Here's something that is probably obvious but I can't seem to see it.

Suppose $(x_n)_{n=1}^\infty$ is a Schauder basis for a Banach space $X$, "seminormalized" in the sense that we have $$0<\inf\|x_n\|\leq\sup\|x_n\|<\infty.$$ Now consider the corresponding normalized basis formed by setting $$y_n=\frac{x_n}{\|x_n\|},\quad n\in\mathbb{N}.$$

Recall that two Schauder bases $(x_n)_{n=1}^\infty$ and $(y_n)_{n=1}^\infty$ are said to be equivalent whenever there is a constant $C\in[1,\infty)$ satisfying the property $$C^{-1}\left\|\sum_{n=1}^\infty a_ny_n\right\| \leq \left\|\sum_{n=1}^\infty a_nx_n\right\| \leq C\left\|\sum_{n=1}^\infty a_ny_n\right\| \qquad \forall\,(a_n)_{n=1}^\infty\in c_{00},$$ where $c_{00}$ denotes the space of all scalar sequences with finitely many nonzero entries.

Question #1. Is $(x_n)_{n=1}^\infty$ always equivalent to $(y_n)_{n=1}^\infty$?

The principle of small perturbations guarantees that they have equivalent basic subsequences, but this is not enough. I want the original bases themselves to be equivalent.

It's also easy to see that they are equivalent whenever they are unconditional. However, again, this is not enough for me. I want equivalence to be guaranteed even when the bases are conditional.

If the answer to Question #1 above is negative, then:

Question #2. What is a specific counter-example to question #1? In other words, can we give an example of a seminormalized basis $(x_n)_{n=1}^\infty$ which is not equivalent to its normalized counterpart $(y_n)_{n=1}^\infty$?

  • 1
    $\begingroup$ Consider the exponential basis $e^{2\pi in t}$ on $L^p(0,1)$: when $p\in (1,\infty)\setminus \{2\}$ it is a normalized conditional Schauder basis. If the answer to your question was "yes", then every sequence $\mu_n$ that is bounded above and below would be a Fourier multiplier on $L^p$. I'm positive this is not the case; something like $\mu_n = \exp(i n^2)$ should be a counterexample. Can't find a source, though. (BTW I think bounded basis is a more common term than "semi-normalized". It does mean having both upper and lower bounds.) $\endgroup$
    – user357151
    Oct 21, 2017 at 4:14

1 Answer 1


Consider the Banach space $c$ of convergent sequences $\mathbb N\to\mathbb R$ with sup norm. This has an unconventional normalized Schauder basis $(y_n)_{n\geq 1}$ defined by

$$y_n(m)=\begin{cases}1&\text{ if $m\geq n,$}\\0&\text{ otherwise.}\end{cases}$$

Let $x_n=\tfrac1{w_n}y_n$ where $w_n=1$ for odd $n,$ and $w_n=2$ for even $n.$ The element $a=(0,\tfrac 32,\tfrac 23,\tfrac 54,\dots)$ defined by $a_n=1+(-1)^n/n$ can be written as a series

$$a = \sum_{n\geq 2} (-1)^n(\tfrac 1 n + \tfrac 1 {n-1})y_n = \sum_{n\geq 2} (-1)^n(\tfrac 1 n + \tfrac 1 {n-1})w_nx_n$$

But the series

$$\sum_{n\geq 2} (-1)^n(\tfrac 1 n + \tfrac 1 {n-1})w_ny_n$$

does not converge in $c$: it's $a$ plus a kind of harmonic series.

All this really needed is a series whose finite (rearranged) sums are unbounded. Then we can amplify a disjoint set of finite indices whose sums have norms tending to infinity. There are conditionally convergent series whose finite sums are bounded.


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