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

[Eduardo]

Homework Statement

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}).

Homework Equations

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).

 

[Office_Shredder]

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 n_k that converges to zero, such that kn_k is unbounded?

 

[Eduardo]

Just, logged into 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.

 

[kinkong]

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