Traces on $K(H)$ 
*

*Are there any traces on $K(H)$ for an infinite dimensional Hilbert space $H$?


I think that the answer is no: viewing $K(H)$ as a direct limit of matrix algebras, if there was a trace we knew what it does on each copy of the matrix algebras (by uniqueness). But then it can not be extended naturally (for example, from $M_2$ to $M_3$, by the connecting maps).


*Is it the motivation for the trace class operators?

*Moreover, I know that there is no trace on $B(H)$ and one proof is by showing that it must be a sum of commutators. If we know that there is no trace on the compacts, doesn't it imply that there is no trace on $B(H)$?
Because if $\tau$ was a trace on $B(H)$ then its restriction was a trace on $K(H)$, thus must be zero. But if it is zero on the compacts it is also zero on $B(H)$ because $K(H)$ is SOT dense in $B(H)$, am I wrong?
Thanks
 A: If by "trace" you mean "tracial state", then the answer is no. Consider matrix units $\{E_{kj}\}$ (take them countable even if $H$ is non-separable). Then, if $f$ is a tracial state, 
$$
f(E_{kk})=f(E_{k1}E_{1k})=f(E_{1k}E_{k1})=f(E_{11}). 
$$
Let $c=f(E_{11})$. Now form the compact operator $$X=\sum_k\frac1k\,E_{kk}.$$ Then, since the series converges in norm, 
$$
f(X)=\sum_k\frac ck,
$$
so $c=0$. But, if $c=0$, then any projection $P$ is equivalent to $\sum_1^nE_{kk}$ for some $n$, so $f(P)=nc=0$. From this one deduces that $f(T)=0$ for all selfadjoint $T$ (via the Spectral Theorem) and thus for all $T$ (thinking of the real and imaginary part). 
I don't have a good "motivation" for the trace-class operators. I don't have one either for the $L^p$ spaces, nor for the continuous functions. We tend to use and study the spaces where we can do the most, where we can achieve characterizations. The trace-class operators have an important role as the dual of $K(H)$, and thus the pre-dual of $B(H)$. 
For your reasoning about relating to a trace in $B(H)$, it doesn't work. Because in principle it could be the case that a trace exists on $K(H)$ (in the end, it doesn't) that cannot be extended to $B(H)$. For such extension to work you would need the trace on $K(H)$ to be normal. So your reasoning discards the existence of a normal trace on $K(H)$  via the non-existence of the trace on $B(H)$ (here is another proof); but it doesn't discard the existence of a non-normal trace. 
