Some refinements of numerical radius inequalities

In this paper, we give some refinements for the second inequality in $\frac{1}{2}\|A\| \leq w(A) \leq \|A\|$, where $A\in B(H)$. In particular, if $A$ is hyponormal by refining the Young inequality with the Kantorovich constant $K(\cdot, \cdot)$, we show that $w(A)\leq \dfrac{1}{\displaystyle {2\inf_{\| x \|=1}}\zeta(x)}\| |A|+|A^{*}|\|\leq \dfrac{1}{2}\| |A|+|A^*|\|$, where $\zeta(x)=K(\frac{\langle |A|x,x \rangle}{\langle |A^{*}|x,x \rangle},2)^{r},~~~r=\min\{\lambda,1-\lambda\}$ and $0\leq \lambda \leq 1$ . We also give a reverse for the classical numerical radius power inequality $w(A^{n})\leq w^{n}(A)$ for any operator $A \in B(H)$ in the case when $n=2$.


Introduction
Suppose that (H, ·, · ) is a complex Hilbert space and B(H) denotes the C * -algebra of all bounded linear operators on H. For A ∈ B(H), let w(A) and A denote the numerical radius and the usual operator norm of A, respectively. It is well known that w(·) defines a norm on B(H), which is equivalent to the usual operator norm · . In fact, for every A ∈ B(H), for each n ∈ N. Many authors have investigated several inequalities involving numerical radius inequalities, see e.g. [1,5,6,8,13,14]. If x, y ∈ H are arbitrary, then the angle between x and y is defined by cos φ x,y = Re x, y x y or by cos ψ x,y = | x, y | x y .
The following inequality for angles between two vectors was obtained by Kreȋn [11] φ x,z ≤ φ x,y + φ y,z (1. 3) for any nonzero elements x, y, z ∈ H. By using the representation and inequality (1.3), he showed that the following triangle inequality is valid for any nonzero elements x, y, z ∈ H.
In section 2 of this paper, we first introduce some new refinements of numerical radius inequality (1.1) by applying the Kreȋn-Lin triangle inequality (1.3) and obtain a reverse of inequality (1.2) in the case when n = 2. In section 3, we obtain some refinements of inequality (1.1) by applying a refinement of the Young inequality.

Some refinements of inequality (1.1) by Kreȋn-Lin triangle inequality
In order to achieve our goals, we need the following lemmas. The first lemma is a simple consequence of the classical Jensen and Young inequalities.
for any r ≥ 1.
The second lemma is a simple consequence of the classical Jensen inequality for convex function f (t) = t r , where r ≥ 1.
Lemma 2.2. If a and b are nonnegative real numbers, then for any r ≥ 1.

Lemma 2.3. [4, Lemma 2.4]
Suppose that x, y ∈ H with y = 1. Then The following lemma is known as a generalized mixed Schwarz inequality.
(ii) If f and g are nonnegative continuous functions on [0, ∞) satisfying f (t)g(t) = t, In the next result, we use some ideas of [3].
If we multiply (2.2) by x y z 2 , then we deduce for any z ∈ H with z = 1 and λ, µ ∈ C.
On the other hand, by applying Lemma 2.4 and the AM-GM inequality, we have Applying again the AM-GM inequality, we get By combining inequalities (2.5), (2.6) and (2.7), we reach By taking the supremum over z ∈ H with z = 1, we deduce for any λ, µ ∈ C.
Finally, taking the infimum over λ, µ ∈ C in the inequality above and utilizing we obtain the desired result (2.1).
Remark 2.6. In Theorem 2.5 if we choose r = 1, f (t) = g(t) = √ t, we get where a 1 , a 2 , b ∈ R. Remark 2.8. If there exists λ 0 ∈ C in which A − λ 0 I ≤ s, then by putting λ = µ = λ 0 and by taking the supremum over z ∈ H with z = 1 in (2.5), we deduce gives the following proposition: Proposition 2.10. Let x, y, z ∈ H with z = 1 and λ, µ ∈ C, a, b > 0, and r ≥ 1 such that Then In particular if x ⊥ y, then for any r ≥ 1.
Proof. Since z is a unit vector, from (2.3) we have By taking supremum over z ∈ H with z = 1, we get and for any r ≥ 1 and a, b > 0.
Proposition 2.10 induces several inequalities as special cases, but here we only focus on the case r = 1, i.e., Remark 2.12. Suppose that the assumptions of Proposition 2.10 are still valid.
• As an application of inequality (2.10) the following reverse of inequality (1.2) for n = 2, i.e., an upper bound for w 2 (A) − w(A 2 ) can be obtained. In fact, by choosing x = Az and y = A * z with z = 1 and taking supremum over z ∈ H with z = 1, we get • By choosing x = Az and y = A −1 z with z = 1, in inequality (2.10) and taking supremum over z ∈ H with z = 1, we have 3. Some refinements of inequality (1.1) by using Young's inequality In this section, we obtain some refinements of inequality (1.1) by applying refinements of the Young inequality. The next lemma is an additive refinement of the scalar Young inequality.
Proof. Let x ∈ H be a unit vector. Then we have By taking supremum over x ∈ H with x = 1, we deduce which is an improvement of inequality (1.1).