Find all $\alpha$ for which $T_\alpha$ is well-defined, injective, $J \circ T_\alpha $ is a contraction, $T_\alpha$ is a homeomorphism. Let $T_\alpha:l_2 \to l_1: (x_k)_{k=1}^\infty \mapsto (\alpha_k x_k)_{k=1}^\infty$ for $ \alpha = (\alpha_k)_{k=1}^\infty \in \omega$ where $\omega$ is the set of all sequences. Find all $\alpha$ for which $T_\alpha$ is well-defined,
injective,
$J \circ T_\alpha $ is a contraction in $l_2$ where $ J : l_1 \to l_2 : y \mapsto y$,
$T_\alpha$ is a homeomorphism.
For the solution, I think that since in $ l_2 $, $\sum x_n^2 < \infty$ then  $\alpha_k = 1/n$ to be well-defined. But I am not sure how to find $\alpha$ for the rest.
 A: $T_\alpha$ is well-defined if and only if $\alpha \in \ell^2$.
It follows from this proposition:

Let $(\beta_n)_{n=1}^\infty$ be a sequence of scalars. Then the series $$\sum_{n=1}^\infty \beta_nx_n$$
  converges for every $(x_n)_{n=1}^\infty \in \ell^p$ if and only if $(\beta_n)_{n=1}^\infty \in \ell^q$, where $p, q \in \langle 1, +\infty\rangle$ are conjugate exponents.

One direction follows from Hölder's inequality, and the other is nontrivial.

$T_\alpha$ is injective if and only if $\alpha_n \ne 0$ for all $n \in \mathbb{N}$.
This is because $T_\alpha$ is injective if and only if $\operatorname{Ker} T_\alpha = \{0\}$. Assume $\alpha_n \ne 0$ for all $n \in \mathbb{N}$:
$$0 = T_\alpha(x) = (\alpha_n x_n)_{n=1}^\infty \implies \alpha_nx_n = 0, \forall n \in \mathbb{N} \implies x_n =0, \forall n \in \mathbb{N} \iff x = 0$$
Hence, $\operatorname{Ker} T_\alpha = \{0\}$. Conversely, if $\alpha_j = 0$, then $e_j \in \operatorname{Ker} T_\alpha = \{0\}$ so $T_\alpha$ is not injective.

$J\circ T_\alpha : \ell^2 \to \ell^2$ is a contraction if and only if it is bounded and $\|J\circ T_\alpha\| < 1$, with $\|\cdot\|$ being the operator norm on $\mathcal{L}(\ell^2)$.
This is because $$\|(J\circ T_\alpha)x - (J\circ T_\alpha)y\| = \|(J\circ T_\alpha)(x-y)\| \le \|(J\circ T_\alpha)\|\|x-y\|$$
and if $\|J\circ T_\alpha\| \ge 1$ then there exists $x \in \ell^2, \|x\|_2 = 1$ such that $\|(J\circ T_\alpha)x\|_2 \ge 1-\varepsilon$ for all $\varepsilon > 0$
Therefore
$$\|(J\circ T_\alpha)x - (J\circ T_\alpha)0\|_2 = \|(J\circ T_\alpha)x\|_2 \ge 1-\varepsilon$$
so it cannot be $\|(J\circ T_\alpha)x - (J\circ T_\alpha)0\|_2 \le C\|x - 0\| = C$ for some $C < 1$.
$\|J\circ T_\alpha\| < 1$ will hold if and only if $\|\alpha\|_\infty < 1$.
We have:
$$\|(J\circ T_\alpha)x\|_2 = \left\|(\alpha_nx_n)_{n=1}^\infty\right\|_2 = \sqrt{\sum_{n=1}^\infty |\alpha_n|^2\|x_n\|_2} \le \|\alpha_n\|_\infty \sqrt{\sum_{n=1}^\infty \|x_n\|_2} = \|\alpha_n\|_\infty \|x\|_2$$
Thus $\|J\circ T_\alpha\| \le \|\alpha_n\|_\infty$.
$$\|J\circ T_\alpha\| \ge \frac{\|(J\circ T_\alpha)e_n\|_2}{\|e_n\|_2} = |\alpha_n|, \forall n \in \mathbb{N} \implies \|J\circ T_\alpha\| = \|\alpha\|_\infty$$
Thus $\|J\circ T_\alpha\| = \|\alpha\|_\infty$, which proves the claim.

$T_\alpha$ cannot be a homeomorphism between $\ell^2$ and $\ell^1$ because they are not homeomorphic! To see this, notice that $\ell^2$ is reflexive, and $\ell^1$ is not. Alternatively, $\ell^1$ has the property that every weakly convergent sequence also converges strongly, while $\ell^2$ hasn't.
