# Bounded Operator with unbounded Inverse

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! :-)

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

$$\|Ax\|^2=\sum n^{-2}|x_n|^2\le \sum |x_n|^2=\|x\|^2$$ 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.