In this section, we recall some basic properties of general \(L_{p}\)-projection bodies.
Theorem 3.1
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), and
\(\tau\in[-1, 1]\), then
$$ \Pi^{\tau}_{p}(-K)=\Pi^{-\tau}_{p}K=- \Pi^{\tau}_{p}K. $$
(3.1)
Proof
From (1.5) it follows that, for all \(u\in S^{n-1}\),
$$\begin{aligned} h^{p}_{-\Pi^{\tau}_{p}K}(u) =& h^{p}_{\Pi^{\tau}_{p}K}(-u) = \alpha_{n,p}(\tau) \int_{S^{n-1}}\varphi_{\tau}(-u\cdot v)^{p}\,dS_{p}(K, v) \\ =&\alpha_{n,p}(\tau) \int_{S^{n-1}}\varphi_{\tau}\bigl(u\cdot (-v) \bigr)^{p}\,dS_{p}(-K, -v) = h^{p}_{\Pi^{\tau}_{p}(-K)}(u). \end{aligned}$$
This gives
$$ \Pi^{\tau}_{p}(-K)=-\Pi^{\tau}_{p}K. $$
(3.2)
In addition, by (1.9) we have that
$$\begin{aligned}& f_{1}(\tau)+f_{2}(\tau)=1, \end{aligned}$$
(3.3)
$$\begin{aligned}& f_{1}(-\tau)=f_{2}(\tau), \qquad f_{2}(- \tau)=f_{1}(\tau). \end{aligned}$$
(3.4)
From (3.3) and (3.4), together with (1.6) and (1.8), we obtain
$$\begin{aligned} \Pi^{-\tau}_{p}K =& f_{1}(-\tau)\cdot \Pi^{+}_{p}K+_{p}f_{2}(-\tau)\cdot\Pi ^{-}_{p}K \\ =& f_{2}(\tau)\cdot\Pi^{-}_{p}(-K)+_{p}f_{1}( \tau)\cdot\Pi ^{+}_{p}(-K)= \Pi^{\tau}_{p}(-K). \end{aligned}$$
(3.5)
Obviously, (3.2) and (3.5) yield (3.1). □
Theorem 3.2
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), \(\tau\in[-1, 1]\), and
\(\tau\neq0\), then
$$\Pi^{\tau}_{p}K=\Pi^{-\tau}_{p}K \quad\Longleftrightarrow\quad \Pi^{+}_{p}K=\Pi^{-}_{p}K. $$
Proof
From (1.8) and (3.4) it follows that, for \(K\in{\mathcal {K}}^{n}_{\mathrm {o}}\), \(p\geq1\), and \(\tau\in[-1,1]\),
$$\Pi^{-\tau}_{p}K=f_{2}(\tau)\cdot \Pi^{+}_{p}K+_{p}f_{1}(\tau)\cdot \Pi^{-}_{p}K, $$
that is,
$$ h^{p}_{\Pi^{-\tau}_{p}K}(u)=f_{2}(\tau)h^{p}_{\Pi^{+}_{p}K}(u)+f_{1}( \tau )h^{p}_{\Pi^{-}_{p}K}(u) $$
(3.6)
for all \(u\in S^{n-1}\). Therefore, by (3.3), (1.7), and (3.6), if \(\Pi^{+}_{p}K=\Pi^{-}_{p}K\), then
$$h^{p}_{\Pi^{\tau}_{p}K}(u)=h^{p}_{\Pi^{-\tau}_{p}K}(u) $$
for all \(u\in S^{n-1}\). This gives \(\Pi^{\tau}_{p}K=\Pi^{-\tau}_{p}K\).
Conversely, if \(\Pi^{\tau}_{p}K=\Pi^{-\tau}_{p}K\), then (1.7) and (3.6) yield
$$\bigl[f_{1}(\tau)-f_{2}(\tau)\bigr]h^{p}_{\Pi^{+}_{p}K}(u)= \bigl[f_{1}(\tau)-f_{2}(\tau )\bigr]h^{p}_{\Pi^{-}_{p}K}(u) $$
for all \(u\in S^{n-1}\). Since \(f_{1}(\tau)-f_{2}(\tau)\neq0\) when \(\tau\neq0\), we get \(\Pi^{+}_{p}K=\Pi^{-}_{p}K\). □
Haberl and Schuster [30] proved the following fact.
