Spectrum of linear operator, essential spectral radius Consider the operator $L:L^1(S^1)\to L^1(S^1)$ given by
$$
(Tf)(x)=\dfrac{1}{2}\left( f\left( \dfrac{x}{2}\mod 1\right)+f\left( \dfrac{x+1}{2} \mod 1 \right)  \right)
$$
where we identified $S^1$ with $\mathbb{R}/\mathbb{Z}$. What is it possible to say about the spectrum of $L$? I know that it should be contained in the disk of radius $1$, and clearly $1$ is an eigenvalue associated to the constant function equal to $1$. I also know that the action of $T$ restricted to smoother functions spaces (Lipschitz) has essential spectral radius strictly less than $1$, but I would like to determine the spectrum when acting in $L^1$. I have the idea that the whole disk should be contained in the spectrum, but couldn't prove it. Thanks in advance.
 A: Let's just try to find an eigenfunction for a given $|\lambda|< 1$: we want an $f$ that solves
$$
f\left( \frac{x+1}{2}\right) = 2\lambda f(x)- f(x/2) . \quad\quad\quad\quad (1)
$$
Start out with an arbitrary (integrable) function on $(0,1/2)$, then use (1) for $0<x<1/2$ to define $f(t)$ for $1/2<t<3/4$, then reenter (1), with $1/2<x<3/4$ to find $f(t)$ on $3/4<t<7/8$ etc.
We must check that this function is integrable (and this is where we use that $|\lambda|< 1$). This is straightforward but mildly tedious to write down. Write $I_0=(0,1/2)$, $I_1=(1/2,3/4)$, and abbreviate $q=|\lambda|$. By substituting $t=(x+1)/2$ and using (1), we find that
$$
\|f\|_{I_1} \le q\|f\|_{I_0} + \|f\|_{(0,1/4)} .
$$
Then just continue in this style. Let $I_2=(3/4,7/8)$. Then
$$
\|f\|_{I_2} \le q \|f\|_{I_1} +\|f\|_{(1/4,3/8)}\le q^2\|f\|_{I_0} +q\|f\|_{(0,1/4)} + \|f\|_{(1/4,3/8)} ,
$$
so
$$
\|f\|_{I_1\cup I_2}\le (q^2+q)\|f\|_{I_0} + (q+1)\|f\|_{(0,1/4)} + \|f\|_{(1/4,3/8)} .
$$
We can now see the general pattern: we will obtain a decomposition of $(0,1/2)$ and over time, the norm over each interval will get multiplied by a geometric sum. We can sum these geometric series and the norms over the disjoint integrals to produce a finite bound on $\|f\|$.
It follows that the spectrum equals the closed unit disk, as claimed.
A: Here is a very explicit way to do it: first write $Uf(x)=f(2x\bmod 1)$. This is a norm 1 operator (the Koopman operator of the doubling map, while $T$ is the Perron-Frobenius operator of the doubling map). One can check that $TU(f)=f$ for all $f$. Next, notice that if $f=\cos(2\pi x)$, then $f$ lies in the kernel of $T$ (any function in the kernel of $T$ will do). 
Then for arbitrary $\lambda$ in the unit disc, an explicit eigenfunction of $T$ with eigenvalue $\lambda$ is given by
$$
h_\lambda=\sum_{n=0}^\infty \lambda^nU^nf
$$
The fact that $\|U\|=1$ ensures $h_\lambda\in L^1$. 
