The following Lemmas will be required to prove our main theorems.
Lemma 1
[5]If K(f ) is the Aleksandrov body associated with f ∈ C^{+}(S^{n 1}), then h_{K(f)}= f almost everywhere with respect to the measure S(K(f ), ·) on S^{n 1}.
Obviously, if K(f ) is the Aleksandrov body corresponding to a given function f ∈ C^{+}(S^{n 1}), its support function has the property that 0 < h_{
K
} ≤ f and V (f ) = V (h_{K(f )}).
Lemma 2
[5]If p ≥ 1, K(f ) is the Aleksandrov body associated with f ∈ C^{+}(S^{n 1}), then V (f ) = V (K(f )) = V_{
p
} (K(f ), f ), i.e.
V\left(f\right)=\frac{1}{n}\underset{{S}^{n1}}{\int}h\left(K\left(f\right),u\right)\mathsf{\text{d}}S\left(K\left(f\right),u\right).
Lemma 3
[5]IfK\in {\mathcal{K}}_{0}^{n}, f ∈ C^{+}(S^{n 1}), then, for p ≥ 1,
\frac{n}{p}{V}_{p}\left(K,f\right)=\underset{\epsilon \to 0}{\mathsf{\text{lim}}}\frac{V\left({h}_{K}{+}_{p}\epsilon \cdot f\right)V\left({h}_{K}\right)}{\epsilon}.
(3.1)
We get the following BrunnMinkowski inequality for the Aleksandrov bodies associated with positive continuous functions.
Lemma 4
If f, g ∈ C^{+}(S^{n 1}), and λ, μ ∈ℝ^{+}, then
V{\left(\lambda f+\mu g\right)}^{1\u2215n}\ge \lambda V{\left(f\right)}^{1\u2215n}+\mu V{\left(g\right)}^{1\u2215n},
(3.2)
with equality if and only if there exist a constant c > 0 and t ≥ 0, such that f = cg + t, almost everywhere with respect to S(K(f ), ·) on S^{n 1}.
Proof
Since f, g ∈ C^{+}(S^{n 1}), from (2.11), (2.12) and the BrunnMinkowski inequality (see [21]), we get
\begin{array}{cc}\hfill V{\left(K\left(\lambda f+\mu g\right)\right)}^{1\u2215n}& \ge V{\left(\lambda K\left(f\right)+\mu K\left(g\right)\right)}^{1\u2215n}\hfill \\ \ge \lambda V{\left(K\left(f\right)\right)}^{1\u2215n}+\mu V{\left(K\left(g\right)\right)}^{1\u2215n}.\hfill \end{array}
(3.3)
The equality condition in (3.3) is that f, g are the support functions of K(f ) and K(g) which are homothetic, respectively.
From Lemma 1 and Lemma 2, we get the following result
V{\left(\lambda f+\mu g\right)}^{1\u2215n}\ge \lambda V{\left(f\right)}^{1\u2215n}+\mu V{\left(g\right)}^{1\u2215n},
(3.4)
with equality if and only if there exist a constant c > 0 and t ≥ 0, such that f = cg + t, almost everywhere with respect to S(K(f ), ·) on S^{n 1}.
An immediate consequence of the definition of a Firey linear combination, and the integral representation (2.13), is that for Q\in {\mathcal{K}}_{0}^{n}, the p mixed volume
{V}_{p}\left(Q,\cdot \right):{C}^{+}\left({S}^{n1}\right)\to \left(0,\infty \right)
is Firey linear.
Lemma 5
If p ≥ 1, Q\in {\mathcal{K}}_{0}^{n}, f, g ∈ C^{+}(S^{n 1}), and λ, μ ∈ ℝ^{+}, then
{V}_{p}\left(Q,\lambda \cdot f{+}_{p}\mu \cdot g\right)=\lambda {V}_{p}\left(Q,f\right)+\mu {V}_{p}\left(Q,g\right).