Theorem 3.A
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), and
p
is not odd integer, then
\(\Pi^{+}_{p}K=\Pi^{-}_{p}K\)
if and only if
\(K\in\mathcal{K}^{n}_{\mathrm{os}}\).
According to Theorems 3.A and 3.2, we get the following:
Theorem 3.3
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), and
p
is not odd integer, then, for
\(\tau\in[-1, 1]\)
and
\(\tau\neq0\), \(\Pi^{\tau}_{p}K=\Pi^{-\tau}_{p}K\)
if and only if
\(K\in\mathcal{K}^{n}_{\mathrm{os}}\).
Theorem 3.4
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), and
\(\tau\in[-1, 1]\), then
$$ \Pi^{\tau}_{p}K+_{p}\Pi^{-\tau}_{p}K= \Pi^{+}_{p}K+_{p}\Pi^{-}_{p}K. $$
(3.7)
Proof
From (1.7) and (3.6), using (3.3), we have that, for any \(u\in S^{n-1}\),
$$h\bigl(\Pi^{\tau}_{p}K, u\bigr)^{p}+ h\bigl( \Pi^{-\tau}_{p}K, u\bigr)^{p}= h\bigl( \Pi^{+}_{p}K, u\bigr)^{p}+ h\bigl( \Pi^{-}_{p}K, u\bigr)^{p}, $$
that is,
$$h\bigl(\Pi^{\tau}_{p}K+_{p}\Pi^{-\tau}_{p}K, u\bigr)^{p}= h\bigl(\Pi^{+}_{p}K+_{p}\Pi ^{-}_{p}K, u\bigr)^{p}. $$
This is the desired relation. □
From Theorem 3.4 we deduce the following:
Corollary 3.1
If
\(K\in\mathcal{K}^{n}_{\mathrm {o}}\), \(p\geq1\), and
\(\tau\in[-1, 1]\), then
$$ \Pi_{p}K=\frac{1}{2}\cdot\Pi^{\tau}_{p}K+_{p} \frac{1}{2}\cdot\Pi ^{-\tau}_{p}K. $$
(3.8)
Proof
Taking \(\tau=0\) in (1.8) and combining with (1.9) yield
$$ \Pi_{p}K=\frac{1}{2}\cdot\Pi^{+}_{p}K+_{p} \frac{1}{2}\cdot\Pi^{-}_{p}K. $$
(3.9)
From (3.9) and (3.7) we immediately get (3.8). □
Theorem 3.5
If
\(K, L\in\mathcal{K}^{n}_{\mathrm{os}}\), \(p\geq1\)
is not an even integer, and
\(\tau \in[-1, 1]\), then
$$\Pi^{\tau}_{p}K=\Pi^{\tau}_{p}L \quad\Longrightarrow\quad K=L. $$
The proof of Theorem 3.5 requires the following two lemmas.
Lemma 3.1
If
\(K, L\in\mathcal{K}^{n}_{\mathrm {o}}\), and
\(p\geq1\)
is not an even integer, then
\(\Pi _{p} K=\Pi_{p} L\)
if and only if
\(V_{p}(K, Q)=V_{p}(L, Q)\)
for any
\(Q\in\mathcal{K}^{n}_{\mathrm{os}}\).
