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Consider the operator $T: \ell^2 \to \ell^2$ defined as $$\begin{cases} (Tx)_1 = 0, \\ (Tx)_n = -x_n + \alpha x_{n+1}, \quad n\ge 2 \end{cases} $$ where $\alpha \in \mathbb{C}$.

I want to find the norm $\|T\|$.


The best constraint I found is $$ \sqrt{1+|\alpha|^2} \le \|T\| \le 1+|\alpha|. $$ To do this I considered that

  1. $$\|T\|^2=\sum_{n=2}^\infty |-x_n + \alpha x_{n+1}|^2 \le 2 \sum_{n=2}^\infty (|x_n|^2+|\alpha|^2 |x_n|^2) \le 2(1+|\alpha|^2) \|x\|^2$$ hence $\|T\| \le \sqrt{2(1+|\alpha|^2)}$. This is actually worst that the bound found below so not very useful.
  2. $$x=(0,0,1+\alpha,0,...) \Longrightarrow Tx=(1+\alpha)(0,\alpha,-1,0,...)$$ hence $\|T\| \ge \sqrt{1+|\alpha|^2}$.

  3. Thinking of $T$ as sum of $T_1$ and $T_2$ defined respectively as $$\begin{cases} (T_1 x)_1 = 0, \\ (T_1 x)_n = -x_n, \quad n \ge 2, \end{cases}$$ and $$\begin{cases} (T_2 x)_1 = 0, \\ (T_2 x)_n = \alpha x_n, \quad n \ge 2, \end{cases}$$ noting that $\|T_1\|=1$ and $\|T_2\|=|\alpha|$, we conclude from the triangle inequality that $$\|T\| \le 1 + |\alpha|.$$ This constraint is better than the above as is easily seen in the plot below (where $x=|\alpha|$): enter image description here

What can I use to obtain the exact result?

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The "geometric" sequence $x_1=0, x_2=1$, $x_{n+1}=qx_n, n\ge2$, is in $\ell^2$ iff $|q|<1$. For this choice we get $(y_n)=T(x_n)$ where for all $n\ge2$ we have $$ y_n=-x_n+\alpha x_{n+1}=x_n(-1+q\alpha). $$ In other words $(x_n)$ is an eigensequence of $T$ belonging to the eigenvalue $\lambda=\lambda(q,\alpha):=-1+q\alpha$. This implies that we can make $|\lambda|$ arbitrarily close to $1+|\alpha|$, which then has to be the norm.

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  • $\begingroup$ Thanks for the answer! Did you mean to write $n \ge 2$? Otherwise the whole sequence seems to be zero... $\endgroup$
    – glS
    Jun 15 '15 at 14:29
  • $\begingroup$ wonderful method by the way! $\endgroup$
    – glS
    Jun 15 '15 at 14:49
  • $\begingroup$ Correct, @glance. It should have been $n\ge2$ all along. Fixed now. Wonderful method? You did half the work in getting the upper bound. Geometric sequences are simply my go-to example with operators like this. Often get eigenvalues, or failing that, can at least calculate the norms explicitly and consequently get a tighter lower bound for the operator norm. $\endgroup$ Jun 15 '15 at 19:55
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Write $\alpha = re^{i\varphi}$ with $r \geqslant 0$ and $\varphi \in [0,2\pi)$. Let $\lambda = -e^{-i\varphi}$. Then consider

$$x_n = \sum_{k=2}^{n+1} \lambda^k e_k.$$

We have $\lVert x_n\rVert = \sqrt{n}$, and

$$T(x_n) = \sum_{k=2}^n (\alpha \lambda^{k+1} - \lambda^k )e_k - \lambda^{n+1}e_{n+1} = - \sum_{k=2}^n (1+r)\lambda^k e_k - \lambda^{n+1}e_{n+1},$$

whence

$$\lVert T(x_n)\rVert > \sqrt{n-1}(1+r).$$

It follows that

$$\lvert T\rVert \geqslant \frac{\lvert T(x_n)\rVert}{\lVert x_n\rVert} > \sqrt{1-\frac{1}{n}}(1+r)$$

for all $n \geqslant 2$, so $\lVert T\rVert = 1+r = 1 + \lvert \alpha\rvert$.

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  • $\begingroup$ Thank you very much! I'll accept the other answer just because I have to choose one of the two $\endgroup$
    – glS
    Jun 15 '15 at 15:19

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