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Let

  • $X,Y$ be infinite-dimensional separable $\mathbb R$-Hilbert spaces
  • $Z$ be a $\mathbb R$-Banach space
  • $(\tilde e_n)_{n\in\mathbb N}$ be an orthonormal basis of $X$
  • $A$ be a nuclear operator from $X$ to $Y$
  • $B$ be a bounded bilinear operator from $X\times Y$ to $Z$

Are we able to show that $\sum_{n=1}^\infty\left\|B(\tilde e_n,A\tilde e_n)\right\|_Z<\infty$?

Let $$A=Q|A|\tag1$$ be the polar decomposition of $A$, i.e. $Q$ is the unique partial isometry from $X$ to $Y$ with $\ker Q=\ker A\tag2$ and $(1)$ and $$|A|:=\sqrt{A^\ast A}\;.$$ Since $A$ is compact, $|A|$ is compact, nonnegative and self-adjoint and hence there is an orthonormal basis $(e_i)_{i\in I}$ of $$\overline{\operatorname{im}|A|}=(\ker A)^\perp\tag3$$ with $I:=\mathbb N\cap[0,\dim(\ker A)^\perp]$ and $$|A|e_i=\lambda_ie_i\;\;\;\text{for all }i\in I\tag4$$ for some $(\lambda_i)_{i\in I}\subseteq(0,\infty)$. $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $U$ by an orthonormal basis of $\ker A$ and $$\sum_{n\in\mathbb N}\left\|B(e_n,Ae_n)\right\|_Z=\sum_{i\in I}\lambda_i\left\|B(e_i,Qe_i)\right\|_Z\le\left\|B\right\|\sum_{i\in I}\lambda_i\tag5\;.$$

Since $A$ is nuclear, $$\sum_{i\in I}\lambda_i<\infty\tag6$$ and hence we obtain the claim at least for the orthonormal basis $(e_n)_{n\in\mathbb N}$.

In general, \begin{equation}\begin{split}\sum_{n\in\mathbb N}\left\|B(\tilde e_n,A\tilde e_n)\right\|_Z&=\sum_{n\in\mathbb N}\left\|\sum_{i\in I}\lambda_i\langle\tilde e_n,e_i\rangle_XB(\tilde e_n,Qe_i)\right\|_Z\\&\le\sum_{n\in\mathbb N}\sum_{i\in I}\lambda_i\left|\langle\tilde e_n,e_i\rangle_X\right|\left\|B(\tilde e_n,Qe_i)\right\|_Z\\&=\sum_{i\in I}\lambda_i\sum_{n\in\mathbb N}\left|\langle\tilde e_n,e_i\rangle_X\right|\left\|B(\tilde e_n,Qe_i)\right\|_Z\;.\end{split}\tag7\end{equation} By the Cauchy-Schwarz inequality, $$\sum_{n\in\mathbb N}\left|\langle\tilde e_n,e_i\rangle_X\right|\left\|B(\tilde e_n,Qe_i)\right\|_Z\le\underbrace{\left(\sum_{n\in\mathbb N}\left|\langle\tilde e_n,e_i\rangle_X\right|^2\right)^{1/2}}_{=\:1}\left(\sum_{n\in\mathbb N}\left\|B(\tilde e_n,Qe_i)\right\|_Z^2\right)^{1/2}\,,\tag8$$ but that doesn't help, since I don't know a helpful estimate of $\sum_{n\in\mathbb N}\left\|B(\tilde e_n,Qe_i)\right\|_Z^2$.

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  • My intuition is that this question reduces to the question of how quickly an orthonormal sequence in $l^2(\mathbb{R})$ must converge to the zero sequence in the product topology on $\mathbb{R}^\mathbb{N}$ (or some suitable metrisation thereof). Taking $X=Y=l^2(\mathbb{R})$ and $Ae_i=\rho_ie_i$, where $e_i$ is the $i$-th standard unit vector and $\sum |\rho_i|<\infty$: if there exists some "fast" lower bound on the convergence rate of $(\tilde{e}n)$ to $\mathbf{0}$, then hopefully $(|A\tilde{e}_n|)$ should be dominated by a multiple of $(|\rho{an+b}|)$ for some constants $a$ and $b$; – Julian Newman Jan 06 '18 at 02:22
  • but if the convergence of $(\tilde{e}_n)$ can be "sufficiently slow", then for a counterexample take $Z=\mathbb{R}$ and $B(u,v)=(Au)\cdot v$. But I haven't managed to find any results on convergence rates of orthonormal sequences to zero (in any relevant metric/topology where this convergence holds). – Julian Newman Jan 06 '18 at 02:36

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