Spectrum of the operator $T \in \mathcal{L}(L^2(\Bbb{R}_+))$ defined by $(Tf)(x)=(1−e^{−x})f(x)$

Question

I'm preparing a mathematical physics exam (the last one) and one of the topic is the spectral theory on operators. I am totally confused on some concept, so please pardon me. I am studying this operator $T:\mathcal{L}(L^2(R^+))$ defined by $(Tf)(x)=(1-e^{-x})f(x)$

So I have proved that this operator $T$ is limited because $||Tf||^2< +\infty$ and beign a multiplication operator, $||T|| = 1$. Now I want to find the spectre.

So for this I am confused. I have $\lambda f(x) = T f(x)$ and I want the $f \ne 0$ so I arrive at $f(x)(\lambda -1+e^{-x})=0$. Here I know that the resolvent $\rho(T) := \{ \lambda \in C:| \lambda|>1 \}$ (out of the spectral radius) and the spectre $\sigma(T)=C/ \rho(T) $. How can I find if there is punctual, continous and residual spectra of this operator? I wish to have some guidelines and explanation because I am very confused, and also I do not have so much time for study since I work too.

Thanks!

Answer 1

Your operator is self-adjoint in Hilbert space, therefore, from theory follows that it's residual spectrum is empty. Moreover, his spectrum is real. Point spectrum also empty, because from the equation $(1-e^{-x})f(x)=\lambda f(x)$ follows that $f(x)=0$ a.e. for any $\lambda$. Notice, that for $x\in[0,\infty)$, $1-e^{-x}\in[0,1]$ and if $\lambda\in[0,1]$, then set $Y_\varepsilon=\{x\in[0,\infty):|1-e^{-x}-\lambda|<\varepsilon\}$ has a positive measure for all $\varepsilon>0$. Then consider the sequence $f_n(x)=\dfrac{\chi_{Y_n}}{\|\chi_{Y_n}\|}$, where $Y_n=\{x\in[0,\infty):|1-e^{-x}-\lambda|<\frac{1}{n}\}$. It's obvious that $\|f_n\|=1$, but $\|(T-\lambda I)f_n(x)\|\leq\frac{1}{n}\to0$. Thus, $\lambda\in[0,1]$ -- is points of spectrum. In view of the above, these are points of the continuous spectrum.

In this argument I used a well-known fact (it's very easy to prove, only by definitions): if for a number $\lambda$ there exists a normalized sequence $\{x_n\}$ such that $Tx_n-\lambda x_n\to0$, then $\lambda\in\sigma(A)$.

For all other real $\lambda$, you can easily check that the operator $T-\lambda I$ is invertible and it's inverse operator is bounded on whole space, so these are points of the resolvent set.