# $p$-nuclear operators from $C(K)$ to $L_{p}$

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?

• 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. – Dirk Werner Jul 8 '19 at 21:23
• You are right, Dirk. the $\|(y_{n})_{n}\|_{p}^{w}$ should be corrected to be $\|(y_{n})_{n}\|_{q}^{w}$. I am sorry. – Dongyang Chen Jul 9 '19 at 1:34

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 $$\|(y_i)\|_q^w\le1$$.

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