4

Show that the mapping $$ \Phi: \ell^{1}\rightarrow \mathcal{L}(c_{0};\mathbb{K}), \phantom{x} \Phi_{ x}(y):=\sum_{n=1}^{\infty}x_{n}y_{n}$$ is well-defined and an isometric isomorphism.

Updated my answer:

The functional

$$ x \mapsto \sum_{n=1}^{\infty}x_{n}y_{n} $$

is bounded, since

\begin{align*} |\Phi_{x}(y)| &= \left|\sum_{n=1}^{\infty} x_{n}y_{n} \right|\leq \sum_{n=1}^{\infty} |x_{n}y_{n}| = \sum_{n=1}^{\infty} |x_{n}||y_{n}| \\ &\leq \|y\|_{\infty}\|x\|_{1}. \end{align*}

Because the functional is bounded by $\|x\|_{\infty}$, the mapping

$$ \Phi: \ell^{1}\rightarrow \mathcal{L}(c_{0};\mathbb{K}),$$ defined by $$y\mapsto\left( x \mapsto \sum_{n=1}^{\infty}x_{n}y_{n} \right)$$

is well-defined. (Question 1. Is this correct?)

Note that if $|x_{j}|=\|x\|_{\infty}$ and if we take $y_{j}=e_{j}$, where $e_{j}$ is the $j$-th unit vector, then $|\Phi_{x}(e_{j}) |=1$. Thus

$$\|\Phi f\|_{\mathcal{L}(c_{0};\mathbb{K})}=\|f\|_{\ell^{1}} \phantom{x}\mathrm{for}\phantom{x}\mathrm{all}\phantom{x}f\in \ell^{1} .$$

(Question 2. Is this correct or am I on the wrong path, is there are a better way to show this? )

We claim that $\Phi$ is also surjective and hence it is an isometric isomorphism. If $\phi$ is a functional on $c_{0}$ let us denote $\phi(e_{j})$ by $x_{j}$. Then

$$\phi_{x}(y) = \sum_{n=1}^{\infty}\phi(e_{n})y_{n}=\sum_{n=1}^{\infty}\phi(y_{n}e_{n})=\phi(y),$$

(the last equality holds because $\sum_{n=1}^{\infty}y_{n}e_{n}$ converges to $y$ in $c_{0}$ and $\phi$ is continuous in $c_{0}$) (Question 3. Is this correct and how do I prove my convergence argument?)

so $\Phi(x)=\phi$.

(Question 4. Struggling the most with this part. Is what I have done correct, have I really showed that $\mathrm{ran}(\Phi)=\mathcal{L}(c_{0};\mathbb{K})$ or is there are a better way to show this? )

Lech121
  • 333
  • 1
  • 8
  • Regarding your improved argument: "well-defined" is not an issue, as there is no equivalence to take care of. For your question 2: note that on $s_N$ you want to consider the sup norm; so it is enough to see that $|\lambda_n|=1$ to convince you that $|s_N|=1$. For the inequality in question 2, you are using that previous fact and that $\psi$ is bounded. Finally, for the sake of being careful as you are trying to be, in the definition of $\lambda_n$ you need to consider the case where $\psi(e_n)=0$ for some $n$. – Martin Argerami Sep 18 '13 at 15:18
  • Thank you for the fast reply and the feedback. I have updated it yet again with your commments. If you could give me some final feedback I would really appreciate it. Did I do it correctly or have I misunderstood something? – Lech121 Sep 18 '13 at 16:38
  • Yes, I think the argument is fine now. Only there is a typo in the new definition of $\lambda_n$, it should be $\psi(e_n)\ne0$. – Martin Argerami Sep 18 '13 at 20:56

1 Answers1

1

Several comments on your arguments:

  1. You switched the roles of $x$ and $y$ in your discussion as opposed to the statement of the problem. That induces confusion.

  2. In your proof of $\|\Phi f\|_{\mathcal{L}(c_{0};\mathbb{K})}=\|f\|_{\ell^{1}} $ it is not clear what you mean with "$y_j=e_j$". Are you mixing entries with vectors? Or are you now using the notation $y_j$ to denote a sequence of elements in $c_0$? The latter is the one you want, but your choice of elements in the sequence should be those with their first $j$ entries equal to $1$ (modulo a phase, because you want absolute values of the entries of $x$) and the rest equal to $0$; that way, when you apply $\phi x$ to it you get the sum of the first entries of $x$.

  3. For the surjectivity, your idea is correct, but you have to argue that the sequence $(\varphi(e_j))_j$ is in $\ell^1$ to guarantee that it makes sense to apply $\Phi$ to it.

Martin Argerami
  • 205,756