# Functional Analysis, Show that the range of a bounded linear operator

1. Aug 2, 2009

### Eduardo

1. The problem statement, all variables and given/known data
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 $$(\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j})$$.

2. Relevant equations

3. The attempt at a 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 can´t find a limit point of $$\mathcal{R}(T)$$ that doesn´t belong to $$\mathcal{R}(T)$$.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 2, 2009

### Office_Shredder

Staff Emeritus
Think from the other direction. Since x needs to be bounded, the elements in Tx must decrease to zero eventually. Which means that any element that's on the boundary of the image set must also have its elements converge to zero. Can you find a sequence nk that converges to zero, such that knk is unbounded?

3. Aug 3, 2009

### Eduardo

Just, logged in to thank you, Office Shredder, for your quick and helpful answer. Wich was very nice because it was precisely what I needed, to unblock my fixed ideas o a change of approach to face the problem and let me think or find an answer instead of just give me the result inmediately. Thanks again.

4. May 24, 2011

### kinkong

who did you find the find a limit point of
R(T)
that doesn´t belong to
R(T)???