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A kind of extension of the famous Young inequality
Journal of Inequalities and Applications volume 2013, Article number: 437 (2013)
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] has been proposed in an integral form by Young in 1912. Suppose that f(x) is a strictly increasing and continuous function defined in [0,c], {f}^{1}(x) is the inverse function of f(x), f(0)=0, a\in [0,c], b\in [f(0),f(c)]. Then
where the equality holds if and only if b=f(a).
Young inequality has an extensive use and a great effort in mathematical analysis. Now Young inequality was extended as follows.
Let M(u) and N(v) be complementary Nfunction with each other (see Definition 2.1 and Definition 2.2), then the kind of Young inequality uv\le M(u)+N(v) holds, and the equality holds if and only if u=q(v)signv or v=p(u)signu for all u,v\in (\mathrm{\infty},+\mathrm{\infty}).
By the kind of Young inequality, we can get the famous Holder inequality and the Minkowski inequality (see references [2] and [3]). But until now, we have not found its strict proof of analysis. In references [2] and [3], only the probable pattern description was found. Some other decisions can be found in [4–8].
In this paper, we will get its strict proof of analysis with the approximation method.
2 Preliminaries
Definition 2.1 [2]
The mapping M:(\mathrm{\infty},+\mathrm{\infty})\to (\mathrm{\infty},+\mathrm{\infty}) is called an Nfunction if it has the following properties:

(i)
M(u) is even, continuous, convex and M(0)=0.

(ii)
M(u)>0 for all u\ne 0.

(iii)
{lim}_{u\to 0}\frac{M(u)}{u}=0 and {lim}_{u\to \mathrm{\infty}}\frac{M(u)}{u}=\mathrm{\infty}.
Lemma 2.1 [2]
M(u) is an Nfunction if and only if there exists p(u):[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) with the following properties:

(i)
p(u) is rightcontinuous and nondecreasing;

(ii)
p(u)>0 whenever u\ne 0;

(iii)
p(0)=0 and p(\mathrm{\infty})=\mathrm{\infty}, M(u)={\int}_{0}^{u}p(t)\phantom{\rule{0.2em}{0ex}}dt.
Record 2.1 [2]
p(u) is the rightderivative of Nfunction M(u).
Lemma 2.2 Let {p}_{}(u) be the leftderivative of Nfunction M(u), then {p}_{}(u)={lim}_{h\to {0}^{+}}p(uh), and {\int}_{0}^{u}{p}_{}(t)\phantom{\rule{0.2em}{0ex}}dt=M(u).
Proof From the proof process of Theorem 1.4 in reference [2], we know {p}_{}(u) is left continuous, and for all 0<u<v, p(u)\le {p}_{}(v)\le p(v).
Hence, for h>0, we have p(vh)\le {p}_{}(v) .
Therefore,
On the other hand, since {p}_{}(v)\le p(v) and {p}_{}(v) is left continuous, we get
Therefore, we have
Since for all h>0, p(vh)\le {p}_{}(v)\le p(v), then we have
That is,
Let h\to 0, by the property (i) of M(u) in Definition 2.1, we have
□
Definition 2.2 [2]
Suppose that M(u) is an Nfunction. Let p(t) be the right derivative of M(u). Let q(s)={sup}_{p(t)\le s}t={inf}_{p(t)>s}t, called the rightinverse function of p(t). By Theorem 1.5 in reference [2], we know that q(s) also satisfies the three properties of Lemma 2.1, and N(v)={\int}_{0}^{v}q(s)\phantom{\rule{0.2em}{0ex}}ds is called the complementary Nfunction of M(u). It is obvious, the left derivative {q}_{}(s) of N(v) satisfies {q}_{}(s)={sup}_{p(t)<s}t={inf}_{p(t)\ge s}t.
Lemma 2.3 [2]
q(p(t))\ge t, p(q(s))\ge s; q(p(t)\epsilon )\le t, p(q(s)\epsilon )\le s.
Lemma 2.4 [2]
M(u) is strictly convex if and only if p(t) is strictly increasing, that is, q(s) is continuous.
Lemma 2.5 [2]
For any Nfunction M(u) and \epsilon >0, there exists a strictly convex Nfunction {M}_{1}(u), such that
where p(t) and {p}_{1}(t) are the right derivatives 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 ‘M(u)\le {M}_{1}(u)\le (1+\epsilon )M(u),’ for the new conclusion ‘(1\epsilon )M(u)\le {M}_{1}(u)\le (1+\epsilon )M(u).’ 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.
Lemma 2.6 Suppose that u\ge 0 and v\ge 0, then u=q(v) or v=p(u) if and only if u\in [{q}_{}(v),q(v)]. By the symmetry, we get another necessary and sufficient condition, that is, v\in [{p}_{}(u),p(u)].
Proof Sufficiency.
Suppose that u\in [{q}_{}(v),q(v)].

