Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let E be a Banach space with a 1-unconditional basis $(e_n)$ (for example, $\ell^p$). Then an operator T on E can be thought of as an infinite matrix, in the obvious way. Clearly each scalar on the diagonal of this matrix is bounded by $\|T\|$. By the unconditionality, it follows that if I have any diagonal matrix with uniformly bounded entries, this will define a bounded operator. So we conclude that the projection taking the matrix of T to the diagonal matrix given by setting all off-diagonal terms zero, is a contraction.

This is absolutely well-known for $\ell^p$ spaces say, and must be well-known in this more general setting, though I don't know a reference.

Now suppose we have two Banach spaces E and F, both with 1-unconditional bases. We can still think of operators as given by infinite matrices. Is this procedure of projecting down to the diagonal still bounded? What's hard is that we cannot expect, in general, the diagonal to be isomorphic to $\ell^\infty$ anymore. So we need an indirect argument. I think I can do this in an adhoc way for maps between $\ell^p$ and $\ell^q$; does anyone know a more systematic treatment, or any references?

share|cite|improve this question
up vote 5 down vote accepted

This is covered by Proposition I.1.16 on page 22 of Lindenstrauss-Tzafriri (in the old Springer Lecture Notes 338 edition):

Let the matrix $A = (\alpha_{i,j})$ represent a bounded linear operator $T$ from $X$ to $Y$ with unconditional bases $\{x_i\}$ and $\{y_j\}$. Then the diagonal operator of $A$ (i.e., the matrix $(\delta_{j}^{i} \alpha_{i,j})$) also represents a bounded linear operator $D$. If the unconditional constants of $\{x_i\}$ and $\{y_j\}$ are $1$ then $\Vert D \Vert \leq \Vert T\Vert$.

This result certainly is in the later editions of that book—it enters in various proofs later on—but Google doesn't let me check at the moment.


For the benefit of the readers, let me quickly summarize the relevant definitions—I'm using the slightly awkward choice of letters by Lindenstrauss-Tzafriri:

  • A sequence $\{x_n\}$ in a Banach space $X$ is called a Schauder basis if every $x \in X$ can be written uniquely as $x = \sum_{n=1}^{\infty} a_n x_n$.

  • If $\{x_n\}$ is a Schauder basis then there is a projection $P_n: X \to \operatorname{span}{\{x_1,\ldots,x_n\}}$. This projection is continuous and the basis constant of $\{x_n\}$ is $\sup_n \|P_n\| \geq 1$.

  • A series $x = \sum_{n=1}^{\infty} x_n$ in a Banach space is said to be unconditionally convergent if for every permutation $\pi$ of the positive integers the series $\sum_{n=1}^{\infty} x_{\pi(n)}$ converges.

  • A (Schauder) basis is called unconditional if the representation $x = \sum_{n=1}^{\infty} a_n x_n$ converges unconditionally. One can show that with an unconditional basis one can omit summands and arbitrarily change their signs and rearrange their order without affecting convergence (hence the term). See e.g. M.M. Day, Normed linear spaces Chapter IV, §3 and §4.

  • Let $\{x_n\}$ be an unconditional basis. If $I$ is a finite set of natural numbes then write $P_I$ for the projection onto the span of $\{x_i\}_{i \in I}$. The unconditional constant of an unconditional basis is $C = \sup{\{ \|P_{I}\|\,:\,\emptyset \neq I \subset \mathbb{N}\text{ finite}\}}$. If $C = 1$ then the basis is called $1$-unconditional.

For your convenience I'm reproducing a picture of the easy but very slick proof:

proof from LT

Added: Lindenstrauss and Tzafriri attribute the result to A. Tong, Diagonal submatrices of matrix maps, Pacific J. Math. Volume 32, Number 2 (1970), 551–559 where you'll find more general and more specialized variants. For a "detailed study of some related and deeper questions" they refer the reader to S. Kwapień, A. Pełczyński, The main triangle projection in matrix spaces and its applications, Studia Mathematica, Vol. 34 (1970), 43–68. See here for the full text of that volume.

share|cite|improve this answer
Great! That's exactly what I wanted (and the proof is very nice...) I knew I should have checked Lindenstrauss and Tzafriri, though in my defence, I'm travelling, so it would have meant a walk to the library, and not just my shelf... – Matthew Daws Aug 1 '11 at 13:16
No problem and no defense needed! Well, that's why I have electronic and OCR'ed versions of many of my reference "bibles". It took me 5 seconds to find the book on the hard drive, then 5 seconds to enter diagonal in the search box hit enter and see the result... I guess it's one of those proofs that you won't forget once you've seen them. Enjoy your journey and we'll encounter each other soon enough again via the internet. Best wishes from Zurich, – t.b. Aug 1 '11 at 13:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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