# Continuous spectral value of right shift operator $\ell^2(\mathbb{N})$

Let $T:\ell^2(\mathbb{N} \to \ell^2(\mathbb{N})$ be the operator that sends$(x_1,x_2,x_3,...) \to (0,x_1,x_2,x_3,....)$. I want to show that $\lambda = 1$ is in the continuous spectrum.

To approach this, I first showed that $\lambda$ cannot be an eigenvalue. If it were, $Tx = x$ for some $x \in \ell^2$, which would imply $(0,x_1,x_2,x_3,....) = (x_1,x_2,x_3,....)$ and hence we have $x_1 = 0$, $x_2 = x_1$, $x_2 = x_3$, .... which implies $x = 0$. So, $\lambda = 1$ is not an eigenvalue. It is not in the residual spectrum, since the range of $T-I$ is dense. If it weren't, then its adjoint should have nonzero kernel, so Ker$(T^*-I)$ should be nonempty. However, $(T^* -I)x = 0$ implies $(x_1-x_2,x_2-x_3,x_3-x_4,... ) = 0$, so $x_1 = x_2 = x_3 = ....$. This will only be in $\ell^2$ if $x_1 = 0$, so $x = 0$, so the kernel is just $x = 0$, so we have the range of $T-I$ must be dense in $\ell^2$.

Now, I get a bit stuck. I know so far that $\lambda = 1$ is neither an eigenvalue nor in the residual spectrum. To show that $\lambda = 1$ is in the continuous spectrum, I need to show that $(T-I)^{-1}$ is unbounded. I considered actually computing explicitly computing the inverse and showing that I can find $y_n \in \ell^2$ such that $||(T-I)^{-1} y_n||$ grows arbitrarily large (with $||y_n|| = 1$). Here was my attempt, but it doesn't seem correct.

Since $(T-I)x = (-x_1, x_1 - x_2, x_2, - x_3, ...)$, if we have $(T-I)^{-1} y = x$, then we know that $y_1 = -x_1$. We next see that $y_2 = x_1 - x_2$, so $x_2 = x_1 - y_2 = -y_ 1- y_2$. Next, we have $y_3 = x_2 - x_3$, so $x_3 = x_2 - y_3 = -y_1 - y_2 - y_3$. This pattern seems to repeat, so it appears that I find that the $j$-th entry of $(T-I)^{-1} y = \sum_{i=1}^j -y_i$. If I try taking the norm of this, I find $||(T-I)^{-1}y||^2 = \sum_{j=1}^{\infty} ( \sum_{i=1}^j -y_i)^2$.

My guess from here was to consider $y_k$ to be the sequence $(\frac{1}{n}$ where $n$ goes from $1$ to $k$ and zeros otherwise. Then, I have $y_k$ is bounded above by $\sqrt{\frac{\pi^2}{6}}$ in the norm (summing the series $\sum \frac{1}{n^2}$), but when I compute $||(T-I)^{-1}y_k||^2 = \sum_{j=1}^k (\sum_{i=1}^j -\frac{1}{i} )^2$, each component has a harmonic series term, so the norm of this thing should be growing very large and get larger as $k \to \infty$. So, even though $||y_k||^2 \leq \frac{\pi^2}{6}$, the norms of $(T-I)^{-1} y_k$ grow arbitrarily large, so $(T-I)^{-1}$ is unbounded.

Am I on the right track at least? If anyone has any suggestions on a better way to approach the problem, that would be very helpful. Is there a "general" approach to showing that a value is in the continuous spectrum? If anyone has a suggestion on a textbook that discusses these problems, that would also be quite helpful.

Thanks!

-
Every number $|\lambda|<1$ is an eigenvalue of $T^*$, and since the spectrum is compact, $T^*-I$ is not invertible. –  user31373 Jul 17 '12 at 1:31
Any elegant argument will require additional knowledge. See for example Leonid's comment –  Norbert Jul 17 '12 at 1:37
You might show that $T-I$ is not surjective because your sequence $y_k=1/k$ is not in the range. –  Jochen Wengenroth Jul 17 '12 at 10:24