On ranges of nuclear operators Consider nuclear (trace class) operators acting on a separable infinite-dimensional Hilbert space. Does there exist a nuclear operator $A$ such that, for any other nuclear operator $B$, $\mathrm{ran}(B) \subset \mathrm{ran}(A)$?
It is known that a nuclear operator cannot have a closed range unless it is finite-dimensional, so if the Hilbert space is infinite-dimensional, there does not exist a nuclear operator whose range is the whole space. But I am not sure if one can still construct an operator with the "largest" possible range. I would appreciate any insight about this.

No. This fails even if you require compact and not trace-class.
Note first that the range of any compact operator has an orthonormal basis. This follows from the polar decomposition, as we can write $T=V|T|$ with $|T|$ positive and $V$ a partial isometry. As $|T|$ is self-adjoint and compact, its range has an orthonormal basis of eigenvectors. And $V$ maps this orthonormal basis to an orthonormal basis of the range of $T$.
Fix any dense subspace $H_0\subset H$ with $\{e_j\}$ an orthonormal basis of $H_0$. Let $f\in H\setminus H_0$ with $\|f\|=1$. Define
$$
Ax=\langle x,f\rangle\,f+\sum_j\frac1j\,\langle x,e_j\rangle\,e_j. 
$$
Then the range of $A$ contains $Af=f+\sum_j\frac1j\langle f,e_j\rangle\,e_j$ which is not in $H_0$.
