Based on the local fractional calculus, we establish some new generalizations of H\"{o}lder's inequality. By using it, some results on the generalized integral inequality in fractal space are investigated in detail.
While the renowned inequality of Hölder [1] is well celebrated for its beauty and its wide range of important applications to real and complex analysis, functional analysis, as well as many disciplines in applied mathematics. The purpose of this work is to establish some generalizations of Hölder inequality on local fractional calculus and other inequality based on it.
Fractal calculus (also called local fractional calculus) has played an important role in not only mathematics but also in physics and engineers [2][3][4][5][6][7][8][9][10][11][12][13][14][15]. Local fractional derivative [6][7][8] were written in the form
where
Local fractional integral of f (x) [6][7]9] was written in the form
with ∆t j = t j+1 -t j and ∆t = max{∆t 1 , ∆t 2 , . . . , ∆t j , . . .}, where for j = 1, 2, . . . , N -1t 0 = a and t N = b, [t j , t j+1 ] is a partition of the interval [a, b]. Aims of this paper are to study the some new generalizations of Hölder’s inequality and some results based on them.
In the section, we give some generalizations of Hölder inequality. In order to prove our results, we first review the Hölder inequality [15]:
equalities holding if and only if f (x) = λg(x), where λ is a constant.
Based on Theorem 2.1, we have the following important result.
equalities holding if and only if f (x) = λg(x), where λ is a constant.
Proof. Set c = 1/p, then we have q = -pd, d = c/(c -1). By (2.1), we obtain
In (2.3), multiplying both sides by
. By Theorem 2.1 and Theorem 2.2, we give the following result.
(1) for p j > 1, we have
(2) for 0 < p 1 < 1, p j < 0, j = 2, . . . m,we have
Proof.
(1)We use induction on m. When m = 2, we are given p 1 , p 2 > 0 with 1/p 1 + 1/p 2 = 1.
In particular, we have p 1 , p 2 > 1 and so (2.5) is reduced to the Höder’s inequality (2.1). Now suppose (2.5) holds for some integer m ≥ 2. We claim that it also holds for m + 1. So let p 1 , p 2 , . . . , p m+1 > 0 be real numbers with
Note that, as above, we must have p i > 1 for j = 1, 2, . . . m, m + 1. In particular, we have
Thus by the Höder’s inequality (2.1),
Next, since p j (p 1 -1)/p 1 > 0, for j = 2, . . . m, m + 1.
(2.9)
(2.10) by induction hypothesis and (2.8), we arrive at
Hence, we arrive at the result.
(2) Similar to the proof of (2.5), we use induction on m. Clearly when m = 2, equation (2.6) reduces to the Hölder’s inequality (2.2). Now suppose that (2.6) holds for some integer m ≥ 2. We claim that it also holds for m + 1. So let p 1 , p 2 , . . . , p m < 0 and p m+1 ∈ R be such that m+1 j=1 1/p j = 1 and let f j (x) ≥ 0, j = 1, 2, . . . m, m + 1. Note that 0 < p m+1 < 1, since
(2.12)
by the Höder’s inequality (2.2), we have
and as in (2.10), m+1 j=2 p 1 /p j (p 1 -1) = 1.
(2.15) by induction hypothesis and (2.13), we obtain
To set the stage, we recall Minkowski inequality [15]:
equalities holding if and only if f (x) = λg(x), where λ is a constant.
equalities holding if and only if f (x) = λg(x), where λ is a constant.
. By Hölder inequality, in view of 0 < p < 1, we have
By inequality (3.3), we arrive to reverse Minkowski’s inequality and the theorem is completely proved.
(2) for 0 < p < 1 ,we have
Proof.
(1) it follows from theorem 2.1 that (2) Similar to the proof of (3.2), we obtain (3.5). (3.7)
(2) for 0 < p < 1, we have
equalities holding if and only if f (x) = λg(x).
Proof. By Theorem 2.1 and Theorem 3.1, We have .
(3.12)
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