 1 +\cos(x - t) \right) f(t) dt. $$ Weh put this into the definition of $T(f)$ again; as, 
\begin{align*} cf(x) = (Tf)(x) &= \int_{-\pi}^\pi \left( 1 +\cos(x - u) \right) f(u) du \\ 
& = \int_{-\pi}^\pi \left( 1 +\cos(x - u) \right) \left(\frac{1}{c} \int_{-\pi}^\pi \left( 1 +\cos(u - t) \right) f(t) dt \right)du .
\end{align*}
Be aware of the use of variable; $\int_a^b f(x)dx = \int_a^b f(u) du$ and I changed the name of variables here. 
This leads us as follows; 
\begin{align*} c^2 f(x) &= \int_{-\pi}^\pi  \int_{-\pi}^\pi \left( 1 +\cos(x - u) \right)\left( 1 +\cos(u - t) \right) f(t) dt du \\
&\stackrel{*}{=} \int_{-\pi}^\pi  \int_{-\pi}^\pi \left( 1 +\cos(x - u) \right)\left( 1 +\cos(u - t) \right) f(t) du dt \\ 
& = \int_{-\pi}^\pi f(t)\int_{-\pi}^\pi \left( 1 +\cos(x - u) \right)\left( 1 +\cos(u - t) \right) du dt \\
& \stackrel{**}{=} \text{constant},
\end{align*}
i.e. $f(x)$ is constant. Here $(*)$ is Fubini's theorem, $(**)$ is the calculation.
So, If the function $f$ on $[-\pi, \pi]$ and $c \ne0$ satisfies $T(f) = cf$, $f$ should be constant function. In other words, the only eigenvector of linear operator $T$ is $1$ (up to constant coefficient).
