We know that for any Banach space $E$ if $\hat{\otimes}$ denotes projective tensor product then $L^1([0,1])\hat{\otimes} E$ is isomorphism isometric with $$L^1([0,1],E)=\{f:[0,1]\to E:\ \ \int_0^1\|f(s)\|ds<\infty\}.$$ (for $f\in L^1([0,1]),x\in E$ define $\phi_{f,x}(s)=f(s)x$ for all $s\in[0,1]$ then $\Phi(f\otimes x)=\phi_{f,x}$ is that isometric isomorphism)
Now let $\omega:[0,1]\to(0,+\infty)$ be a continuous function such that $\omega(0)=1$ and $\omega(s+t)\leq\omega(s)\omega(t)$.Define $$L^1([0,1],\omega)=\{f:[0,1]\to \mathbb{C}:\ \int_0^1|f(s)|\omega(s)ds<\infty\}$$ and $$L^1([0,1],\omega,E)=\{f:[0,1]\to E:\ \int_0^1\|f(s)\|\omega(s)ds<\infty\}.$$ Could we conclude that there is an isometric isomorphism of Banach spaces such that $L^1([0,1],\omega)\hat{\otimes} E\cong L^1([0,1],\omega,E)$?
I define $\psi_{f,x}(s)=f(s)\omega(s)x$ and $\Psi(f\otimes x)=\psi_{f,x}$. Is this true? If it is true, how can I prove that it is an isometric?Is it surjective?