why this limit is equal to $\sqrt{2}\;?$ Consider the matrices
$
F_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad
F_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
$ on $(\mathbb{C}^2,\|\cdot\|_2)$.

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

Note that $F_1^{p}=F_1, \;\forall p\in \mathbb{N}$ and $F_2^{q}=F_2, \;\forall q\geq5$.
Thanks!
 A: The limit is indeed $\sqrt2$. 
As you say, $F_1^p=F_1$ for all $p\geq1$. For $F_2$, we have 
$$
F_2^{4k+r}=F_2^r=\begin{cases} (-1)^r F_2,&\ r\ \text{ odd}\\ \ \\ (-1)^{r/2}I,&\ r\ \text{ even}\end{cases}
$$
If $\|x\|=1$, we have $|x_1|^2+|x_2|^2=1$. So 
$$
F_1^pF_2^{(4k+r)}x=\begin{cases}(-1)^{r+1}x_2,&\ r\ \text{ odd}\\ \ \\  (-1)^{r/2+1}x_1,&\ r\ \text{ even}\end{cases}
$$
and
$$
\|F_1^pF_2^{(4k+r)}x\|_2^2=\begin{cases}1-|x_1|^2,&\ r\ \text{ odd}\\ \ \\  |x_1|^2,&\ r\ \text{ even}\end{cases}
$$
Now
\begin{align}
\sum_{q=0}^n\frac{n!}{q!(n-q)!}\,\|F_1^pF_2^qx\|_2^2&=\sum_{q\ \text{ even}}^n{n\choose q}\,|x_1|^2+\sum_{q\ \text{ odd}}{n\choose q}(1-|x_1|^2)\\ \ \\
&=\sum_{q=0}^n(-1)^q|x_1|^2+\sum_{q\ \text{ odd}}{n\choose q}\\ \ \\
&=(1-1)^n+2^{n-1}\\ \ \\
&=2^{n-1}.
\end{align}
Thus
$$
\lim_{n\longrightarrow +\infty}\sup_{\|x\|_2=1}\bigg(\displaystyle\sum_{\substack{p+q=n\\ p,q\in \mathbb{N}}}\frac{n!}{p!q!}\|F_1^{p}F_2^{q}x\|_2^2\bigg)^{\frac{1}{2n}}
=\lim_n (2^{n-1})^{\frac1{2n}}=\sqrt2.
$$
