Let $M \subseteq B(H)$ be a *-subalgebra containing the identity on H. If there is a unitary T in the unit ball of the SOT-closure of $M$, is there a net of unitary oprators in the unit ball of $M$ that strongly converges to $T$? I know that there is, if $M$ is a $C^*$-subalgebra, but what if it is not? If there is, how do I prove it?
Thank you
Suppose that $M$ is the algebra of finite-rank operators on a countably infinite dimensional Hilbert space $H$. Then its SOT-closure is $B(H)$ (which contains unitaries). However $M$ does not contain any unitary operators.
As for your updated question, where one insists that $M$ is unital, the answer is also no.
If $A$ is an algebra, then let $U(A)$ denote the set of unitaries in $A$. Let $M$ be the $*$-algebra of polynomials inside $C[0,1]$ (which acts on $L_2[0,1]$ by multiplication). Then $M$ is unital and has SOT-closure $L_\infty[0,1]$. The only unitaries in $M$ are of the form $\mathbb C\cdot 1$ and so the SOT-closure of $U(M)=\mathbb C \cdot 1$. However, $$ U(L_\infty[0,1])=\{f \in L_\infty[0,1] : |f(x)|=1, \forall x \in [0,1] \}. $$