Want to show that an operator is not surjective So here is my problem,
Let $$M_1:L^1\rightarrow L^1$$
$$f(x)\mapsto \arctan(x)f(x)$$
In order to compute the spectrum of $M_1$ I am investigating for which $\lambda\in\mathbb C$ the following map is surjective,
$$M_1-\lambda\operatorname{id}_{L^1}$$
I think it is never surjective, therefore I tryied to compose a proof. The idea was to assume for contradiction that there exists some $\lambda$ such that the upper map is surjective. Hence there must exist an integrable function $f$ s.t $M_1f(x)=\mathcal X_{[0,1]}(x)$ is the indicator function of $[0,1]$. By assumption it follows then that $f(x)=\frac{1}{\arctan(x)-\lambda}$ which is not $L^1$. A contradiction!
Since I am not sure if my proof is correct and I wanted to ask if there is a different way to show this?
Thanks in advance!
EDIT
My hypothesis was wrong, $M_1-\lambda\operatorname{id}_{L^1}$ is bijective for all $\lambda\in\mathbb C\setminus[-\pi/2,\pi/2]$
 A: I hope this is correct... 
The goal of the exercise was compute the spectrum of 
$$M_2^n:L^1\rightarrow L^1$$
$$f\mapsto arctan(x)^n\cdot f(x)$$
So here is my proof.
Claim: $spec(M_2^n)=[-\frac{\pi^n}{2^n},\frac{\pi^n}{2^n}]$
Remark 1
In the following we will use the that the multiplication operator $M_{\phi}:L^1\rightarrow L^1$ defined as $M_{\phi}(f)=\phi\cdot f$ where $\phi$ is an arbitrary bounded function is well defined and has norm
$$||M_{\phi}||=||\phi||_{\infty}$$
Proof:
Since $||g\cdot f\||_1\leq||g||_{\infty}||f||_1$ the operator $M_g$ is well defined. Furthermore
note that $(L^1)^*$ is isometric isomorphic to $L^{\infty}$. Hence $\int_{\mathbb R}g(x)f(x)dx=:\psi_g(f)$  is an linear bounded operator mapping into the field with $||\psi_g||=||g||_{\infty}$.
Moreover by the following
$$||g||_{\infty}=||\psi_g||=\sup_{||f||_1\leq 1}|\int_{\mathbb R}g(x)f(x)dx|\leq \sup_{||f||_1\leq 1}\int_{\mathbb R}|g(x)||f(x)|dx=||M_{g}||\leq ||g||_{\infty}$$
we can conclude $||M_{\phi}||=||\phi||_{\infty}$.
$\square$
Proof:
Since for all $\lambda\in\mathbb C$ we have
$$\arctan(x)^nf(x)-\lambda f(x)=f(x)(\arctan(x)^n-\lambda)=0$$
iff $f(x)=0$ or  $(\arctan(x)^n-\lambda)=0$. But since  $(\arctan(x)^n-\lambda)\neq 0$ a.e it follows that $f=0$ a.e. Hence $(M_2^n-\lambda\operatorname{id})(f)=0$ iff $f\sim_{L^1}0$.
Thus we can conclude that $M_2^n-\lambda\operatorname{id}$ is injective for all $\lambda\in\mathbb C$.
In order to compute $\operatorname{spec}(M^n)$ it is left to find all $\lambda\in\mathbb C$ such that $M_2^n-\lambda\operatorname{id}$ is surjective. Obviously $\operatorname{spec}(M^n)\subset\{z\in\mathbb C:|z|\leq||M_2^n||=\frac{\pi^n}{2^n}\}$
Furthermore since $\arctan(x)^n$ is real for all $x$, the function $\frac{1}{\arctan(x)^n-\lambda}$ is not singular for $\lambda\in\mathbb C\setminus\mathbb R$ and consequently 
$$\frac{1}{|\arctan(x)^n-\lambda|}\leq\frac{1}{|\Im(\lambda)|}$$
Hence $\frac{1}{\arctan(x)^n-\lambda}\in L^{\infty}$. Therefore $((M_2^n-\lambda\operatorname{id})(f))^{-1}:=f(x)\frac{1}{\arctan(x)^n-\lambda}$ is well defined as showed in Remark 1.
Moreover as for all $\lambda\in\mathbb C\setminus\mathbb R$ the inverse exists we can conclude that $\operatorname{spec}(M^n_2)\subset[-\frac{\pi^n}{2^n},\frac{\pi^n}{2^n}]$.
In order to show the equality, assume for contradiction that there exist $\lambda\in[-\frac{\pi^n}{2^n},\frac{\pi^n}{2^n}]$ such that $(M_2^n-\lambda\operatorname{id})(f)$ is bijective. Then 
$((M_2^n-\lambda\operatorname{id})(f))^{-1}:=f(x)\frac{1}{\arctan(x)^n-\lambda}$ is a bounded linear operator by the inverse operator Theorem. Moreover we have a "new" multiplication operator given by
$$Mf=f(x)\frac{1}{\arctan(x)^n-\lambda}$$
Hence by Remark 1 $||M||=||\frac{1}{\arctan(x)^n-\lambda}||_{\infty}=\infty$.   A contradiction!
And finally we can conlude $\operatorname{spec}(M^n_2)=[-\frac{\pi^n}{2^n},\frac{\pi^n}{2^n}]$.