(3.5)
Proof
From (2.13), (2.14), we obtain
\begin{array}{cc}\hfill {V}_{p}\left(Q,\lambda \cdot f{+}_{p}\mu \cdot g\right)& =\frac{1}{n}\underset{{S}^{n1}}{\int}{\left(\lambda \cdot f{+}_{p}\mu \cdot g\right)}^{p}h{\left(Q,u\right)}^{1p}\mathsf{\text{d}}S\left(Q,u\right)\hfill \\ =\frac{1}{n}\underset{{S}^{n1}}{\int}\left(\lambda {f}^{p}+\mu {g}^{p}\right)h{\left(Q,u\right)}^{1p}\mathsf{\text{d}}S\left(Q,u\right)\hfill \\ =\lambda {V}_{p}\left(Q,f\right)+\mu {V}_{p}\left(Q,g\right).\hfill \end{array}
In the following, we will prove the p Minkowski inequality for the Aleksandrov bodies associated with positive continuous functions.
Proof of Theorem 1.
Firstly, let p = 1 in Lemma 3, we get
n{V}_{1}\left(Q,f\right)=\underset{\epsilon \to 0}{\mathsf{\text{lim}}}\frac{V\left({h}_{Q}+\epsilon f\right)V\left({h}_{Q}\right)}{\epsilon},
let \epsilon =\frac{t}{1t}, we have
\begin{array}{cc}n{V}_{1}\left(Q,f\right)\hfill & =\underset{t\to 0}{\mathsf{\text{lim}}}\frac{V\left(\left(1t\right){h}_{Q}+tf\right){\left(1t\right)}^{n}V\left({h}_{Q}\right)}{t{\left(1t\right)}^{n1}}\hfill \\ =\underset{t\to 0}{\mathsf{\text{lim}}}\frac{V\left(\left(1t\right){h}_{Q}+tf\right)V\left({h}_{Q}\right)}{t}+\underset{t\to 0}{\mathsf{\text{lim}}}\frac{\left(1{\left(1t\right)}^{n}\right)V\left({h}_{Q}\right)}{t}\hfill \\ =\underset{t\to 0}{\mathsf{\text{lim}}}\frac{V\left(\left(1t\right){h}_{Q}+tf\right)V\left({h}_{Q}\right)}{t}+nV\left({h}_{Q}\right).\hfill \end{array}
Let
f\left(t\right)=V{\left(\left(1t\right){h}_{Q}+tf\right)}^{1\u2215n},\phantom{\rule{1em}{0ex}}0\le t\le 1,
we see that
{f}^{\prime}\left(0\right)=\frac{{V}_{1}\left(Q,f\right)V\left({h}_{Q}\right)}{V{\left({h}_{Q}\right)}^{\frac{n1}{n}}}.
From Lemma 4, we know that f is concave, i.e.
\frac{{V}_{1}\left(Q,f\right)V\left({h}_{Q}\right)}{V{\left({h}_{Q}\right)}^{\frac{n1}{n}}}\ge V{\left(f\right)}^{\frac{1}{n}}V{\left({h}_{Q}\right)}^{\frac{1}{n}}.
Thus,
{V}_{1}\left(Q,f\right)\ge V{\left(Q\right)}^{\frac{n1}{n}}V{\left(f\right)}^{\frac{1}{n}}.
(3.6)
According to the equality condition in inequality (3.3), and using Lemma 1 and Lemma 2, we have the equality holds in inequality (3.6), if and only if there exist a constant c > 0 and t ≥ 0, such that h_{
Q
} = cf + t, almost everywhere with respect to S(Q, ·) on S^{n 1}.
