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For bounded positive operators satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Self-adjointness and the Hilbert projection theorem decompose the whole Hilbert space as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.

Commutativity and equality of squares giveThus $S-T$ vanishes on $\mathrm{Ran}(S+T)$, and by continuity on its closure. The preceding kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Since $S+T$ is self-adjoint, part (i) and the Hilbert projection theorem give the orthogonal direct sumTherefore $S-T$ vanishes on all of $X$, provingThis proves commuting positive operators with equal squares are equal without assuming a square-root uniqueness theorem.

By the Cauchy-Schwarz inequality for a positive operator, for every $y$,Taking $y=Tx$ gives $Tx=0$, and $(S+T)x=0$ then gives $Sx=0$. ThusThe reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.