Let A be the infinite Hankel matrix with the coefficient $$A_{kj}=e^{(-t(k+j)^2)}-e^{(-t(k+j+2)^2)},$$ with $t$ a nonnegative real number.

Is $A$ in trace class with a norm bounded by an absolute constant?

It is not hard to see A is in trace class with a constant depending on t by either a result of J. S. Howland (MR0288630) or V. Peller (MR0602274). But the mystery is wether we can get rid of t?

  • $\begingroup$ Hi Tao. I am unable to access this article of Bonsall (ams.org/mathscinet-getitem?mr=849553 or dx.doi.org/10.1016/S0924-6509(09)70256-9 ) but since it obtains estimates of the trace class in terms of coefficients it might help. $\endgroup$ – Yemon Choi Apr 4 '14 at 19:18
  • $\begingroup$ Actually, what happens when you compute the Hilbert-Schmidt norm of $\Gamma$? Some back-of-the-envelope calculations seem to indicate (assuming I have not made a mistake) that $\Vert\Gamma\Vert_{\rm HS}$ is bounded away from zero by a uniform constant as $t\to 0^+$ but I have not checked these calculations carefully. $\endgroup$ – Yemon Choi Apr 4 '14 at 21:36
  • $\begingroup$ This question was the starting point for a joint work with Tao. The positive answer to the question implies that the heat semigroup is bounded (although not positive) on the free group von Neumann algebras. We also realized that the same holds for every hyperbolic group, for the simple reason that (essentially) every cb radial multipliers on the free group von Neumann algebras are cb multipliers on the von Neumann algebra of every hyperbolic group. Our paper contains a detailed study of other radial multipliers (eg Bochner-Riesz), and some computations on abelian groups... $\endgroup$ – Mikael de la Salle Sep 19 '15 at 12:03
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    $\begingroup$ (continued) Our paper "Complete boundedness of heat semigroups on the von Neumann algebra of hyperbolic groups" is available on arxiv.org/abs/1405.5178 , and will soon appear in Transactions of the AMS. $\endgroup$ – Mikael de la Salle Sep 19 '15 at 12:04

I think that the answer is yes.

Edit: as noticed in the comments, the answer would be obviously yes if the matrix $A$ was positive, since in this case its trace norm (denoted $\|A\|_1$) would be equal to its trace which is less than $1$. But $A$ is not positive: when $t$ goes to $0$, $A_{i,j}/4t$ goes to to $1+i+j$, which is not a positive matrix since every $2\times 2$ submatrix has determinant $(1+2i)(1+2j)-(1+i+j)^2<0$.

One therefore has to be more careful, and the fact that $\sup_t \|A\|_1 <\infty$ follows from the following claim: (end of Edit)

If $f:[0,\infty)\to \mathbf R$ is smooth function with first derivatives vanishing fast enough at infinity (say for simplicity that $f$ is in the Schwarz class), then the matrix $$\widetilde A_{k,j} = \varepsilon f(\varepsilon(k+j))$$ belongs to the trace class with norm independant from $\varepsilon$.

How is this related to your question? Well, take $f(x) = 4x e^{-x^2}$ and $\varepsilon = \sqrt t$. Then using that $e^{-t n^2} - e^{-t (n+2)^2} = f(\sqrt n) + 0((t+t^2n^2) e^{-tn^2})$, you see that your matrix $A$ and the matrix $\widetilde A$ have a difference uniformly bounded in the trace class (for example because the $\ell^1$-norm of the coefficients of the difference is bounded). And by the claim $\widetilde A$ belongs to the trace class uniformly in $\varepsilon$.

Why is the claim true? It follows from Peller's theorem quoted in the question, and from the elementary inequality $\| \varphi \|_{L^1(\mathbf T)} \leq c \sqrt{ \|\varphi\|_{L^2} \|(1-z)\varphi\|_{L^2}}$ for every function $\varphi$ on the unit circle $\mathbf T$ with Lebesgue measure.

Indeed, Peller's theorem characterizes, up to uniform constants, the trace class norm of a Hankel matrix $A_{k,j} =\widehat \varphi(k+j)$ by the norm of $\varphi = \sum_{k \geq 0} \widehat \varphi(k)z^k$ in the Besov space $B_{1,1}^1$. Namely if $w \in C_c(0,\infty)$ is a smooth function satisfying $\sum_{n \in \mathbf Z} w(2^n x)=1$ (say that $w$ is supported in $[1/2,2]$ for the sequel) and $W_n(z) = \sum_k w(k/2^n) z^k$ for $n>0$ and $W_0(z) =1+z$, $$ \|\widetilde A\|_{S^1} \simeq \sum_{n \geq 0} 2^n \| W_n \ast \varphi\|_{L^1(\mathbf T)}$$

Now if $\widehat \varphi(k) = \varepsilon f(\varepsilon k)$ then $\|W_n \ast \varphi\|_2$ is dominated, up to a constant and small error that I forget here, by $\sqrt \varepsilon (\int_{2^{n-1}\varepsilon}^{2^n \varepsilon} |f|^2)^{1/2}$. Similarly $\|(1-z)W_n \ast \varphi\|_2$ is dominated by $2^{-n} \sqrt \varepsilon (\int_{2^{n-1}\varepsilon}^{2^n \varepsilon} |f|^2)^{1/2} + \varepsilon^{3/2} (\int_{2^{n-1}\varepsilon}^{2^n \varepsilon} |f'|^2)^{1/2}$. (the first term comes from the derivative of $w(\cdot/2^n)$ and the second from the derivative of $f(\varepsilon \cdot)$). By the "elementary inequality", we get the bound $$\|W_n \ast \varphi\|_{L^1} \lesssim \sqrt \varepsilon (\sqrt \varepsilon + \sqrt{2^{-n}}) \left(\int_{2^{n-1}\varepsilon}^{2^n \varepsilon} |f|^2+|f'|^2\right)^{1/2}+\textrm{ small error}.$$

Then at least if one forgets the error, one gets for $\delta>0$ $$ \sum_n 2^n \|W_n \ast \varphi\|_{L^1} \lesssim \left(\int (1+x^{2+\delta})(|f(x)|^2+|f'(x)|^2)dx\right)^{1/2}.$$

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  • $\begingroup$ It the original Hankel matrix is PSD, then we'll immediately get $\|H\| \le e^{-4t} \le 1$, but I haven't checked if it is psd (but clearly my reasoning seems naive to me).. $\endgroup$ – Suvrit Apr 8 '14 at 16:13
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    $\begingroup$ You are right. It also occured to me that my answer (and the question...) was absurd it there was an easy way to prove that $A$ is positive. And the question whether $A$ is positive is even more natural now that we known (from my complicated answer) that the trace of $A$ and its trace norm are comparable. $\endgroup$ – Mikael de la Salle Apr 8 '14 at 18:30
  • $\begingroup$ Thanks for catching the bug; my error lay in calling $e^{4t(i+j+1)}-1$ a rank-1 matrix --- how silly can one be! $\endgroup$ – Suvrit Apr 8 '14 at 21:17
  • $\begingroup$ Thanks for the comments, Yemon and Suvrit.Hi, MiKael, Thanks for the answer. $\endgroup$ – tao mei Apr 10 '14 at 4:54
  • $\begingroup$ Thanks for the comments, Yemon and Suvrit. Hi, MiKael, Thanks for the answer. I like your clever use of "the elementary inequality". Your answer implies a very interesting result (to me), see the email I am sending to you. $\endgroup$ – tao mei Apr 10 '14 at 5:17

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