No trace on $B(H)$ if $H$ is infinite dimensional Let $H$ be an infinite dimensional Hilbert space and $B(H)$ the bounded linear operators on $H$. Then thre is no ultra weakly continous non-zero positve trace $tr:B(H)\rightarrow \mathbb{C}$. 
I talked to my professor about it and he showed me how he would do it. But he used without further clarifcation that $tr$ is SOT-contionous, which I don't see right now.  
 A: No continuity is needed, nor positivity (which would imply continuity, but that's beside the point). Only linearity and traciality. The argument below uses $H$ separable, but only for notational simplicity.
Let $\phi:B(H)\to\mathbb C$ be a linear and tracial functional.
We can write $I=P+Q$ for two infinite projections, both equivalent to $I$. Explicitly, for matrix units $E_{kj} $ we can take
$$
P=\sum E_{2n-1,2n-1},\ \ \ Q=\sum E_{2n,2n},\ \ \ V=\sum E_{2n,2n-1}, \ \ \ W=\sum E_{2n-1,n}.
$$
Then $V^*V=P$, $VV^*=Q$, $W^*W=I$, $WW^*=P$. So
$$
\phi (P)=\phi (I)=\phi (P)+\phi (Q)=2\phi (P).
$$ It follows that $\phi (P)=0$, and so $\phi (I)=\phi (P)=0$.
For a finite projection $P$, it is easy to construct a sequence of projections $P_n$, each equivalent to $P$ (i.e., there exist $W_n$ such that $W_n^*W_n=P$, $W_nW_n^*=P_n$) and with $\sum P_n=I$. It is also easy to show that $\sum_{n>s}P_n$ and $\sum_{n>t}P_n$ are equivalent for all $s,t$. Then, applying $\phi$ to
$$
\sum_1^mP_n+\sum_{m+1}^\infty P_n=I,
$$
we obtain
$$
m\phi (P)+\phi (\sum_{m+1}^\infty P_n)=0.
$$
As the second term is the same for all $m$, this forces $\phi (P)=0$.
So $\phi$ is zero on all projections. As every operator in $B(H)$ is a linear combination of projections, $\phi (T)=0$ for all $T\in B(H)$.

As mentioned by Ali in the comments, another argument goes by using Halmos' result that every $T\in B(H)$ is a sum of two commutators. Corollary in Page 198 of Halmos, P. R. American Journal of Mathematics
Vol. 76, No. 1 (Jan., 1954), pp. 191-198 (8 pages)
