Example: Check if the operator converges strongly: Let $X=L_p(\mathbb{R})\:,\:p\geqslant1$
$A_{h_k}x(t)=x(t+h_k)$ If $h_k\to h\:\:k\to\infty$ then $A_{h_k}x\to A_n x$
$\forall \epsilon>0,\exists\delta>0\\\int_{-\infty}^{\infty}|x(t+h_k)-x(t+h)|^p dt<\epsilon^p$ we have strong convergence.
Then the author states:
$||A_{h_k}-A_h||\nrightarrow 0\:\:k\to\infty$ and $h_k\neq h$.
Consider $|h_k-h|=d>0$
$x_0(t)=\begin{cases} \frac{1}{d}, & \mbox{ if }0<t<d\\ 0, & \mbox{ if }t\leqslant0,t\geqslant d \end{cases}$ $||x_0||_p=1$
$||(A_{h_k}-A_h)x_0||_{L_p(\mathbb{R})}=(\int_{-\infty}^{\infty}|x(t+h_k)-x(t+h)|^p dt<\epsilon^p)^{\frac{1}{p}}=2^\frac{1}{p}>1$
$||A_{h_k}-A_h||=\sup_{||x||=1}||A_{h_k}x-A_hx||$.
Question:
1) How does $||(A_{h_k}-A_h)x_0||_{L_p(\mathbb{R})}=(\int_{-\infty}^{\infty}|x(t+h_k)-x(t+h)|^p dt)^{\frac{1}{p}}=2^\frac{1}{p}>1$? I tried to compute this integral and I do not get this value.
2) Is the conclusion of this proof that $A_{h_k}$ does not converge strongly?
Thanks in advance!
1) I'm fairly certain that we should use
$$x_0(t)=\frac{1}{d^{1/p}}\chi_{(0,d)}(t)=\begin{cases} \frac{1}{d^{1/p}}, & \mbox{ if }0<t<d\\ 0, & \mbox{ if }t\leqslant0,t\geqslant d \end{cases}$$ instead of what's stated. Then you get $\|x_0\|_p=1$ and $$\|(A_{h_k}-A_h)x_0\|_{L_p(\mathbb{R})}^p=\frac{1}{d}\int|\chi_{(-h_k,d+h_k)}-\chi_{(-h,d+h)}|^p=\frac{1}{d}\int\chi_{(-d,0]}+\chi_{[d,2d)}=2$$
2) No this does not disprove strong convergence. What it shows is that for each $k$ we can find some $x_0$ such that $\|(A_{h_k}-A_h)x_0\|_{L_p(\mathbb{R})}>1,$ so that $$\|A_{h_k}-A_h\|_{L_p(\mathbb{R})}\geq\|(A_{h_k}-A_h)x_0\|_{L_p(\mathbb{R})}>1,$$ and thus $A_{h_k}$ does not converge to $A_h$ in the norm topology of $\mathcal B(L_p(\mathbb{R}))$.