Parity transformation: mistake or puzzle in Sakurai's "Modern Quantum Mechanics"? In Sakurai's Modern Quantum Mechanics, p.270, he wrote an equation the parity transformation $\pi$ (where  $\pi = \pi^\dagger = \pi^{-1}$) as
$$\pi \left(1- \frac{i p \cdot d x'}{\hbar}\right) \pi^\dagger 
= \left(1+ \frac{i p \cdot d x'}{\hbar}\right).  \tag{0}$$

How is this consistent with these two equations (4.2.3) and (4.2.10) also derived:
$$ \pi  x' \pi^\dagger = -x' \tag{1} $$
$$ \pi  p \pi^\dagger = -p  \tag{2} $$
Is that
$$\pi  dx' \pi^\dagger = dx'  \tag{3} $$
$$\pi  (p \cdot dx') \pi^\dagger = -( p \cdot dx' )  \tag{4} $$
How to understand the Eq. 1 versus Eq. 3?
but how to understand the Eqs. 1, 2, versus Eq. 4?
Naively, it seems that
$$ \pi  dx' \pi^\dagger = - dx'  \tag{5},$$
because say $dx'=(x_A- x_B)$ is the spatial interval difference between two points on $A$ and $B$, then
$$\pi  dx' \pi^\dagger=\pi  \Delta x \pi^\dagger=\pi  (x_A- x_B) \pi^\dagger
=(-x_A- (-x_B))=-(x_A- x_B)=-dx'.$$
Also I thought:
$$ \pi  (p \cdot dx') \pi^\dagger = +( p \cdot dx' )  \tag{6} $$
Could you correct me why Eqs. 3 and 4 are correct, but  Eqs. 5 and 6 are not?
 A: With due respect to the formidable JunJohn S, his proof is doomed to unleash dyslexia demons of this kind, which I'd rather not deal with, and neither should you.
From his (4.2.5),
$$
\pi | x\rangle= |-x\rangle, 
$$
and
$$
\hat p = \int\!\! dx ~|x\rangle (-i\hbar\partial_x)\langle x| ,
$$
you immediately see
$$
\pi \hat p \pi^\dagger =  \int\!\! dx |-x\rangle (-i\hbar\partial_x)\langle -x|= \\ - \int\!\! dx |x\rangle (-i\hbar\partial_x)\langle x| =- \hat p,
$$
by bland change of dummy variables of integration (and integration limits at infinity).
And that's that.
A: Infinitesimal translation operator $\hat{T}(dx):=\left(1-i\frac{\hat{P}.\hat{dx}}{\hbar}\right)$ and it satisfies the following relation $T(dx)|x\rangle = |x + dx\rangle$. Then $$\pi T(dx)|x\rangle=\pi |x+dx\rangle=e^{i\delta}|-x-dx\rangle$$ Similarly $$T(-dx)\pi |x\rangle=e^{i\delta}T(-dx)|-x\rangle=e^{i\delta}|-x-dx\rangle$$ This proves eqn (4.2.8). Similarly for eqn (4.2.9) we have: $$\pi T(dx) \pi^{\dagger} |x\rangle=\pi T(dx) (e^{i\delta}|-x\rangle)=\pi (e^{i\delta}|-x+dx\rangle)=|x-dx\rangle=T(-dx)|x\rangle$$ Hence $$\pi \left(1-i\frac{\hat{P}.\hat{dx}}{\hbar}\right)\pi^{\dagger}=\left(1+i\frac{\hat{P}.\hat{dx}}{\hbar}\right)$$
