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Ostrowski's Type Inequalities for the Complex Integral on Paths

In this paper we extend the Ostrowski inequality to the integral with respect to arc-length by providing upper bounds for the quantity |f(v)l(γ)-∫_{γ}f(z)|dz|| under the assumptions that γ is a smooth path parametrized by z(t), t∈[a,b] with the length l(γ), u=z(a), v=z(x) with x∈(a,b) and w=z(b) while f is holomorphic in G, an open domain and γ⊂G. An application for circular paths is also given. Several applications for circular paths and for some special functions of interest such as the exponential functions are also provided.


INTRODUCTION
In 1938, A. Ostrowski [8], proved the following inequality concerning the distance between the integral mean 1 b−a b a f (t) dt and the value f (x), x ∈ [a, b]. Theorem 1.1 (Ostrowski, 1938 [8]). Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) such that f : (a, b) → R is bounded on (a, b), i.e., f ∞ := sup t∈(a,b) |f (t)| < ∞. Then, for all x ∈ [a, b] and the constant 1 4 is the best possible. In [6], S. S. Dragomir and S. Wang, by the use of the Montgomery integral identity [7, p. 565], where p : [a, b] 2 → R is given by Silvestru Sever Dragomir gave a simple proof of Ostrowski's inequality and applied it for special means (identric mean, logarithmic mean, etc.) and to the problem of estimating the error bound in approximating the Riemann integral b a f (t) dt by one arbitrary Riemann sum (see [6,Section 3]). For extensions of Ostrowski's inequality in terms of the p-norms of the derivative, see [1], [2] and [3]. For a recent survey on Ostrowski's inequality, see [4].
In order to extend this result for the complex integral, we need some preparations as follows. Suppose γ is a smooth path parametrized by z (t) , t ∈ [a, b] and f is a complex function which is continuous on γ. Put z (a) = u and z (b) = w with u, w ∈ C. We define the integral of f on We observe that the actual choice of parametrization of γ does not matter. This definition immediately extends to paths that are piecewise smooth. Suppose γ is parametrized by z (t), t ∈ [a, b], which is differentiable on the intervals [a, c] and [c, b], then assuming that f is continuous on γ we define where v := z (c) . This can be extended for a finite number of intervals. We also define the integral with respect to arc-length and the length of the curve γ is then Let f and g be holomorphic in G, an open domain and suppose γ ⊂ G is a piecewise smooth path from z (a) = u to z (b) = w. Then, we have the integration by parts formula We recall also the triangle inequality for the complex integral, namely We also define the p-norm with p ≥ 1 by If p, q > 1 with 1 p + 1 q = 1, then by Hölder's inequality, we have In the recent paper [5], we obtained the following result for functions of complex variable: Motivated by the above results, in this paper, we extend the Ostrowski inequality to the complex integral, by providing upper bounds for the quantity an open domain and γ ⊂ G. An application for circular paths is also given.

OSTROWSKI TYPE RESULTS
We have the following result for functions of complex variable: Proof. Using the integration by parts formula, we have If we add these two equalities, we get which gives the following equality of interest By taking the modulus in (2.9) and using the properties of modulus, we get and by (2.10), we get the first inequality in (2.8). The second part follows by Hölder's inequalities and if s ∈ γ such that γu,s |f (z)| |dz| = γs,w |f (z)| |dz|, then We have also the following result for p-norms: Theorem 2.4. With the assumption of Theorem 2.3, we have for p, q > 1 with 1 1/q f γu,w;p , Proof. Using the weighted Hölder integral inequality for p, q > 1 with 1 ) . If we add these two inequalities, we get which proves the first inequality in (2.13). We also have By Hölder's discrete inequality, we have which prove the last part of (2.13).
which by addition produce and by utilising the inequality (2.10), we get the first part of (2.16). The second part is obvious and we omit the details.
and |z (t)| = R for t ∈ [a, b] giving that .
and v = R exp (ix) , then by (2.8), we have In particular, if we take x = a+b 2 in (3.17), then we get By making use of (2.13), we get that is equivalent to If we take x = a+b 2 in (3.20), then we get then by (3.20), we get Further, if we use (2.16), then we have |f (R exp (it))| , and by the inequality (3.17), we get