Determine a bounded linear operator.

Question

Let $\alpha = \{ \alpha_{j} \}_{j=1}^{\infty} \in l^1 $. Consider $A: l^{1} \rightarrow l^{2}, A\alpha =\alpha $. Please show $A\alpha \in l^2$ when $\alpha \in l^1$. Determine whether or not $A$ is a bounded linear operator. If so, prove it. If not, please give a counter example. Let $R_{A}$ be the range of $A$. Determine if $A^{-1}: R_{A} \rightarrow l^1$ is a bounded operator. If so prove it, if not, please also give a conter example.

It is is easy to show that $A$ is a bounded linear operator. However, for the second $A_{-1}$, I think it is not, but how could I give a counter example?

Answer 1

$A$ is bounded. Let $\alpha\in \ell^1$ be such that $\|\alpha\|_1=1$， this means $|\alpha_j|\leq 1$ for all $j\in\mathbb{N}$, thus $|\alpha_j|^2\leq |\alpha_j|$, which yields $$\|\alpha\|_2^2\leq\|\alpha\|_1=1\Rightarrow \|\alpha\|_2\leq 1=\|\alpha\|_1$$ For general non zero sequence $\alpha\in \ell^1$, consider the sequence $\alpha':=\frac{\alpha}{\|\alpha\|_1}$, then we have $\|\alpha'\|_1=1$ and by previous argument: $$\|\alpha'\|_2\leq\|\alpha'\|_1,$$ and thus yields $$\|A\alpha\|_2=\|\alpha\|_2\leq\|\alpha\|_1,\quad\forall \alpha\in\ell^1$$

$A^{-1}$ is unbounded. Let $e^{(n)}\in\ell^1\cap\ell^2$ be defined as $$e^{(n)}_k=\begin{cases}1, &\text{if }n\geq k\\ 0,&\text{otherwise}.\end{cases}$$ Then we have $\|e^{(n)}\|_1=n$ and $\|e^{(n)}\|_2=\sqrt{n}$. Assume $A^{-1}$ is bounded, then there exists a universal constant $C>0$ such that $$\|\alpha\|_1=\|A^{-1}\alpha\|_1\leq C\|\alpha\|_2,\text{ for all }\alpha\in R_A$$ in particular we have $$\|e^{(n)}\|_1\leq C\|e^{(n)}\|_2,\quad\forall n\in\mathbb{N}$$ which yields $$\sqrt{n}\leq C,\quad\forall n\in\mathbb{N}$$ which is clearly impossible. Thus $A^{-1}$ cannot be bounded.

Answer 2

Thanks guys.

1. $A$ is bounded, lets start directly form the definition. $||A\alpha||_{2} = ||\alpha||_{2} $, since $A\alpha = \alpha$. where $||\alpha||_{2} = \{\sum_{j=1}^{\infty} ||\alpha_{j}||^{2}\}^{\frac{1}{2}}\le\sum_{j=1}^{\infty} ||\alpha_{j}|| = ||\alpha||_{1}$, obviously, $A$ is bounded.
2. $A^{-1}$ is not bounded, given a manifest counter example. Regard the famous harmonic series: $\alpha =\sum_{j=1}^{\infty} \frac{1}{n}$. Therefore $||\alpha||_{2} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}$. This is bounded. Whereas the harmonic series is not bounded. Problem solved.