(i)
If {q}_{}(v)=q(v), it is clear that u={q}_{}(v)=q(v).

(ii)
If {q}_{}(v)\ne q(v), then {q}_{}(v)<q(v). If u=q(v), then the conclusion holds.
If {q}_{}(v)\le u<q(v), we need only to prove that p(u)=v.
From Definition 2.2, we have
Since {q}_{}(v)\le u\Rightarrow {sup}_{p(t)<v}t\le u, then for any \frac{1}{n}, we get
Let n\to \mathrm{\infty}, since p(t) is right continuous, then we have
On the other hand, from u<q(v)={sup}_{p(t)\le v}t={inf}_{p(t)\ge v}t, we get
So, we have
Necessity.
If u=q(v), it is clearly established.
If p(u)=v, then from
and
We have
□
The next two lemmas are about the change of variable of integral and distribute integral.
Lemma 2.7 [9]
Suppose thatf(x) and g(x) are defined on the interval [a,b], and the Stieltjes integral of f(x) about g(x) exists. Suppose that x(t) is a strictly increasing and continuous function on the interval [\alpha ,\beta ], and x(\alpha )=a and x(\beta )=b, then
Lemma 2.8 [9]
Suppose that f(x) and g(x) are defined on the interval [a,b], and the Stieltjes integral of f(x) about g(x) exists, then
3 Main result
Theorem 3.1 Suppose that M(u) is an Nfunction, and N(v) is the complementary Nfunction of M(u), then Young inequality uv\le M(u)+N(v) holds, and uv=M(u)+N(v) holds if and only if u=q(v)signv or v=p(u)signu.
Proof Suppose that u\ge 0 and v\ge 0.
Firstly, we will prove the necessity of the equality.
Suppose that there exist {u}_{0}\ge 0 and {v}_{0}\ge 0 satisfying
Let
From Young inequality, we have learned that for all u and v, F(u,v)\ge 0.
From M({u}_{0})+N({v}_{0})={u}_{0}{v}_{0}, we have F(u,{v}_{0})=M(u)+N({v}_{0})u{v}_{0} and we can get the minimum 0 in {u}_{0}.
If {u}_{0}=0, from M({u}_{0})+N({v}_{0})={u}_{0}{v}_{0}, we get that {v}_{0}=0, then {u}_{0}=q({v}_{0})=0 or {v}_{0}=p({u}_{0})=0, that is, the necessity of the equality holds.
If {u}_{0}\ne 0, then F({u}_{0},{v}_{0}) is the minimum of the F(u,{v}_{0}) on the interval (0,+\mathrm{\infty}).
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,
Then
From Lemma 2.6, we get {u}_{0}=q({v}_{0}) or {v}_{0}=p({u}_{0}).
That is, the necessity of the equality holds.
Secondly, we will get the proof of the Young inequality and the sufficiency of the equality in three steps.
Step I. Suppose that M(u) and N(v) are all strictly convex. From Lemma 2.4, the right derivative p(t) and q(s) are all strictly increasing, continuous, and are the right inversefunction of each other. From the reference [9], we have that the Stieltjes integral {\int}_{0}^{q(v)}t\phantom{\rule{0.2em}{0ex}}dp(t) exists.
From Lemma 2.7 and Lemma 2.8, we have