Secondly, from the Hölder inequality (see [22]), together with the integral representations (2.13) and (2.6), we obtain
\begin{array}{cc}\hfill {V}_{p}\left(Q,f\right)& =\frac{1}{n}\underset{{S}^{n1}}{\int}f{\left(u\right)}^{p}h{\left(Q,u\right)}^{1p}\mathsf{\text{d}}S\left(Q,u\right)\hfill \\ \ge {V}_{1}{\left(Q,f\right)}^{p}V{\left(Q\right)}^{1p},\hfill \end{array}
when this combined with inequality (3.6), we have
{V}_{p}\left(Q,f\right)\ge V{\left(Q\right)}^{\frac{np}{n}}V{\left(f\right)}^{\frac{p}{n}}.
(3.7)
To obtain the equality conditions, we note that there is equality in Hölder's inequality precisely when V_{1}(Q, f )h_{
Q
} = V (Q)f, almost everywhere with respect to the measure S(Q, ·) on S^{n 1}. Combining the equality conditions in (3.6), and using Lemma 1, it shows that the equality holds if and only if there exists a constant c > 0 such that h_{
Q
} = cf, almost everywhere with respect to S(Q, ·) on S^{n 1}.
Using the above Lemmas and Theorem 1, we can get the following Corollaries describing the uniqueness results.
Corollary 1
Suppose K, L\in {\mathcal{K}}_{0}^{n}, and\mathcal{F} ⊂ C^{+} (S^{n 1}) is a class of functions such that h_{
K
}, h_{
L
} ∈ \mathcal{F}. (i) If n ≠ p > 1, and V_{
p
} (K, f ) = V_{
p
} (L, f ), for all f ∈ \mathcal{F}, then K = L. (ii) If p = n, and V_{
p
} (K, f ) ≥ V_{
p
} (L, f ), for all f ∈ \mathcal{F}, then K and L are dilates, and hence
{V}_{p}\left(K,f\right)={V}_{p}\left(L,f\right),for\phantom{\rule{0.3em}{0ex}}all\phantom{\rule{0.3em}{0ex}}f\in \phantom{\rule{2.77695pt}{0ex}}{C}^{+}\left({S}^{n1}\right).
Proof
If n ≠ p > 1, take f = h_{
K
} , and from (2.13), Lemma 2 and Theorem 1, we get
{V}_{p}\left(K,f\right)={V}_{p}\left(K,{h}_{K}\right)={V}_{p}\left(L,{h}_{K}\right)\ge V{\left(L\right)}^{\frac{np}{n}}V{\left({h}_{K}\right)}^{\frac{p}{n}}.
Hence,
V\left(K\right)\ge V\left(L\right).
Similarly, take f = h_{
L
} , we get
V\left(L\right)\ge V\left(K\right).
In view of the equality conditions of Theorem 1, we obtain that K = L.
If n = p, the hypothesis together with Theorem 1, we have
{V}_{p}\left(K,f\right)\ge {V}_{p}\left(L,f\right)\ge V{\left(L\right)}^{\frac{np}{n}}V{\left(f\right)}^{\frac{p}{n}},
with equality in the right inequality implying that L and K(f ) are dilates. Take f = h_{
K
} , since n = p, the terms on the left and right are identical, and thus, K and L must dilates; hence,
{V}_{p}\left(K,f\right)={V}_{p}\left(L,f\right),\phantom{\rule{1em}{0ex}}\mathsf{\text{forall}}\phantom{\rule{1em}{0ex}}f\in {C}^{+}\left({S}^{n1}\right).
Corollary 2
Suppose f, g ∈ C^{+}(S^{n 1}), and\mathcal{F} ⊂ C^{+} (S^{n 1}) is a class of functions such that f, g ∈ \mathcal{F}. If p > 1, and
{V}_{p}\left(Q,f\right)={V}_{p}\left(Q,g\right),\phantom{\rule{1em}{0ex}}for\phantom{\rule{0.3em}{0ex}}all\phantom{\rule{0.3em}{0ex}}{h}_{Q}\in \mathcal{F},
then f = g almost everywhere on S^{n 1}.
