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!


My hypothesis was wrong, $M_1-\lambda\operatorname{id}_{L^1}$ is bijective for all $\lambda\in\mathbb C\setminus[-\pi/2,\pi/2]$

  • 1
    $\begingroup$ You want to show that $M_1 - \lambda\operatorname{id}_{L^1}$ is not surjective, but you say that there'd be an $f$ with $M_1 f = \chi_{[0,1]}$. In fact, for most $\lambda$, $M_1-\lambda\operatorname{id}_{L^1}$ is invertible. $\endgroup$ – Daniel Fischer May 17 '14 at 13:10
  • $\begingroup$ @Daniel Fischer Thanks, I just found out why. $\endgroup$ – Thorben May 17 '14 at 18:36
  • $\begingroup$ Good. You can post an answer then. $\endgroup$ – Daniel Fischer May 17 '14 at 18:39
  • $\begingroup$ @DanielFischer I posted my answer. I hope this is correct... $\endgroup$ – Thorben May 17 '14 at 19:38

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 $||M_{\phi}||=||\phi||_{\infty}$.



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 $\operatorname{spec}(M^n_2)=[-\frac{\pi^n}{2^n},\frac{\pi^n}{2^n}]$.

  • $\begingroup$ Generally, the spectrum of $M_\phi$, for $\phi\in L^\infty$ is the essential range of $\phi$, that is the $z\in\mathbb{C}$ such that $m(\phi^{-1}(B_\varepsilon(z)) > 0$ for all $\varepsilon > 0$, where $m$ is the Lebesgue measure. $\endgroup$ – Daniel Fischer May 17 '14 at 19:47
  • $\begingroup$ Thanks for the very helpful comment! $\endgroup$ – Thorben May 17 '14 at 19:49

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