por qué $r(F_1,F_2)= \sqrt{2}\;?$

Question

Considere los operadores $ F_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad F_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, $ en $(\mathbb{C}^2,\|\cdot\|_2)$ .

Por qué $$r(F_1,F_2):=\displaystyle\lim_{n\longrightarrow +\infty}\sup_{\|x\|_2=1}\bigg(\displaystyle\sum_{\substack{p+q=n\\ p,q\in \mathbb{N}}}\frac{n!}{p!q!}\|F_1^{p}F_2^{q}x\|_2^2\bigg)^{\frac{1}{2n}}= \sqrt{2}\;?$$

Tenga en cuenta que $F_1^{p}=F_1, \;\forall p\in \mathbb{N}$ y $F_2^{q}=F_2, \;\forall q\geq5$ .

Gracias.

Answer 1

El límite es efectivamente $\sqrt2$ .

Como tú dices, $F_1^p=F_1$ para todos $p\geq1$ . Para $F_2$ tenemos $$ F_2^{4k+r}=F_2^r=\begin{cases} (-1)^r F_2,&\ r\ \text{ odd}\\ \ \\ (-1)^{r/2}I,&\ r\ \text{ even}\end{cases} $$ Si $\|x\|=1$ tenemos $|x_1|^2+|x_2|^2=1$ . Así que $$ F_1^pF_2^{(4k+r)}x=\begin{cases}(-1)^{r+1}x_2,&\ r\ \text{ odd}\\ \ \\ (-1)^{r/2+1}x_1,&\ r\ \text{ even}\end{cases} $$ y $$ \|F_1^pF_2^{(4k+r)}x\|_2^2=\begin{cases}1-|x_1|^2,&\ r\ \text{ odd}\\ \ \\ |x_1|^2,&\ r\ \text{ even}\end{cases} $$ Ahora \begin {align} \sum_ {q=0}^n \frac {n!}{q!(n-q)!}\,\|F_1^pF_2^qx\|_2^2&= \sum_ {q\ \text { par}}^n{n \choose q}\N,|x_1|^2+ \sum_ {q\ \text { impar}}{n \choose q}(1-|x_1|^2) \\ \ \\ &= \sum_ {q=0}^n(-1)^q|x_1|^2+ \sum_ {q\ \text { impar}}{n \choose q} \\ \ \\ &=(1-1)^n+2^{n-1} \\ \ \\ &=2^{n-1}. \end {align} Así $$ \lim_{n\longrightarrow +\infty}\sup_{\|x\|_2=1}\bigg(\displaystyle\sum_{\substack{p+q=n\\ p,q\in \mathbb{N}}}\frac{n!}{p!q!}\|F_1^{p}F_2^{q}x\|_2^2\bigg)^{\frac{1}{2n}} =\lim_n (2^{n-1})^{\frac1{2n}}=\sqrt2. $$