where S is the right-shift operator. The von Neumann inequality proves it true for p = 2 {\displaystyle p=2} and for p = 1 {\displaystyle p=1} and p = ∞ {\displaystyle p=\infty } it is true by straightforward calculation. S.W. Drury has shown in 2011 that the conjecture fails in the general case.[2]