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problem 5.5.36 - Let $\mathscr{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{1}^{\infty}$ is a countable dense subset of the unit ball of $\mathscr{X}$, and define $T:L^{1}(\mu)\rightarrow \mathscr{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_n$.

a.) $T$ is bounded.

b.) $T$ is surjective.

c.) $\mathscr{X}$ is isomorphic to a quotient space of $L^{1}(\mu)$.

Attempted proof a.) We have that $Tf = \sum_{1}^{\infty}f(n)x_n$, hence $$\lVert Tf\rVert = \lVert \sum_{1}^{\infty}f(n)x_n\rVert \leq \sum_{1}^{\infty}|f(n)|\lVert x_n\rVert \leq \sum_{1}^{\infty}|f(n)| = \lVert f\rVert_{1}$$ hence $T$ is bounded.

Attempted proof b.) My thoughts here is to assume that $\{x_n\}$ converges to $x$ and $x$ is in $\mathscr{X}$. Then go from there to prove that $T$ is surjective.

For c.) I have no clue, any suggestions is greatly appreciated.

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Wolfy
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  • Your thought on b) seems perfectly fine. For c) think of which quotient space it can be, and use Open Mapping Theorem. To be more clear, what would your guess be if $T$ was a basic real matrix. – Emre Apr 13 '16 at 21:43
  • @user159517 your about 3 years late ;) – Wolfy Feb 18 '19 at 19:21
  • @Wolfy it's never too late to close a question ;) – user159517 Feb 18 '19 at 19:41

1 Answers1

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Your proof for (a) is correct. For (b), since $T$ is linear, it's enough to show that every $x\in X$ with $||x||\leq 1$ is in the image of $T$.

To do this, proceed inductively: there exists $x_{n_1}$ such that $||x-x_{n_1}||<\frac{1}{2}$. If $y=2(x-x_{n_1})$ then $||y||<1$, so there exists $x_{n_2}\neq x_{n_1}$ such that $||y-x_{n_2}||<\frac{1}{2}$, hence $||x-x_{n_1}-\frac{1}{2}x_{n_2}||<\frac{1}{4}$.

In general, if $x_{n_1},\dots,x_{n_k}$ have been chosen such that $$ \Big|\Big|x-\sum_{j=1}^k2^{1-j}x_{n_j}\Big|\Big|<2^{-k}$$ then $y=2^k(x-\sum_{j=1}^k2^{1-j}x_{n_j})$ is in the unit ball, hence there exists $x_{n_{k+1}}\not\in\{x_{n_1},\dots,x_{n_k}\}$ such that $||y-x_{n_{k+1}}||<\frac{1}{2}$, i.e. $$ \Big|\Big|x-\sum_{j=1}^{k+1}2^{1-j}x_{n_j}\Big|\Big|<2^{-k-1}$$

Finally, define $f\in \ell^1(\mathbb{N})$ by $f(n)=2^{1-k}$ if $n=n_k$, $f(n)=0$ otherwise. Then $Tf=x$.

For part (c), recall that if $X_1,X_2$ are Banach spaces and $T\in L(X_1,X_2)$ with $T(X_1)$ closed, then $X_1/\ker(T)\simeq T(X_1)$. In this case $T(\ell^1(\mathbb{N}))=X$ by part (b), hence is closed.

carmichael561
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