Consider the following situation: Let $H, K$ be Hilbert spaces and let $\Phi$ be some mapping defined on simple tensors in $H\otimes K$ taking values in $B(H\otimes K)$ with the property that each simple tensor (of vectors) is transformed to a simple tensor (of operators) and moreover $$\|\Phi(x\otimes y)\|\le \|x\otimes y\|.$$ (where the norm on the left hand side is the operator norm on $B(H\otimes K)$ and the left hand side is the norm on $H\otimes K$.)
Is it true that then $\Phi$ linearly extends [uniquely] to a [continuous] mapping $H\otimes K\to B(H\otimes K)$? Is the inequality $$\|\Phi(\xi)\|\le \|\xi\|$$ preserved?
What if instead of taking two Hilbert spaces we have three or more? Does the answer change in that case?
Let $H = K = \ell_2$. Define a function $\Phi$ by $\Phi( x,y ) = \tfrac{1}{2}\|x\|\|y\|I_{H\otimes K}$.
If there were an extenstion of $\Phi$ to a bounded linear operator $\ell_2(\mathbb{N}\times \mathbb{N})\to B(H\otimes K)$ (still dented by $\Phi$), it would map the weakly null sequence $(e_n\otimes e_n)_{n=1}^\infty$ in the domain to a weakly null sequence in the codomain. However $\Phi ((e_n\otimes e_n)_{n=1}^\infty ) = (\tfrac{1}{2}I_{H\otimes K})_{n=1}^\infty$ is not weakly null.
Moral. Such problems frequently pop up in every-day practice but there is no chance for such a result in general due to too large arbitrariness of choice. (Maybe apart from some cases where your domain is $\ell_1$ rather than $\ell_2$.) Of course the tensor-product ornaments have been redundant here. In order to prove that some function extends to a linear map you really need to get your hands dirty.