(i)
If u>q(v), then p(u)>v.
Hence, by expression (1), we have

(ii)
If u<q(v), then p(u)\le v.
Hence, by expression (1), we have

(iii)
If u=q(v), then v=p(u).
From expression (1), we have uv=M(u)+N(v).
That is, the sufficiency of the equality holds.
Step II. Suppose that M(u) is strictly convex, then from Lemma 2.4, the right derivative p(t) is strictly increasing, and the rightinverse function q(s) is continuous and nondecreasing.
From Lemma 2.5 and Record 2.2, \mathrm{\forall}0<\epsilon <\frac{1}{2}, we can construct a function strictly increasing and continuous {q}_{1}(s) such that
Hence,
Let {p}_{1}(t) be the rightinverse 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 expression (2), let s={p}_{1}(t), we have
That is,
From Lemma 2.3 and expression (3), we get
Since q(s) is nondecreasing, by expression (4), we get
From the result in Step I, we get
Therefore,
Let \epsilon \to 0, we have
In the following, we will prove the sufficiency of the equality.
If v=p(u), from Lemma 2.3 and expression (3), for 0<\epsilon <\frac{1}{2} above, we have
In expression (6), let \epsilon \to 0, by Lemma 2.2, we get
On the other hand, in expression (5), let \epsilon \to 0, we get
Therefore,
By Lemma 2.2, we get
Now we need to prove that
In fact, if s=p(u), from Definition 2.2, since p(u) is strictly increasing, then we have q(s)={sup}_{p(t)\le s}t={sup}_{p(t)\le p(u)}t=u. If s\in [{p}_{}(u),p(u)), from Lemma 2.6, we get q(s)=u. Therefore, we have {\int}_{{p}_{1}(u)}^{p(u)}q(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{{p}_{1}(u)}^{p(u)}u\phantom{\rule{0.2em}{0ex}}ds=u(p(u){p}_{1}(u)).
By the result in Step I, we have
From expressions (9) and (11), we get
Let \epsilon \to 0, we have
On the other hand, we have got the inequality uv\le M(u)+N(v) .
Let v=p(u), we have
Therefore, together with expression (12), we have
That is, the sufficiency of the equality holds.
Step III for any Nfunction M(u), suppose that its complementary Nfunction is N(v), p(t) is the rightinverse function of M(u), and q(s) is the rightinverse function of N(v). From Lemma 2.5, for 0<\epsilon <\frac{1}{2} above, we can find a strictly convex Nfunction {M}_{1}(u) and its rightderivative {p}_{1}(t) such that
Suppose that {N}_{1}(v) is the complementary Nfunction of {M}_{1}(u), {q}_{1}(s) is the right derivative of {N}_{1}(v).
In the following, we will get the relation of {q}_{1}(t) and q(t) for 0<\epsilon <\frac{1}{2} above. In expression (13), let t={q}_{1}(s)\epsilon, we have
From Lemma 2.3, we have that
Therefore, by expressions (14) and (15), we have
Then, by Lemma 2.3, together with expression (16), we have
Since p(t) is nondecreasing, then by expression (17), we get
From the result in Step II, we get
Let \epsilon \to 0, we have
In the following, we will prove sufficiency of the equality.
By the result in Step II, we have
Therefore,
Let \epsilon \to 0, together with expression (13), we get {p}_{1}(u)\to p(u), {M}_{1}(u)\to M(u), and N(\frac{{p}_{1}(u)}{1+2\epsilon})\to N(p(u)) since N(v) is continuous.
Therefore,
On the other hand, we have got the inequality uv\le M(u)+N(v).
Let v=p(u), we have
Therefore, together with expression (20), we have
That is, the sufficiency of the equality holds. □
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Acknowledgements
This research was partially supported by the National Natural Science Foundation of China (Grant No: 11271245, and Grant No: 11301397), and the Natural Science Foundation Guangdong Province of China (2012KJCX0101).
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LX conceived of the study, XZ participated in its design and study. All authors read and approved the final manuscript.
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Xianqiang, L., Zhiping, X. A kind of extension of the famous Young inequality. J Inequal Appl 2013, 437 (2013). https://doi.org/10.1186/1029242X2013437
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DOI: https://doi.org/10.1186/1029242X2013437
Keywords
 Young inequality
 Nfunction
 strictly convex function