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Let us recall that an operator $T$ from a Banach space $X$ to a Banach space $Y$ is called $p$-nuclear if $T$ can be written as $$T=\sum_{n=1}^{\infty}x^{*}_{n}\otimes y_{n},$$ where $\|(x^{*}_{i})_{i=1}^{\infty}\| _{p}:=(\sum_{i=1}^{\infty}\|x^{*}_{i}\|^{p})^{\frac{1}{p}}<\infty$ and $$\|(y_{n})_{n}\|_{q}^{w}:=\sup_{y^{*}\in B_{Y^{*}}}(\sum_{n=1}^{\infty}|\langle y^{*},y_{n}\rangle|^{q})^{\frac{1}{q}}<\infty.$$

The $p$-nuclear norm of $T$ is defined by $$\nu_{p}(T)=\inf\{\|(x^{*}_{i})_{i=1}^{\infty}\| _{p}\|(y_{n})_{n}\|_{q}^{w}\},$$ where the infimum is taken over all $p$-nuclear representations of $T$.

Let $K$ be a compact Hausdorff space and $\mu$ be a Borel probability measure on $K$. Let $\tau=\{A_{i}\}_{i=1}^{n}$ be a partition of $K$ into finitely many Borel sets $A_{i}$ of positive measure. Define the operator $P_{\tau}:C(K)\rightarrow L_{p}(\mu)$ by $P_{\tau}(f)=\sum\limits_{i=1}^{n}\frac{\int_{A_{i}}fd\mu}{\mu(A_{i})}\chi_{A_{i}}$ for all $f\in C(K)$.

Question. Is the $p$-nuclear norm of $P_{\tau}$ equal to 1?

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  • $\begingroup$ There seems to be a typo in the definition of a $p$-nuclear operator: one should have $\|(y_n)_n\|_q^w<\infty$, where $1/p+1/q=1$. Likewise, the expression for $\nu_p(T)$ should be corrected. $\endgroup$ – Dirk Werner Jul 8 '19 at 21:23
  • $\begingroup$ You are right, Dirk. the $\|(y_{n})_{n}\|_{p}^{w}$ should be corrected to be $\|(y_{n})_{n}\|_{q}^{w}$. I am sorry. $\endgroup$ – Dongyang Chen Jul 9 '19 at 1:34
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With the usual definition of a $p$-nuclear operator (see comment above), $\nu_p(P_\tau)\le1$: Let $x_i^*(f)= \int_{A_i} f / \mu(A_i)^{1/q}$ and $y_i= \chi_{A_i}/ \mu_(A_i)^{1/p}$. Then $P_\tau= \sum x_i^* \otimes y_i$, $\|(x_i^*)\|_p=1$ und $\|(y_i)\|_q^w\le1$.

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  • $\begingroup$ Thanks, Dirk. You are right. $\endgroup$ – Dongyang Chen Jul 9 '19 at 1:36

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