Folland, "Real Analysis", Chapter 5.3, Exercise 36:

Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbf{N}$. Suppose that $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$ by $Tf = \sum_1^\infty f(n) x_n$. (a) $T$ is bounded. (b) $T$ is surjective.

I have proved (a). I would like help on (b). Here are my ideas so far.

Say we want to show $y \in \mathcal{X}$ is in the image of $T$. Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$. My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| < 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$. Then use the continuity of $T$ to conclude $T(f) = y$.

The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m > M_N$, $||y - x_{n_m}|| < 1/N$. Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m > M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m > M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m > M_N}^\infty f_N(m) || y - x_{n_m} || < \frac{1}{N}. $$ Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.

However, getting all these hypotheses to hold simultaneously has been difficult. In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function. If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge. Perhaps I am missing something cleaner. If so, a simpler solution would be much appreciated.


  • 1
    $\begingroup$ See this for one way to show your result. $\endgroup$ – David Mitra May 25 '15 at 21:08

The following version of the open mapping theorem holds: If $T:Y\to X$ is a bounded operator between Banach spaces with $B_X \subseteq c \,\overline{T(B_Y)}$ for some constant $c>0$ (and the unit balls $B_X$ and $B_Y$) then $T$ is surjective (and open). Sometimes such a $T$ is called almost open. (The proof is in every book on functional analysis, the first step in the proof of the classical formulation of the OMT is to show that $T$ is almost open.)

In your case, $T(B_Y)$ contains the dense subset $\lbrace x_n:n\in \mathbb N\rbrace$ of $B_X$.

  • $\begingroup$ Thank you. This helped a great deal. Folland's proof of the Open Mapping Theorem essentially uses, as you predicted, the statement you presented; however, not so explicitly. Therefore, I am going to write an answer that basically recapitulates your answer with respect to Folland's presentation. $\endgroup$ – Doug May 28 '15 at 16:38
  • $\begingroup$ One nitpicky thing: In your last sentence, I presume you are realizing $x_n = Tf_n$, where $f_n$ is 1 on $n$ and 0 everywhere else. Then since $||f_n|| = 1$, I believe you meant to say $$ T((1+\epsilon)B_y) \supset \left\{x_n \colon n \in \mathbf{N}\right\}$$ for any $\epsilon > 0$. $\endgroup$ – Doug May 28 '15 at 16:56
  • $\begingroup$ For the open unit balls you would need the $\epsilon$, for the closed balls it works without. $\endgroup$ – Jochen May 30 '15 at 12:08

(See my comment under Jochen's answer, which led to my writing this answer as a clarification. The first part is intended to be viewed in the context of Folland's presentation.)

Folland's version of the Open Mapping Theorem is:

Let $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces. If $T \in L(\mathcal{X}, \mathcal{Y})$ is surjective, then $T$ is open.

In Folland's proof (using his notation without explaining here what the symbols represent), he proves in the second paragraph that if $$||y|| < r \Longrightarrow y \in \overline{T(B_1)}, $$ then, in fact, there exists $r^\prime < r$ such that $$||y|| < r^\prime \Longrightarrow y \in T(B_1).$$

I now relate this to Jochen's answer and the solution of Exercise 36. Let $\mathcal{Y}$ and $\mathcal{X}$ be Banach spaces, $T \colon \mathcal{Y} \to \mathcal{X}$ a bounded linear map, and denote by $B_\mathcal{Y}(r)$ and $B_\mathcal{X}(r)$ the balls of radius $r$ in $\mathcal{Y}$ and $\mathcal{X}$, respectively. Essentially, what Folland is proving here is the statement that if $$B_\mathcal{X}(r) \subset c \overline{T(B_\mathcal{Y}(1))}, $$ where $c$ is a positive constant, then there exists $r^\prime < r$ such that $$B_\mathcal{X}(r^\prime) \subset T(B_\mathcal{Y}(c)). $$ As I mention in my comments to Jochen, regarding Exercise 36, we know $$ E := \left\{ x_n \colon n \in \mathbf{N} \right\} \subset (1+\epsilon)T(B_{\ell^1}(1)), $$ for any $\epsilon > 0$, hence $$ B_\mathcal{X}(1) \subset \overline{E} \subset (1+\epsilon)\overline{T(B_{\ell^1}(1))}. $$ Hence, by the statement, there exists $r^\prime < 1$ such that $$ B_\mathcal{X}(r^\prime) \subset T(B_{\ell^1}(1 + \epsilon)). $$ It follows immediately that $T$ is surjective, finishing the solution of the exercise.


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, we 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 $Tf=x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.