Let $S$ and $T$ be bounded operators over a Hilbert space $\mathcal{H}$. Define their tensor product $S\otimes T$ as acting on $\mathcal{H}\otimes\mathcal{H}$ by $S\otimes T(x\otimes y):=Sx\otimes Ty$ and then extended by linearity to the dense subspace of finite linear combination of such $x\otimes y$ and after proving boundedness by continuity to all of $\mathcal{H}\otimes\mathcal{H}$.
How do I prove there that it is indeed bounded on the dense subspace?
The problem is that one has to consider finite linear combinations of these $x\otimes y$. I remember from the proof of boundedness for the creation and annihilation operators over Fock space that the trick was to reduce quadratic expressions to linear ones via $\|x\|=\sup_{\|\hat{z}\|=1}|\langle\hat{z},x\rangle|$. However I don't know how and wether at all this works here.
Now, let $A$ and $B$ be selfadjoint operators but not necessarily bounded and define $A\otimes B$ on an appropriate subspace.
How do I prove that it is essentially selfadjoint?
To show symmetry is a quite straightforward but selfadjointness...
To prove boundedness of $S\otimes T$, it suffices to prove it when $S$ and $T$ are unitaries (since the unitaries span $\mathcal{B}(H)$. Now if $$ z = \sum_{i=1}^n x_i\otimes y_i $$ where $\{y_1, y_2,\ldots, y_n\}$ are orthogonal. So $$ \|(S\otimes T)(z)\|^2 = \|\sum_{i=1}^n S(x_i)\otimes T(y_i)\|^2 $$ Since $\{y_i\}$ are orthogonal $$ = \sum_{i=1}^n \|S(x_i)\otimes T(y_i)\|^2 $$ $$ = \sum_{i=1}^n \|S(x_i)\|^2 \|T(y_i)\|^2 = \sum_{i=1}^n \|x_i\|^2 \|y_i\|^2 = \|z\|^2 $$ Hence, $\|S\otimes T\| = 1$