Proof
Since f, g ∈ C^{+}(S^{n 1}), according to (2.11), we denote two Aleksandrov bodies K(f ) and K(g). From the hypothesis, taking Q = K(f ), and using Lemma 2 and Theorem 1, we get
{V}_{p}\left(K\left(f\right),f\right)=V\left(f\right)={V}_{p}\left(K\left(f\right),g\right)\ge V{\left(K\left(f\right)\right)}^{\frac{np}{n}}V{\left(g\right)}^{\frac{p}{n}},
then,
V\left(f\right)\ge V\left(g\right).
Similarly, take Q = K(g), we get
V\left(g\right)\ge V\left(f\right).
From the equality conditions of Theorem 1, we obtain
K\left(f\right)=K\left(g\right).
In view of the definition of Aleksandrov body, and using Lemma 1, then
f=g,\phantom{\rule{1em}{0ex}}\mathsf{\text{almost}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{everywhere}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}{S}^{n1}.
Corollary 3
Suppose n ≠ p > 1, and f, g ∈ C^{+}(S^{n 1}), such that S_{
p
} (K(f ), ·) ≤ S_{
p
} (K(g), ·).

(i)
If V (f ) ≥ V (g), and p < n, then f = g almost everywhere on S^{n 1}.

(ii)
If V (f ) ≤ V (g), and p > n, then f = g almost everywhere on S ^{n 1}.
Proof
Suppose a function φ ∈ C^{+}(S^{n 1}), and n ≠ p > 1, since S_{
p
} (K(f ), ·) ≤ S_{
p
} (K(g), ·), it follows from the integral representation (2.13) and (2.8) that
{V}_{p}\left(K\left(f\right),\varphi \right)\le {V}_{p}\left(K\left(g\right),\varphi \right),\mathsf{\text{forall}}\varphi \in {C}^{+}\left({S}^{n1}\right).
As before, take φ = h_{K(g)}, from Lemma 1, Lemma 2, and Theorem 1, we get
V{\left(f\right)}^{\frac{np}{n}}\le V{\left(g\right)}^{\frac{np}{n}}.
Applying the hypothesis, and from the definition of the Aleksandrov body and Lemma 1, we obtain the desired results.
Corollary 4
Suppose n ≠ p ≥ 1, f, g ∈ C^{+}(S^{n 1}), and\mathcal{F} ⊂ C^{+} (S^{n 1}) is a class of functions such that f, g ∈ \mathcal{F}. If
\frac{{V}_{p}\left(K,f\right)}{V\left(f\right)}=\frac{{V}_{p}\left(K,g\right)}{V\left(g\right)},\phantom{\rule{1em}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}all\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{h}_{K}\in \mathcal{F},
then f = g almost everywhere on S^{n 1}.
Proof
According to (2.11), we denote two Aleksandrov bodies K(f ) and K(g). From the hypothesis, taking K = K(f ) and K = K(g), and combining with Lemma 2 and Theorem 1, respectively, we obtain
V\left(g\right)\ge V\left(f\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}V\left(f\right)\ge V\left(g\right),
Hence, in view of the equality conditions of Theorem 1, the definition of Aleksandrov body, and Lemma 1, we get the desired result
f=g,\phantom{\rule{1em}{0ex}}\mathsf{\text{almost}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{everywhere}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}{S}^{n1}.
Now, the p BrunnMinkowski inequality for the p Aleksandrov bodies and the Aleksandrov bodies associated with positive continuous functions is established as following.
Proof of Theorem 2.
From Lemma 5 and Theorem 1, we get
\begin{array}{cc}\hfill {V}_{p}\left(Q,\lambda \cdot f{+}_{p}\mu \cdot g\right)& =\lambda {V}_{p}\left(Q,f\right)+\mu {V}_{p}\left(Q,g\right)\hfill \\ \ge V{\left(Q\right)}^{\frac{np}{n}}\left[\lambda V{\left(f\right)}^{\frac{p}{n}}+\mu V{\left(g\right)}^{\frac{p}{n}}\right],\hfill \end{array}
with equality if and only if K(f) and K(g) are dilates of Q.
