I have to solve this problem.
Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by
$$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\}
$$
Prove that $T$ is compact.
My attempt is the following.
I have considered the operator $$
T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}.
$$
$T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact.
Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have:
$$
||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}
$$
and, given $\epsilon>0$, there exists $\bar n$ such that $|\lambda_n|<\epsilon$, for every $n>\bar n$. So if we take $N>\bar n$, we get
$$
\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon
$$
Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact.
My questions are:
(1) is the proof of the compactness of the operator $T_N$ correct?
(2) is the conclusion correct?
Thanks
Best Answer
Yes, it is correct. The operators $T_N$ are of finite rank (hence compact) and converge to $T$. Thus, $T$ is compact as well.