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We consider three sequences $(x_n),(y_n),(z_n)$, where $(x_n) \in \ell^1$ is positive and the other two sequences are merely assumed to be positive, i.e. $y_n,z_n \ge 0$ where $0<z_n<z_{n+1}$ is strictly monotone.

I wonder about the following:

Let us assume that $\langle (x_n),(y_n)\rangle_{\ell^2} <\infty$ and that $\lim_{\varepsilon \downarrow 0} \varepsilon \log(\sum_{n} e^{-\varepsilon y_n}) =0.$

The first property somehow tells us that $(y_n)$ does not grow too fast relative to $x_n$, the second property tells us that $(y_n)$ grows fast enough for this sum to be small such that the factor $\varepsilon$ wins over the log sum.

If we now assume that $\lim_{\varepsilon \downarrow 0} \varepsilon \log(\sum_{n} e^{-\varepsilon z_n}) >0,$ i.e. the $z_n$ grow not fast enough for the $\varepsilon$ to win against the log sum, does this implies that $$ \langle (x_n),(z_n) \rangle <\infty \text{ as well }?$$

The question is essentially whether gauging the growth rate by this limit, which shows in a way that $y_n$ grows faster than $z_n$, is enough to conclude that $$ \langle (x_n),(y_n)\rangle_{\ell^2} <\infty \Rightarrow \langle (x_n),(z_n) \rangle <\infty.$$

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  • $\begingroup$ @ChristianRemling Thanks, I added as a constraint that $(z_{n})$ is strictly increasing. Notice that $y$ cannot vanish too often, since otherwise the sum is infinite right away. Does this address your concern? $\endgroup$
    – António Borges Santos
    Commented Jan 14 at 23:18
  • $\begingroup$ well, $y$ also has to have the property that $\lim_{\varepsilon \downarrow 0} \varepsilon \log(\sum_{n} e^{-\varepsilon y_n})=0$, right? $\endgroup$
    – António Borges Santos
    Commented Jan 14 at 23:26
  • $\begingroup$ @ChristianRemling yes, but remember also that $x_n \to 0$ and $y_n \to \infty$ are necessary. So you can rearrange but the tail behaviour is somehow determined... $\endgroup$
    – António Borges Santos
    Commented Jan 14 at 23:52
  • $\begingroup$ Okay, let's say $z_n=\log(n)^2$ then $z_n$ satisfies the assumptions since $\log(\int_0^{\infty} e^{-\varepsilon \log(x)^2} dx) $ behaves like $\frac{1}{4\varepsilon}.$ Thus $\lim_{\varepsilon \downarrow 0}\varepsilon \log( \sum_n e^{-\varepsilon z_n}) = \frac{1}{4}$ if I am not mistaken. $\endgroup$
    – António Borges Santos
    Commented Jan 15 at 0:15
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    $\begingroup$ I of course fell into the same little trap as before, this time with $y_n=0$. $\endgroup$
    – Christian Remling
    Commented Jan 15 at 1:15

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As I tried to point out in various not so well thought out comments, this is immediately suspicious because the conditions we impose on $x$ and $y$ exclusively allow us to rearrange, so that what was supposed to be the crucial assumption, namely $\sum x_ny_n<\infty$, doesn't have much force.

Counterexample: Let's take $z_n=\log^2 n$, as you suggested, but it doesn't matter much what exactly $z$ does.

I first claim that we can make $\log\sum e^{-\epsilon z_n^{1/2}} \le \epsilon^{-1/2}$ for $0<\epsilon\le 1$, if the sum is now taken only over a sufficiently sparse subsequence $n=N_1,N_2,\ldots$. To do this, simply pick $N_1<N_2<\ldots$ such that $e^{-\epsilon z_{N_k}^{1/2}}\le 2^{-k}e^{\epsilon^{-1/2}}$ for all $0<\epsilon\le 1$; we only need $z_{N_k}$ to be large enough for this to work for a fixed $k$. In particular, we can also insist that $z_{N_k}\ge 2^k$.

Now we define $y_n=z_n^{1/2}$ for $n=N_k$ and $y_n=n$ otherwise and $x_n=1/z_n$ for $n=N_k$ and $x_n=1/n^3$ otherwise. Then $x,y$ have the required properties, but $\limsup x_nz_n=1$.

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