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For an exercise I need to investigate the boundedness for an operator and its inverse. Concretely, the exercise states :

Consider the linear mapping $A : l^2 \rightarrow l^2$ defined by

$A : \{x_1, x_2, x_3, ...\} \rightarrow \left \{ \frac{x_1}{1}, \frac{x_2}{2}, \frac{x_3}{3} \right\}$ i.e. $A : \{x_n\}_{n \in \mathbb{N}} \rightarrow \left \{ \frac{x_n}{n} \right \}_{n \in \mathbb{N}}$

  1. Show that $A$ is bounded with $\left\| A \right\| = 1$ .

  2. Show that $A$ is injective and determine its inverse, i.e. the operator $A^{-1} : \mathcal{R}(A) \rightarrow l^2$ such that $A^{-1} A x = x$, $\forall x \in l^2$

  3. Show that $A^{-1} : \mathcal{R}(A) \rightarrow l^2$ is unbounded.

My thoughs so far, please correct me where I am wrong.

Task 1 : In order to show boundedness, we need to show that $\left\| Ax \right\|_{l^2} \leq C \cdot \left\| x \right\|_{l^2}$. To this, I would proceed as follows

We can write that

$\left\| Ax \right\|_{l^2}^2 = \left\| x_n / n\right\|_{l^2}^2 = \sum_{n = 1}^{\infty} |x_n / n|^2 = \sum_{n = 1}^{\infty}( |x_n|^2 \cdot 1/n^2) \leq \sup 1/n^2 \cdot \sum_{n = 1}^{\infty} |x_n|^2$

Accordingly, $\left\| Ax \right\|_{l^2} \leq C \cdot \left\| x \right\|_{l^2}$ and hence $A$ is bounded with $C = 1$.

For the operator norm I would compute

$\left\| A \right\| = \sup \frac{\left\| Ax \right\|_{l^2}}{\left\| x \right\|_{l^2}} = \frac{C \cdot \left\| x \right\|_{l^2}}{\left\| x \right\|_{l^2}} = C = 1$

Is this correct?

Task 2 : In order to show injectivity, I would simply show that $\mathcal{N}(A) = {0}$. This was a theorem in our class. In fact, I would say that this is obvious by inspection since only the zero-vector will be mapped to zero and nothing else.

Accordingly, we now know that $A$ is invertible and the inverse I would compute as

$A^{-1} A x = \{n\}_{n \in \mathbb{N}} \cdot \{x_n/n\}_{n \in \mathbb{N}} = \{x_n\}_{n \in \mathbb{N}}$

Which means that the inverse is simply the continuous sequence of $n$'s.

Task 3 : In order to show unboundedness, we could now generally show that $\left\| A^{-1}x \right\|_{l^2} > C \cdot \left\| x \right\|_{l^2}$.

If my version of $A^{-1}$ is correct, then I would assume that for every $n > 0$ we can find a $C_n > 0$ s.t. this holds. Or in other words, for every $C$ there exists an index $n$ s.t. we can show unboundedness.

But how can we do that?

Remark : I am not sure whether my conclusions are right for Tasks 1 and 2. In particular though, I am interested in how we can solve Task 3.

Any help is much appreciated! :-)

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    $\begingroup$ Oh dear. $ab+cd\ne (a+c)(b+d)$; in particular $\sum_{n = 1}^{\infty}( |x_n|^2 \cdot 1/n^2) \ne \sum_{n = 1}^{\infty} 1/n^2 \cdot \sum_{n = 1}^{\infty} |x_n|^2$. $\endgroup$ Commented Jun 26, 2022 at 15:37
  • $\begingroup$ Hint: If $a_n,b_n\ge0$ then $\sum a_nb_n$\le(\sum a_n)(\sup b_n)$. $\endgroup$ Commented Jun 26, 2022 at 15:38
  • $\begingroup$ OMG. I am so sorry :/ All this typing in LaTex messed up my mind! Will correct right away, thanks! $\endgroup$
    – aladin
    Commented Jun 26, 2022 at 15:44
  • $\begingroup$ You showed $\|A\| \le 1$ but I do not see that you showed $\|A\|\ge 1 $(to conclude $\|A\|=1$) although it's easy. $\endgroup$ Commented Jun 26, 2022 at 19:27

1 Answer 1

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$$\|Ax\|^2=\sum n^{-2}|x_n|^2\le \sum |x_n|^2=\|x\|^2$$ Thus $ \|A\|\le 1.$ Next $Ae_1=e_1$ hence $\|A\|\ge 1.$ Here $e_1$ denotes the first element of the standard basis in $\ell^2.$

As $Ae_n=n^{-1}e_n$ the range of $A$ contains all the elements $e_n$ and their finite linear combinations. Actually the range of $A$ is equal $$\left\{ x\in\ell^2\,:\, \sum n^2|x_n|^2<\infty \right\}$$ The inverse operator satisfies $A^{-1}e_n=ne_n.$ We thus have $\|A^{-1}e_n\|=n=n\|e_n\|.$ Hence the inverse operator $A^{-1}$ is unbounded.

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    $\begingroup$ Thank you very much! The standard basis element example has made it perfectly clear to me! I have edited your solution since there was a small typo in the last sentence :-) $\endgroup$
    – aladin
    Commented Jun 26, 2022 at 17:19

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