1. ## Bounded linear operator

Question
Show that the range $\mathcal{R}(T)$ of a bounded linear operator $T: X \rightarrow Y$ is not necessarily closed.
Hint: Use the linear bounded operator $T: l^{\infty} \rightarrow l^{\infty}$ defined by $(\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j})$.

Attempt to solution

My idea was to find an element $y \in l^{\infty}$ that does not belong to the range and then try to build a convergent sequence in $\mathcal{R}(T)$ that has limit $y$. The element $y = (1, 1, \ldots)$ satisfy the criteria because $T^{-1}y = \{ x\}$, with $x = (\xi_{j}), \xi_{j} = j$, but, clearly, $x \not\in l^{\infty}$, therefore, $y \not\in \mathcal{R}(T)$. The problem arise when I try to build the sequence, because $(T x_{m})$ with $x_{m} \in l^{\infty}$ cannot converge to $y$. Briefly, my problem is that I canīt find a limit point of $\mathcal{R}(T)$ that doesnīt belong to $\mathcal{R}(T)$.

2. Originally Posted by kinkong
Question
Show that the range $\mathcal{R}(T)$ of a bounded linear operator $T: X \rightarrow Y$ is not necessarily closed.
Hint: Use the linear bounded operator $T: l^{\infty} \rightarrow l^{\infty}$ defined by $(\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j})$.

Attempt to solution

My idea was to find an element $y \in l^{\infty}$ that does not belong to the range and then try to build a convergent sequence in $\mathcal{R}(T)$ that has limit $y$. The element $y = (1, 1, \ldots)$ satisfy the criteria because $T^{-1}y = \{ x\}$, with $x = (\xi_{j}), \xi_{j} = j$, but, clearly, $x \not\in l^{\infty}$, therefore, $y \not\in \mathcal{R}(T)$. The problem arise when I try to build the sequence, because $(T x_{m})$ with $x_{m} \in l^{\infty}$ cannot converge to $y$. Briefly, my problem is that I canīt find a limit point of $\mathcal{R}(T)$ that doesnīt belong to $\mathcal{R}(T)$.
That's the right idea. How about something like defining the $m^{\text{th}}$ element of your sequence has $1$ in the first $m$ positions and zero afterwards-- i.e. $(1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots, 0,\cdots)$. This is clearly in $\text{im}(T)$ since it is equal to $T(1,\cdots,m,0,\cdots)$ and it converges to $(1,\cdots,1,\cdots)$

3. Originally Posted by Drexel28
How about something like defining the $m^{\text{th}}$ element of your sequence has $1$ in the first $m$ positions and zero afterwards-- i.e. $(1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots, 0,\cdots)$. This is clearly in $\text{im}(T)$ since it is equal to $T(1,\cdots,m,0,\cdots)$ and it converges to $(1,\cdots,1,\cdots)$
The trouble there is that $(1,\ldots,1,0,\ldots)$ does not converge to $(1,\ldots,1,\ldots)$ in the space $l^\infty$ (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking $y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr).$ Then $(1,\ldots,\tfrac1{\sqrt m},0,\ldots,0,\ldots)$ is in $\text{im}(T)$ since it is equal to $T(1,\ldots,\sqrt m,0,\ldots)$, and it does converge to $(1,\ldots,\tfrac1{\sqrt m},\ldots)$ in the $l^\infty$ norm.

4. Originally Posted by Opalg
The trouble there is that $(1,\ldots,1,0,\cdots)$ does not converge to $(1,\ldots,1,\ldots)$ in the space $l^\infty$ (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking $y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr).$ Then $(1,\cdots,\tfrac1{\sqrt m},0,\cdots,0,\cdots)$ is in $\text{im}(T)$ since it is equal to $T(1,\ldots,\sqrt m,0,\ldots)$' and it does converge to $(1,\ldots,\tfrac1{\sqrt m},\ldots)$ in the $l^\infty$ norm.
You're right, that was a stupid mistake.