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Let $(\lambda_n)$ be a sequence of non-zero scalers and let

$D(T)= \{x=(\epsilon_j) \in l^2 : \sum^\infty_{j=1} |\lambda _j |^2 |\epsilon _j |^2 <\infty \}$

We define a linear operator $D(T) \to Ran(T)$, $D(T) \in l^2$ and $Ran(T) \in l^2$ , as

$$Tx= T(\epsilon_j)^\infty_1=(\lambda _j \epsilon _j)^\infty _1$$ where $x=(\epsilon_j) \in D(T)$

I am trying to show that $T$ is bounded if and only if $\sup|\lambda_j| \leq \infty$

This is the proof I have: For any $x \in l^2$ we have $||Tx||^2 = \sum_{j} |\lambda _j | |\epsilon_j|^2 \leq (\sup_{j} |\lambda_j|)^2 ||x||^2 $

So $$||T|| \leq \sup_j |\lambda_j|$$

I understand the above but I dont understand the remaining proof below. In the other direction $||T || \geq \sup_j ||Te_j|| = \sup _j |\lambda _j |$

Where does this last bit come from?

What is $e_j$? and why do we require it?

Al jabra
  • 2,331

1 Answers1

4

For $j\in \mathbb{N}$ we write

$$ e_j= (\underbrace{0, \dots, 0}_{\text{j-1 times}}, 1, 0, \dots)\in D(T) .$$

Then

$$ \Vert e_j \Vert_{l^2} = \vert 1 \vert =1 $$

and hence

$$ \Vert T \Vert_{l^2\rightarrow l^2} = \sup_{\Vert x \Vert_{l^2}\leq 1, \ x\in D(T)} \Vert T(x) \Vert_{l^2} \geq \sup_{j\in \mathbb{N}} \Vert T(e_j)\Vert_{l^2} = \sup_{j\in \mathbb{N}} \Vert \lambda_j e_j\Vert_{l^2} = \sup_{j\in \mathbb{N}} \vert \lambda_j \vert.$$

Severin Schraven
  • 19,179