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?



$p$-nuclear operators from $C(K)$ to $L_{p}$
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 $\|(y\_i)\|\_q^w\le1$.