Now, take Q = K_{
p
} (λ · f + _{
p
} μ · g), use (2.10), and recall V (f ) = V (K(f )) = V_{
p
} (K(f ), f ), we have
V{\left(\lambda \cdot f{+}_{p}\mu \cdot g\right)}^{\frac{p}{n}}\ge \lambda V{\left(f\right)}^{\frac{p}{n}}+\mu V{\left(g\right)}^{\frac{p}{n}}.
Also, we note that the equality holds, if and only if K(f ) and K(g) are dilates. Using Lemma 1, we get the condition of equality holds if and only if there exists a constant c > 0 such that f = cg, almost everywhere with respect to S(K(f ), ·) on S^{n 1}.
Then, we will prove Theorem 3 by using the generalized Blaschke linear combination.
Proof of Theorem 3.
Suppose a function φ ∈ C^{+}(S^{n 1}), and n ≠ p ≥ 1, from the integral representation (2.13), (2.8), and (2.4), it follows that for K(f ), K\left(g\right)\in {\mathcal{K}}_{e}^{n},
{V}_{p}\left(K\left(f\right){+}_{p}K\left(g\right),\varphi \right)={V}_{p}\left(K\left(f\right),\varphi \right)+{V}_{p}\left(K\left(g\right),\varphi \right),
(3.8)
which together with Theorem 1, yields
{V}_{p}\left(K\left(f\right){+}_{p}K\left(g\right),\varphi \right)\ge V{\left(\varphi \right)}^{\frac{p}{n}}\left[V{\left(K\left(f\right)\right)}^{\frac{np}{n}}+V{\left(K\left(g\right)\right)}^{\frac{np}{n}}\right],
(3.9)
with equality if and only if K(f); K(g) and K(φ ) are dilates.
Now, take \varphi ={h}_{K\left(f\right){+}_{p}K\left(g\right)}, recall Vp(K, h_{
K
} ) = V (K ), and from Lemma 2, we get
V{\left(K\left(f\right){+}_{p}K\left(g\right)\right)}^{\frac{np}{n}}\ge V{\left(f\right)}^{\frac{np}{n}}+V{\left(g\right)}^{\frac{np}{n}}.
In view of (2.16), we have
V\left({K}_{p}\left(f{+}_{p}g\right)\right)\ge V\left(K\left(f\right){+}_{p}K\left(g\right)\right).
Hence, we get
\begin{array}{cc}\hfill V{\left({K}_{p}\left(f{+}_{p}g\right)\right)}^{\frac{np}{n}}& \ge V{\left(K\left(f\right){+}_{p}K\left(g\right)\right)}^{\frac{np}{n}}\hfill \\ \ge V{\left(f\right)}^{\frac{np}{n}}+V{\left(g\right)}^{\frac{np}{n}}.\hfill \end{array}
From Lemma 2 again, we obtain
V{\left(f{+}_{p}g\right)}^{\frac{np}{n}}\ge V{\left(f\right)}^{\frac{np}{n}}+V{\left(g\right)}^{\frac{np}{n}}.
In view of the equality condition (3.9), and from Lemma 1, we get the equality holds if and only if there exists a constant c > 0 such that f = cg, almost everywhere with respect to S(K(f ), ·) on S^{n 1}.
Remark 1
The case p = 1 of the inequality of Theorem 3 is
V{\left(f+g\right)}^{\frac{n1}{n}}\ge V{\left(f\right)}^{\frac{n1}{n}}+V{\left(g\right)}^{\frac{n1}{n}},
(3.10)
with equality if and only if there exists a constant c > 0 such that f = cg, almost everywhere with respect to S(K(f ), ·) on S^{n 1}.
The above inequality (3.10) is just the KneserSüss inequality type for the Aleksandrov bodies associated with positive continuous functions.
Actually, from these above proofs, we see BrunnMinkowski inequality, Minkowski inequality, and KnesserSüss inequality are equivalent.