For a sequence of operators $A_n$, determine the type of convergence (strong, uniform, weak, or absent) and find the limit For sequences of operators $A_n \in B(l_p), p \in [1,\infty)$, $A_n(x_1,x_2,\ldots) = (\lambda_{1,n}x_1, \lambda_{2,n}x_2, \ldots)$, , where $|\lambda_{k,n}| < C$ and $\lambda_{k,n}\rightarrow\lambda_{k} $  if $n\rightarrow\infty$ for any $k \in \mathbb{N}$( that is, the vectors $\lambda_n = (\lambda_{1,n}, \lambda_{2,n}, ...)$ *-weakly converge to $\lambda_n = (\lambda_{1}, \lambda_{2}, ...)$ in $l_{\infty}$) determine the type of convergence and find the limit.
I proved strong convergence. And I got $A_n\rightarrow A$ where $A(x_1,x_2,\ldots) = (\lambda_{1}x_1, \lambda_{2}x_2, \ldots)$.
From the strong convergence followed the weak convergence.
I assume that there is no uniform convergence. But I can't show it.
 A: I'm assuming you mean to disprove convergence in the operator norm.
For that, it is enough to find a sequence of non-zero vectors $\{x^{(n)}\}_{n=1}^\infty \subseteq \ell^p$ such that
$$ \big\Vert A_n(x^{(n)})- A_\infty(x^{(n)}) \big \Vert_p \geq \Vert x^{(n)} \Vert_p \quad \text{for all} \quad n\in \mathbb{N}. $$
If you take $x^{(n)}=e_n$, the sequence with $1$ only at index $n$ and $0$ otherwise, you get that
$$ \big\Vert A_n(e_n)- A_\infty(e_n) \big \Vert_p = \vert \lambda_{n,n} -\lambda_n \vert. $$
Just find sequences $\{ \lambda^{(n)} \}_{n=1}^\infty \subseteq \ell^p$  converging pointwise to $(\lambda_k)_{k=1}^\infty \subseteq \ell^p$  such that $\vert \lambda_{n,n} -\lambda_n \vert \geq 1$.
A: Let $$A_\lambda(x)=\{\lambda_nx_n\},\quad x\in \ell^p,\ \lambda\in\ell^\infty$$ Then $$\|A_\lambda-A_\mu\|=\|A_{\lambda-\mu}\|=\|\lambda-\mu\|_\infty$$ Let $$\lambda^{(n)}_k=\begin{cases} 1& 1\le k\le n\\ 0 & k> n
\end{cases}$$ Then $\lambda^{(n)}$ tends pointwise  to the sequence with all terms equal $1.$ The sequence of operators $A_{\lambda^{(n)}}$ does not satisfy the Cauchy condition as $$\|A_{\lambda^{(n+1)}}-A_{\lambda^{(n)}}\|=\|\lambda^{(n+1)}- \lambda^{(n)}\|_\infty=1$$ Hence it is not convergent with respect to the operator norm.
