$\DeclareMathOperator{Img}{Im}$ In Warner's "Foundations of differentiable manifolds and Lie groups" we read that $$E^p\stackrel{(1)}{=}\Delta(E^p)\oplus H^p\stackrel{(2)}{=}d\delta(E^p)\oplus\delta d(E^p)\oplus H^p\stackrel{(3)}{=}d(E^{p-1})\oplus\delta(E^{p+1})\oplus H^p,$$ Where $E^p$ are smooth $p-$forms on given compact oriented Riemannian manifold $M.$
He states that it is sufficient to prove the decompositon in $(1),$ for $(2)$ and $(3)$ follow from three facts:
$$\Delta=\delta d+d\delta\hspace{10pt}<d\alpha,\beta>=<\alpha,\delta\beta>\hspace{10pt}\Delta\alpha=0\iff d\alpha=0\wedge\delta\alpha=0$$
So I conclude, that he means $$\Delta(E^p)=d\delta(E^p)\oplus\delta d(E^p)=d(E^{p-1})\oplus\delta(E^{p+1})\hspace{20pt}(*)$$ follows from above three facts. However I do not see how to prove it. I only have following inclusions based on the knowledge about non-degenerated bilinear forms: $$\Img\Delta\subset(\ker\Delta)^\perp=(\ker d\delta\cap\ker\delta d)^\perp\supset(\ker d\delta)^\perp+(\ker\delta d)^\perp\supset(\Img d\delta)^\perp+(\Img\delta d)^\perp$$ I used above fact that $\Delta,d\delta$ and $\delta d$ are self adjoint.
So I ask how to prove $(*)?$
(1) $d\delta E^p,\ \delta dE^p$ are orthogonal :
$$ (d\delta x, \delta d y)=(dd \delta x,dy) =0 $$
In further we have $ \Delta E^p = V_1\oplus V_2,\ V_1:=d\delta E^p,\ V_2:= \delta d E^p $
(2) Note that $V_1\subset W_1:=dE^{p-1},\ V_2\subset W_2:= \delta E^{p+1}\ \ast$ are orthogonal
(3) Hodge proved first equality So second is followed. We will use $E_p=V_1\oplus V_2\oplus H^p\ \ast\ast$ : If $dx\in W_1,\ y\in H^p$, then $$ (dx,y)=(x,\delta y)=0 $$
(Note that $H^p=\{ x| dx=\delta x=0 \}$) That is, $W_1\oplus W_2\oplus H^p\subset E_p$ So from this and $\ast,\ \ast\ast$ we complete the proof