2

I want to solve the following problem, but I am stuck...

Let $a=(a_n)_n \in \ell_\infty $. Define the linear operator $ \displaystyle T : \ell_1 \to \ell_1 $ by $\displaystyle T(x) = a \cdot x , \quad \text{i.e.} \quad T(x)= (a_nx_n)_n $, for $\displaystyle x=(x_n)_n \in \ell_1 $. Prove that:

(i) If $\displaystyle \text{Ker}(T) =\{0\}$ then $T(\ell_1)$ is dense in $\ell_1$

(ii) $\displaystyle T(\ell_1) = \ell_1 $ if and only if $\displaystyle \inf_{n \in \mathbb N} |a_n | > 0 $

I have prove that $\displaystyle \text{ker}(T)=\{0\} $ if and only if $ a_n \neq 0 \quad \forall n \in \mathbb N$.

Any solution or hints would be appreciated!

Thank you!

passenger
  • 3,913
  • 4
  • 27
  • 44
  • If $a_n=0$, then both the sequences $(0,0,....1,0,0...)$ (1 in the nth place) and $(0,0,...0,0...)$ are both mapped to the same thing by $T$.. So what can you say about the kernel in this case? – Arkady Dec 10 '13 at 17:39
  • 1
    @BaronVT: No, I prefer a solution without such tools.. – passenger Dec 10 '13 at 17:44
  • @user30535: I don't understand about which question youare talking.. – passenger Dec 10 '13 at 17:46

1 Answers1

3

To prove that $\mbox{ker}(T) = \{0\}$ if and only if $a_n \neq 0$ for all $n$, you can argue as in the finite-dimensional case:

  • Suppose first $a_k = 0$ for some $k$ and let $e_k = (0,...,0,1,0,...)$ be the sequence whose $k$-th term is $1$ and whose other terms are $0$. Then $Te_k = (0,...,0,a_k,0,...) = 0$ and thus $\mbox{ker}(T) \neq \{0\}$.
  • On the other hand, suppose $a_n \neq 0$ for all $n$. Let $x = (x_k), y = (y_k) \in \ell^1$. If $Tx = Ty$, then $a_kx_k = a_ky_k$ for all $k$. But $a_k \neq 0$ so $x_k = y_k$ for all $k$ and hence $x=y$. Thus $T$ is injective and its kernel is trivial.

Knowing that $a_n \neq 0$ for all $n$, we see that $\frac{1}{k}e_k \in \ell^1$ and $T(\frac{1}{k}e_k) = e_k$. Thus the range of $T$ contains the set $$\mbox{span} \{e_k: k \in \mathbb{N} \}$$ which is a dense set. This proves the first assertion.

For the second, notice that $$\|Tx\|_1 \geq \inf_{k} |a_k| \|x\|_1$$ so that $T$ is bounded from below if $\inf_{k} |a_k| > 0$. Therefore, it has closed range. Since the range is also dense, it is equal to all of $\ell^1$.

For injective $T$ (which is the case here), the answer linked to above also shows that if the range is closed, then the operator is bounded from below. Thus if $T$ is surjective, there exists a $c > 0$ such that $$|a_k| = \|a_ke_k\|_1 = |\|Te_k\|_1 \geq c\|e_k\|_1 = c$$ for all $k$. If follows that $\inf_{k} |a_k| \geq c > 0$.

Community
  • 1
  • Thank you for your answer, but I know this. As I wrote in the begging I have a proof for this. I am stuck with the questions in the black box... – passenger Dec 10 '13 at 17:47
  • @passenger Sorry, I misread your "I have prove" as "I have to prove." I will edit the answer to address your concern. –  Dec 10 '13 at 17:48
  • Thank's! I have just two questions about (ii): (1) for the direction $(\leftarrow)$: the fact that the range is closed doesn't follow from the fact that $T$ is continuous (as bounded linear operator) ? (2) In the other direction, could explain me the fist equality in the equation $|a_k| = |Te_k|_1 \geq c|e_k|_1 = c$ ? And how this equation follws from the fact that $T$ is surjective ? – passenger Dec 10 '13 at 18:15
  • As for your question (1): no it does not. There are counterexamples in many books. For (2): regarding the equality, I've edited in an intermediate step. Regarding the second question, see the link. It shows that if $T$ has closed range (so in particular if it is surjective) then there exists some $c$ such that $|Tx| \geq c|x|$. –  Dec 10 '13 at 18:21
  • O.k Now it was me who misread closed graph insted of closed range...! Yes of course you are right... Thank you very much for your time! Your explanation, was very clear! – passenger Dec 10 '13 at 18:30