Proof
From (1.1) we know that, for any \(u\in S^{n-1}\),
$$\begin{aligned} h^{p}_{\Pi_{p}(-K)}(u) =&\alpha_{n,p} \int_{S^{n-1}} | u\cdot v|^{p}\,dS_{p}(-K, v) \\ =&\alpha_{n,p} \int_{S^{n-1}} \bigl| u\cdot(-v)\bigr|^{p}\,dS_{p}(K, -v)=h^{p}_{\Pi_{p}K}(u), \end{aligned}$$
which implies \(\Pi_{p}(-K)=\Pi_{p}K\). Thus, for any \(u\in S^{n-1}\),
$$\begin{aligned} h^{p}_{\Pi_{p}K}(u) =&\frac{1}{2}h^{p}_{\Pi_{p}K}(u)+ \frac{1}{2}h^{p}_{\Pi _{p}(-K)}(u) \\ =&\frac{1}{2}\alpha_{n,p} \int_{S^{n-1}} | u\cdot v|^{p} \bigl[dS_{p}(K, v)+dS_{p}(-K, v)\bigr]. \end{aligned}$$
Thus, if \(\Pi_{p} K=\Pi_{p} L\), then, for any \(u\in S^{n-1}\),
$$\int_{S^{n-1}} | u\cdot v|^{p} \bigl[dS_{p}(K, v)+dS_{p}(-K, v)-dS_{p}(L, v)-dS_{p}(-L, v) \bigr]=0. $$
Letting \(\mu(v)=S_{p}(K, v)+S_{p}(-K, v)-S_{p}(L, v)-S_{p}(-L, v)\), we have
$$ \int_{S^{n-1}} | u\cdot v|^{p} \,d\mu(v)=0. $$
(3.10)
Since \(\mu(v)\) is an even Borel measure on \(S^{n-1}\) and \(p\geq1\) is not an even integer, it follows from (3.10) that \(\mu(v)=0\) (see, e.g., [30]), that is,
$$ S_{p}(K, \cdot)+S_{p}(-K, \cdot)=S_{p}(L, \cdot)+S_{p}(-L, \cdot). $$
(3.11)
Since \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\), we have \(h_{Q}(-v)=h_{Q}(v)\) for all \(v\in S^{n-1}\). Therefore, by (2.6) we get
$$V_{p}(K, Q)=\frac{1}{n} \int_{S^{n-1}}h^{p}_{Q}(-v)\,dS_{p}(K, -v)=\frac{1}{n} \int_{S^{n-1}}h^{p}_{Q}(v)\,dS_{p}(-K,v). $$
This and (2.6) yield
$$V_{p}(K, Q)=\frac{1}{2n} \int_{S^{n-1}}h^{p}_{Q}(v)\bigl[dS_{p}(K, v)+dS_{p}(-K, v)\bigr] $$
for any \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\). By (3.11) we see that if \(\Pi_{p} K=\Pi_{p} L\), then \(V_{p}(K, Q)=V_{p}(L, Q)\) for any \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\).
Conversely, if \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\), let \(Q=[-u, u]\) (\(u\in S^{n-1}\)). Then \(h_{Q}(v)=| u\cdot v|\) for any \(v\in S^{n-1}\). This, together with (2.6), yields
$$\begin{aligned} V_{p}(K, Q) =&\frac{1}{n} \int_{S^{n-1}}h^{p}_{Q}(v)\,dS_{p}(K, v) \\ =&\frac{1}{n} \int _{S^{n-1}}| u\cdot v|^{p}\,dS_{p}(K, v)=\frac{1}{n\alpha _{n,p}}h^{p}(\Pi_{p}K, u). \end{aligned}$$
Hence, if \(V_{p}(K, Q)=V_{p}(L, Q)\) for any \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\), then \(\Pi_{p}K=\Pi_{p}L\). □
Lemma 3.2
If
\(K, L\in\mathcal{K}^{n}_{\mathrm{os}}\)
and
\(p\geq1\)
is not an even integer, then
$$\Pi_{p}K=\Pi_{p}L\quad\Longrightarrow\quad K=L. $$
Proof
By Lemma 3.1, if \(\Pi_{p} K=\Pi_{p} L\) and p is not an even integer, then, for any \(Q\in\mathcal{K}^{n}_{\mathrm{os}}\),
$$ V_{p}(K, Q)=V_{p}(L, Q). $$
(3.12)
Taking K for Q in (3.12) and using (2.7) and (2.8), we obtain \(V(K)\geq V(L)\) with equality for \(p>1\) if and only if K and L are dilates (for \(p=1\), if and only if K and L are homothetic). Similarly, taking L for Q in (3.12) yields \(V(K)\leq V(L)\), and equality holds for \(p>1\) if and only if K and L are dilates (for \(p=1\), if and only if K and L are homothetic). Therefore, \(V(K)=V(L)\), and K and L are dilates when \(p>1\) (K and L are homothetic when \(p=1\)). Since \(K, L\in\mathcal{K}^{n}_{\mathrm{os}}\), we have that, for \(p\geq1\), \(K=L\). □
Proof of Theorem 3.5
If \(K\in\mathcal{K}^{n}_{\mathrm{os}}\), then by (3.5) and Corollary 3.1 we have that
$$\Pi_{p}K=\Pi^{\tau}_{p}K=\Pi^{-\tau}_{p}K. $$
Therefore, if \(K, L\in\mathcal{K}^{n}_{\mathrm{os}}\), then, for \(\tau\in[-1, 1]\),
$$\Pi^{\tau}_{p}K=\Pi^{\tau}_{p}L\quad\Longleftrightarrow\quad \Pi_{p}K=\Pi_{p}L. $$
This, together with Lemma 3.2, completes the proof of Theorem 3.5. □