# Showing that the range of a linear operator is not necessarily closed

1. Apr 15, 2012

### SP90

1. The problem statement, all variables and given/known data

Let $T: \ell^{2} \rightarrow \ell$ be defined by

$T(x)=x_{1},\frac{1}{2}x_{2},\frac{1}{3}x_{3},\frac{1}{4}x_{4},...} Show that the range of T is not closed 3. The attempt at a solution I figure that I need to find some sequence of [itex]x_{n} \rightarrow x$ such that $T(x_{n})$ is in $\ell^{2}$ but $T(x)$ is not.

I figured that if $x=(\sqrt{1}, \sqrt{2}, \sqrt{3},...)$ then $T(x)$ is the harmonic series, which is not square summable. The problem is, I can't think of any square summable $x_{n}$ which converge to that $x$.

I have a feeling that I'm taking the wrong approach to this. The thing is, I think the operator $T(x)$ must be bounded since $k_{i}^{2} \geq \frac{1}{i^{2}}k_{i}^2$. Maybe I should be looking for some $x_{n} \rightarrow x$ such that $T(x_{n})$ converges to something other than $T(x)$.

2. Apr 15, 2012

### SP90

I thought if I took $x_{n}=(\frac{n}{1},\frac{n}{2^{2}},\frac{n}{3^{2}}...)$, that would work, since $x_{n} \rightarrow \frac{n\pi^{2}}{6}$. It's clear that $T(x_{n})$ exists for all n, but $x_{n}$ itself diverges. But then obviously the definition of the range being closed is that when $T(x_{n})$ has a limit, that limit is in the range. Obviously since $T(x_{n})$ diverges there's no criteria broken by this.

3. Apr 15, 2012

### SP90

I've solved this now, no need for a response, but if anyone ever comes across a similar problem here is the solution:

Take $x_{n}=(1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, ..., \frac{1}{\sqrt{n}}, 0, 0, 0 ,0,...)$

Obviously, as a series with a finite number of non-zero terms, it is square summable. Likewise $T(x_{n})$ also exists and is the sequence $x_{n}=(1, \frac{1}{2\sqrt{2}}, \frac{1}{3\sqrt{3}}, ..., \frac{1}{n\sqrt{n}}, 0, 0, 0 ,0,...)$

Then $T(x_{n}) \rightarrow y$ where y is the infinite sequence of these. When each term is squared and the sum taken, we get the sum of reciprocal cubes. This is bounded by the sum of reciprocal squares, so it is square summable, so the sequence $T(x_{n})$ is convergent. But $T^{-1}(y)$ is the infinite sequence of reciprocal square roots. When each term is squared, the sum of these is the harmonic series, which is not convergent. So $y$ doesn't belong to the range of $T$ (since there is no $x$ in $\ell^{2}$ such that $T(x)=y$)