selleck inhibitor H?lland et al. [9] have shown that Theorem B reduces to Theorem A if we deal with Ces��ro means of order��and consider a functionf Lip��,0 < �� �� 1. Working in same direction we prove the following theorem.Theorem 1 ��Iff : R �� Ris a2��periodic, Lebesgue integrable and belonging toLip(��(t), p)for > 1and ift1/p)pdt1/p=O(��(1n)),(16)t1/p+2)pdt1/p=O(��(1n)n)(17)conditions (16) and (17) hold uniformly inx, then degree of approximation of??f~(x), conjugate off Lip��(t), p, by Euler (E, q) mean��~n(f;x)=(1+q)?n��k=0n(nk)qn?kS~k,(18)of the conjugate series (2) is given by||��~n(f;x)?f~(x)||p=O(��(1n)(n)1/2p).(19)In order to prove our theorem, we need the following lemma.Lemma 2 ��If0 < t �� ��, then(1+q)?n(1+q2+2qcos?t)n/2��e?2qt2n/��(1+q)2.

(20)Proof ��We ��e?4qt2/��(1+q)2,(21)sinceex(1?��1?4qt2??��2(1+q)2?=1?4qsin2(t/2)(1+q)2?have(1+q)?2(1+q2+2qcos?t) ? x) < 1when0 < x < 1. Therefore,(1+q)?n(1+q2+2qcos?t)n/2��e?2qt2n/��(1+q)2.(22)Proof of Theorem 1 ��Thekthpartial sum of the conjugate series of the Fourier series (2) is given =12��0��cos?(k+1/2)tsin?(t/2)��(t)dt.(23)Taking??+12��0��cos?(k+1/2)tsin(t/2)��(t)dtS~k(f;x)?(?12��0��cot?(t2)��(t)dt)?byS~k(f;x)=?12��0��cot(t2)��(t)dt Euler (E, q) means, we get(24)(25)using Lemma +|f(u+x)?f(u?x?t)|.(26)Hence, by?2.Clearly,|��(x+t)?��(x)|��|f(u+x+t)?f(u+x)| Minkowski's =O(��(t)).(27)Then,f?+pdx1/p??��pdx1/p?inequality,(��(x+t)?��(x))1/p Lip(��(t), p)�� Lip(��(t), p).

Using H?lder’s inequality,��(t) Lip(��(t), p), condition (16),sint �� (2t/��), lemma, and second mean value theorem for integrals, Entinostat we ��??t(e?2qt2n/��(1+q)2)dt].(30)Using???��(1+q2+2qcos?t)n/2dt}=O��(t)=O[1n��1/n�Ц�(t)sin(t/2)???��sin?(t/2)1/p��=O(��(1n))��01/nt?p��dt1/p��=O(��(1n)(n)1/2p).(29)Now,|K~2(x)|=O[��1/n��|��(t)|sin(t/2)(1+q)?n(1+q2+2qcos?t)n/2|R(t)|dt]=O{��1/n��|��(t)|sin(t/2)(1+q)?n?����01/n((1+q)?nS~(t)sin?(t/2))p��dt1/p��,(28)wherep��=pp?1|K~1(x)|=O��01/n(1/p?have|K~1(x)|=O��01/n(1/p H?lder’s inequality,��(t) Lip(��(t), p), and condition (17), we ��[��1/n��??t(e?2qt2n/��(1+q)2)p��dt]1/p��=O(��(1n)(n)1/2p).(31)Combining?have|K~2(x)|=O[1n��1/n��(��(t)t1/p+2)pdt]1/p (24) with (31), we have||��~n(f;x)?f~(x)||p=O(��(1n)(n)1/2p),(32)which completes the proof of the theorem.3. CorollariesThe following corollaries may be derived from our theorem.Corollary 3 ��If��(t) = t��, then the degree of approximation of a function??f~(x), conjugate off Lip(��, p),1/p < �� < 1, by Euler’s means(E, q)of the conjugate series of the Fourier series (2) is given by||��~n(f;x)?f~(x)||p=O(1n(��p?1)/2p).