Suppose that $\ell^2 = \biggl\{(x_n)_n \in \mathbb{K}^{\mathbb{N}_0} \biggm| \sum_{n=1}^{\infty}|{x_n}^2| < +\infty \biggr\}$ is a Hilbert-space with the inproduct $\langle\cdot,\cdot\rangle_2: \ell^2 \to \ell^2: (x,y) \mapsto \sum_{n=1}^\infty \overline{x_n}y_n$.

Consider the operator $f: \ell^2 \to \ell^2: (x_0, x_1, \ldots) \mapsto (x_0, 0, x_1, 0, \ldots)$.

I'm supposed to give the spectrum $\sigma(f) = \{\lambda \in \mathbb{K} \mid f-\lambda I \text{ not invertible}\}$, where $I$ is the identical function. I've already shown that 1 is the only eigenvalue of $f$, so 1 should be part of $\sigma(f)$, because $f-I$ isn't injective. I also think that $f$ isn't surjective, so 0 should be in the spectrum too.

Unfortunately, I didn't find a way to calculate the whole spectrum, although I'm having the feeling that it shouldn't be that difficult. How can I do this?


Let us show that the spectrum $\sigma (T)$ of $T$ is the closed unit disk $\overline{\mathbb D}$.

As suggested by user52733, it is a good idea to look at the (Hilbertian) adjoint operator $T^*$. Since $\sigma(T^*)=\{ \bar\lambda ;\; \lambda\in\sigma (T)\}$, it is enough to show that $\sigma (T^*)=\overline{\mathbb D}$.

Let us denote by $(e_n)_{n\geq 0}$ the "canonical basis" of $\ell^2$. By the definition of $T$, you have $Te_0=e_0$, $Te_1=e_2$, $Te_2=e_4$ and so on, i.e. $Te_n=e_{2n}$ for all $n\geq 0$. It follows that $$\langle T^*e_n,e_j\rangle=\langle e_n, Te_j\rangle=\langle e_n, e_{2j}\rangle=\delta_{n,2j}$$ for all $n,j\geq 0$; in other words, $T^*e_n=0$ if $n$ is odd and $T^*e_n=e_{\frac{n}2}$ if $n$ is even. Phrased differently, this means that $T^*$ is given by the formula $$ T^*(x_0,x_1,x_2,\dots )=(x_0,x_2,x_4, \dots)\, .$$

From this, it is easy to show that any complex number $\lambda$ with $\vert \lambda\vert<1$ is an eigenvalue of $T^*$. Indeed, if you set $$x(\lambda):=\sum_{j=0}^\infty \lambda^j e_{2^j}\, ,$$ which is a well defined vector in $\ell^2$ because $\vert\lambda\vert<1$, then $$T^*x(\lambda)=\underbrace{T^*(e_1)}_{=0}+\sum_{j\geq 1} \lambda^jT^*(e_{2^j})=\sum_{j\geq 1} \lambda^j e_{2^{j-1}}=\lambda x(\lambda)\, .$$ (Note that the formula for $x(\lambda)$ does not come from nowhere: you find it when you try to solve the equation $Tx=\lambda x$).

So, the spectrum of $T^*$ contains the open unit disk $\mathbb D$. But $\sigma(T^*)$ is a closed set, so it must in fact contain the closed unit disk $\overline{\mathbb D}$. Finally, $\sigma (T^*)$ is also contained in $\overline{\mathbb D}$ because $\Vert T^*\Vert= 1$.

Altogether, one can conclude that $\sigma(T^*)=\overline{\mathbb D}$ and hence that $\sigma (T)=\overline{\mathbb D}$.


Edit: As noted by @Branimir, I can remove the conjugations from the earlier draft. $T^*$ is the Banach-space adjoint conjugated by a conjugate-linear isomorphism, so constants end up being preserved.

As this is a homework problem, I will attempt to give a hint rather than the complete solution (which is elegant). I here adapt p. 192-194 of (the 1972 edition of) Reed and Simon, Functional Analysis, so that may be a good reference [in particular, I exchange $T$ and $T$'s adjoint, and change the underlying spaces, from that book].

