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?



1 Answer 1


Yes, it is correct. The operators $T_N$ are of finite rank (hence compact) and converge to $T$. Thus, $T$ is compact as well.

  • 1
    $\begingroup$ Intuitively, I think it can be said that this is an if and only if statement, i.e. T is compact iff $lim_{n \rightarrow \infty} \rightarrow 0$. Do you have any thoughts on how such a thing could be proved? I'm at a loss trying to come up with an adequate proof $\endgroup$ Commented May 9, 2018 at 19:36

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