# Data Processing equality variation

Let $$\rho_{AB}$$ be a state and $$T: B \rightarrow C$$ be a CPTP map with $$\sigma_{AC}= T(\rho_{AB})$$. It is well known that $$H_{\infty}(A \vert B)_{\rho} \geq H_{\infty}(A \vert C)_{\sigma}$$ (aka data processing.) Furthermore, equality holds when $$T=V$$ is an isometry giving $$H_{\infty}(A \vert B)_{\rho} = H_{\infty}(A \vert C)_{V\rho V^{\dagger}}$$.

My question is, does the result hold if we consider $$V^{\dagger}$$ instead of $$V$$, that is, is the following equality true?

$$H_{\infty}(A \vert C)_{\sigma} = H_{\infty}(A \vert B)_{V^\dagger \sigma V}$$.

No, the co-isometry map $$\sigma \to V^\dagger \sigma V$$ is not trace preserving. In the worst case you can have something like this. Take an isometry $$V: \mathbb{C}^2 \rightarrow \mathbb{C}^3$$ which just embeds $$\mathbb{C}^2$$ into $$\mathbb{C}^3$$ in the standard way i.e., $$V = |0\rangle \langle 0 | + |1 \rangle \langle 1|.$$ This is an isometry $$V^\dagger V = \mathbb{I}_{\mathbb{C}^2}$$. But imagine now that we have a state $$\sigma = |3 \rangle \langle 3|$$ on $$\mathbb{C}^3$$, then $$V^\dagger \sigma V = 0.$$ Which would result in an infinite min-entropy.