This operator is qualitatively similar to the "stretching" shift operator $T: \ell^2 \to \ell^2$, where $$T(x_0, x_1, x_2, \dotsc) = (0, x_0, x_1, x_2, \dotsc).$$

It helps for this operator to look at its adjoint, which we can determine by just using the definition of adjoint and looking at its effects on the $j$th basis vector $e_j$, where

$$e_j := (0, 0, \dotsc, 0, \overbrace{1}^{j\text{th slot}}, 0, 0, \dotsc ).$$

Note that for any $x$, $$ \begin{align} \left\langle T^* e_j, x \right\rangle & = \left\langle e_j, T x \right\rangle\\ & = (Tx)_j\\ & = \begin{cases} 0, & j = 0, \\ x_{j - 1}, & j \neq 0 \end{cases} \end{align}. $$

Therefore, $\left\langle T^* e_0, x \right\rangle = 0$ for all $x$, so $T^* e_0 = 0$, and letting $x$ range over $\left\lbrace e_k \right\rbrace$, we see that $T^* e_j = e_{j-1}$ if $j \geq 1$. Since the adjoint is a linear operator, we have $$ T^* (x_0, x_1, x_2, \dotsc ), = ( x_1, x_2, x_3, \dotsc )$$ which is a "squishing" shift operator.

Why do we like squishing shift operators? I can hope to describe some eigenvectors. Here, take $\lambda$ with $\vert \lambda \vert < 1$, and define $$x_{\lambda} : = (1, \lambda, \lambda^2, \lambda^3, \lambda^4, \dotsc)$$ By $\vert \lambda \vert < 1$, and the $j$-th entry having norm $\vert \lambda \vert^j$, this is an $\ell^2$ (indeed, $\ell^1$ !) vector. Then convince yourself that $T^*(x_{\lambda}) = \lambda x_{\lambda}$. So $\sigma(T^*)$ contains the open unit disk. Yet spectra are closed, so it contains the closed unit disk. Since $\lambda \in \sigma(T^*)$ if and only if $\overline{\lambda} \in \sigma(T)$, and since the closed unit disk is its own image under conjugation, $\sigma(T)$ contains the closed unit disk.

Now argue with norms and the spectral radius rules that $\sigma(T)$ is contained in the closed unit disk, and you're done.

Try this idea with your operator.

  • $\begingroup$ The adjoint of a linear operator is still linear, not conjugate-linear; in particular, the adjoint of the right shift operator $S : (a_0,a_1,a_2,\cdots) \mapsto (0,a_0,a_1,\cdots)$ is the left shift operator $S^\ast : (a_0,a_1,a_2,\cdots) \mapsto (a_1,a_2,a_3,\cdots)$. $\endgroup$ – Branimir Ćaćić Jun 11 '14 at 23:33
  • $\begingroup$ @BranimirĆaćić, you are right, I will fix this. $\endgroup$ – Charles Baker Jun 11 '14 at 23:45
  • $\begingroup$ @Lost1 , As far as I can tell, your equation for $i > 0$ even is $x_{i/2} - \lambda x_i = y_i$, (not $y_{i - 1}$). Therefore, you get $x_1 - \lambda x_2 = y_2$, so $x_2 = (x_1 - y_2 )/\lambda$. Yet $x_1 = - y_1 / \lambda$, so $x_2 = - y_1 / \lambda^2 - y_2 / \lambda$. Similarly, I think you get $x_4 = - \frac{y_1}{\lambda^3} - \frac{y_2}{\lambda^2} - \frac{y_4}{\lambda_1}$. In general, I think you should separate the indices $\left\lbrace 1, 2, 4, 8, ... \right\rbrace$, $\left\lbrace 3 * 1, 3 * 2, 3 * 4 \right\rbrace$, etc. $\endgroup$ – Charles Baker Jun 12 '14 at 0:26
  • $\begingroup$ contd. and then you can use Cauchy-Schwartz on each piece to get a decent bound, I think, assuming only $\lambda > 1$. [Now that I think about it, though, my method may not be "enough."] $\endgroup$ – Charles Baker Jun 12 '14 at 0:32
  • 1
    $\begingroup$ @user52573: Dear user, It is a general principal that if $|\lambda| > || T||$ (the operator norm of the operator $T$), then $\lambda I - T$ is invertible. (Just take the usual geometric series for $\lambda^{-1}(I - \lambda^{-1} T).$) In the case in hand, we have $||T|| = 1$, and so certainly if $|\lambda| > 1$ then $\lambda$ is not in the spectrum. Regards, $\endgroup$ – Matt E Jun 12 '14 at 1:01

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