Came across this exercise in a book I'm reading. The full exercise is as such:
For a sequence of real numbers $(\lambda_k)_{k=1}^\infty$, define a linear operator $T:l_2\rightarrow l_2$ by $Tx=(\lambda_kx_k)_{k=1}^\infty$. For what multiplier sequences $(\lambda_k)$ is the operator $T$: (a) well defined? (b) bounded? (c) compact?
Here, $l_2$ is the space of all square-summable sequences, such that if $\sum_{k=1}^\infty x_k^2<\infty$, then $x=(x_1,x_2,\ldots)\in l_2$ and $||x||=\sqrt{\sum_{k=1}^\infty x_k^2}$.
I'm not sure what is meant here by "well defined", my best guess is that it means to ask "when does the application of $T$ actually result in a sequence of the space $l_2$". Am I correct?
About boundedness, I noticed that if the sequence $(\lambda_k)$ is unbounded, then $T$ is also unbounded. This is easy to see: since $(\lambda_k)$ is unbounded, there exists a subsequence $(\lambda_{k_n})$, such that $\lambda_{k_n}>n$. But then for the sequence $x=(x_k)\in l_2$, $x_k=\frac{1}{k}$, we see that $||Tx||=||\sum_{k=1}^\infty\lambda_kx_k||<\sqrt{\sum_{k=1}^\infty 1}=\infty$. And so, $Tx\not\in l_2$. Similarly, we can show that when the sequence $(\lambda_k)$ is bounded, so is $T$.
I have no idea what to do about compactness. I know that an operator is compact if and only if it maps the open unit ball to a precompact (such that its closure is compact) set. I don't see how that helps.
If anyone has any ideas about well definedness and compactness, please share them. Also please check my ideas about boundedness. Thank you!