A kind of extension of the famous Young inequality

*Correspondence: luoxq1978@126.com 1Department of Mathematics, Shanghai University, Shanghai, 200444, China 2Mathematics College, Wuyi University, Guangdong, 529020, China Abstract Young inequality, extended in (Geometry of Orlicz Spaces, 1986; Geometry of Orlicz Spaces, 1996), has extensive use and great effort in mathematical analysis. By the kind of extended Young inequality, we can get the famous Holder inequality and the Minkowski inequality. But until now, we have not found its strict proof of analysis. In (Geometry of Orlicz Spaces, 1986; Geometry of Orlicz Spaces, 1996), only the probable pattern description was found. In this paper, we will get the strict proof of analysis of a kind of extension of Young inequality with the approximation method. MSC: 46B20; 46B02; 46A22


§1 Introduction
The original Young Inequality [1]  Inequality(see references [2] and [3]). But until now we have not found its strict analysis proof.
In the references [2] and [3], only the probable pattern description was found.
In this paper, we will get its strict analysis proof with the approximation method. §2 Preliminaries Therefore, we have That is,    Lemma 2.5 [2] For any N-function M (u) and ε > 0 §there exists a strictly convex N- where p(t) and p 1 (t) is the right derivative of M (u) and M 1 (u) respectively.
Record 2.2 Lemma 2.5 is Theorem 1.10 in reference [2] , but it reverses the old conclusion From the construction process of p 1 (t) in the proof in reference [3], we know if p(t) is continuous, then p 1 (t) is also continuous.
. By the symmetry we get another necessary and sufficient condition is v ∈ Proof Sufficiency.
t ≤ u, then for any 1 n , we get Let n → ∞, since p(t) is right continuous, then we have On the other hand, from u < q(v) = sup we get So, we have Necessity.
We have The next two lemmas are about change of variable of integral and distribute integral.
Lemma 2.8 [4] Suppose f (x) and g(x) are defined on the interval [a, b], and the Stieltjes Proof Suppose u ≥ 0 and v ≥ 0.
Firstly, we will prove the necessity of the equality.
Suppose there exist u 0 ≥ 0 and v 0 ≥ 0 satisfying From Young Inequality we have known that for all u and v, F (u, v) ≥ 0.
, that is, the necessity of the equality holds.
Therefore, the left derivative of F (u, v 0 ) is less than or equal to zero on the point u 0 , and the right derivative of F (u, v 0 ) is more than or equal to zero on the point u 0 .
That is, the necessity of the equality holds.
Secondly, we will get the proof of Young Inequality and the sufficiency of the equality in three steps: Step Hence, by the expression (1), we have Hence, by the expression (1), we have From the expression (1), we have uv = M (u) + N (v).
That is, the sufficiency of the equality holds.
Step II. Suppose M (u) is strictly convex, then from Lemma 2.4, the right derivative p(t) is strictly increasing, and the right-inverse function q(s) is continuous and nondecreasing.
From Lemma 2.5 and Record 2.2, ∀0 < ε < 1 2 , we can construct a function strictly increasing and continuous q 1 (s), such that Hence, Let p 1 (t) is the right-inverse function of q 1 (s), then p 1 (t) is strictly increasing and continuous.
In the following we will get the relation of p 1 (t) and p(t) .
In the expression (2), let s = p 1 (t), we have That is, From Lemma 2.3 and the expression (3), Since q(s) is nondecreasing, by the expression (4), we get From the result in step I, we get Therefore, In the following, we will prove the sufficiency of the equality.
If v = p(u), from Lemma 2.3 and the expression (3), for the above 0 < ε < 1 2 , we have In the expression (6), let ε → 0 §by Lemma 2.2, we get On the other hand, in the expression (5), let ε → 0, Therefore, Now we need prove In fact, if s = p(u), from Definition 2.2, since p(u) is strictly increasing, then we have q(s) = sup

By the result in
Step I, we have From the expressions ( 9) and (11), we get On the other hand, we have got the inequality uv ≤ M (u) + N (v) .
Let v = p(u), we have Therefore, together with the expression (12), we have That is, the sufficiency of the equality holds.
Step III §for any N-function M (u), suppose its complementary N-function is N (v), p(t) is the right-inverse function of M (u), and q(s) is the right-inverse function of N (v). From Lemma 2.5, for the above 0 < ε < 1 2 , we can find a strictly convex N-function M 1 (u) and its right-derivative p 1 (t) , such that Suppose N 1 (v) is the complementary N-function of M 1 (u), q 1 (s) is the right derivative of In the following we will get the relation of q 1 (t) and q(t).
for the above 0 < ε < 1 2 . In the expression (13), let t = q 1 (s) − ε, we have From Lemma 2.3, we have that Therefore, by the expressions (14) and (15), And then, by Lemma 2.3, together with the expression (16), we have Since p(t) is nondecreasing, then by the expression (17), we get From the result in Step II, we get Let ε → 0, we have uv ≤ M (u) + N (v).
In the following we will prove sufficiency of the equality .

By the result in
Step II, we have On the other hand, we have got the inequality uv ≤ M (u) + N (v) .
That is, the sufficiency of the equality holds.