MathOverflow is a question and answer site for professional mathematicians. 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

I am trying to understand the properties of square variation, namely, the possibility of preserving it under certain operations. I am following Albiac & Kalton's book: Let $J$ stand for the usual definition of the James space (see Definition 3.4.1 p. 62).

Let $\mathcal{P}$ be the family of all non-increasing sequences of non-negative real numbers, convergent to 0.

Suppose that we are given a real function $f$ on positive integers taking non-negative values only. Assume, moreover, that:

  • $f(0)=0$

  • $(f(x_k))_{k=1}^\infty \in J\cap \mathcal{P}$

  • $\sum_{k=1}^n f(x_k+y_k)\leq \sum_{k=1}^n f(x_k)+\sum_{k=1}^n f(y_k)$

for every $(x_k)_{k=1}^\infty$ and $(y_k)_{k=1}^\infty$ in $c_0$.

Does the inequality hold $$\|(f(x_k+y_k))_{k=1}^\infty\|_J\leq \|(f(x_k))_{k=1}^\infty\|_J + \|(g(x_k))_{k=1}^\infty\|_J$$

In particular, must $(f(x_k+y_k))_{k=1}^\infty\in J$?

If the answer is negative, are there any sufficient conditions for $f$ to satisfy this inequality?

share|cite|improve this question
@Sellapan Nathan: Does "non-increasing" mean: $x_{k+1}\le x_k$ for all $k$ ? – Pietro Majer Jun 20 '11 at 7:07
also note that the third condition is just $f(a+b)\le f(a)+f(b)$ as it follows taking $x:=(a,0,\dots)$ and $y=(b,0,\dots)$ and using $f(0)=0$. – Pietro Majer Jun 20 '11 at 21:21
@Pietro: That's how I would say it, and I can't think of anything else that could mean. – Ricky Demer Oct 28 '11 at 0:11

Your Answer


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

Browse other questions tagged or ask your own question.