Bounded linear operator

• Jun 1st 2011, 08:54 PM
kinkong
Bounded linear operator
Question
Show that the range $\displaystyle \mathcal{R}(T)$ of a bounded linear operator $\displaystyle T: X \rightarrow Y$ is not necessarily closed.
Hint: Use the linear bounded operator $\displaystyle T: l^{\infty} \rightarrow l^{\infty}$ defined by $\displaystyle (\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j})$.

Attempt to solution

My idea was to find an element $\displaystyle y \in l^{\infty}$ that does not belong to the range and then try to build a convergent sequence in $\displaystyle \mathcal{R}(T)$ that has limit $\displaystyle y$. The element $\displaystyle y = (1, 1, \ldots)$ satisfy the criteria because $\displaystyle T^{-1}y = \{ x\}$, with $\displaystyle x = (\xi_{j}), \xi_{j} = j$, but, clearly, $\displaystyle x \not\in l^{\infty}$, therefore, $\displaystyle y \not\in \mathcal{R}(T)$. The problem arise when I try to build the sequence, because $\displaystyle (T x_{m})$ with $\displaystyle x_{m} \in l^{\infty}$ cannot converge to $\displaystyle y$. Briefly, my problem is that I canīt find a limit point of $\displaystyle \mathcal{R}(T)$ that doesnīt belong to $\displaystyle \mathcal{R}(T)$.
• Jun 1st 2011, 09:25 PM
Drexel28
Quote:

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

Attempt to solution

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

That's the right idea. How about something like defining the $\displaystyle m^{\text{th}}$ element of your sequence has $\displaystyle 1$ in the first $\displaystyle m$ positions and zero afterwards-- i.e. $\displaystyle (1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots, 0,\cdots)$. This is clearly in $\displaystyle \text{im}(T)$ since it is equal to $\displaystyle T(1,\cdots,m,0,\cdots)$ and it converges to $\displaystyle (1,\cdots,1,\cdots)$
• Jun 1st 2011, 11:50 PM
Opalg
Quote:

Originally Posted by Drexel28
How about something like defining the $\displaystyle m^{\text{th}}$ element of your sequence has $\displaystyle 1$ in the first $\displaystyle m$ positions and zero afterwards-- i.e. $\displaystyle (1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots, 0,\cdots)$. This is clearly in $\displaystyle \text{im}(T)$ since it is equal to $\displaystyle T(1,\cdots,m,0,\cdots)$ and it converges to $\displaystyle (1,\cdots,1,\cdots)$

The trouble there is that $\displaystyle (1,\ldots,1,0,\ldots)$ does not converge to $\displaystyle (1,\ldots,1,\ldots)$ in the space $\displaystyle l^\infty$ (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking $\displaystyle y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr).$ Then $\displaystyle (1,\ldots,\tfrac1{\sqrt m},0,\ldots,0,\ldots)$ is in $\displaystyle \text{im}(T)$ since it is equal to $\displaystyle T(1,\ldots,\sqrt m,0,\ldots)$, and it does converge to $\displaystyle (1,\ldots,\tfrac1{\sqrt m},\ldots)$ in the $\displaystyle l^\infty$ norm.
• Jun 2nd 2011, 12:04 AM
Drexel28
Quote:

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

You're right, that was a stupid mistake.