In this paper, we give characterizations of orthogonality conditions in certain classes of normed spaces. We first consider Range-Kernel orthogonality in norm-attainable classes then we characterize orthogonality conditions for Jordan elementary operators.
Studies in normed spaces have been carried out with very interesting results obtained as shown in [1,2,3]. Several properties of operators in these spaces have been studied including norms, orthogonality, spectra, among others in Hilbert spaces and Banach spaces in general [4]. Regarding norm-attainable classes, a lot has been done in terms of structural properties [5,6,7,8,9,10,11,12,13]. Elementary operators have also been of considerable attention on many aspects particularly on their orthogonality, (see [14] and the references therein).
Let \(H\) be an infinite dimensional complex Hilbert space and \(B(H)\) the algebra of all bounded linear operators on \(H\). We say \(T\in B(H)\) is said to be norm-attainable if there exists a unit vector \(x_{0}\in H\) such that \(\|Tx_{0}\|=\|T\|.\) We denote by \(NA(H)\) the class of all norm-attainable operators on \(H.\) Benitez [15] gave a detailed description of several types of orthogonality which have been studied in real normed spaces namely: Robert’s orthogonality, Birkhoff’s orthogonality, Orthogonality in the sense of James, Isosceles, Pythagoras, Carlsson, Diminnie, Area among others as described in [16].
For \(x\in \mathcal{M}\) and \(y\in \mathcal{N},\) where \( \mathcal{M}\) and \( \mathcal{N}\) are subspaces of \(E\) which is a normed linear space, we have:
Let \(\mathcal{A}=B(H)\). For \(A,\,B\in B(H)\) we define specific elementary operators as follows [21]:
Definition 1. Let \(W\) be a complex normed space, then for any elements \(x, y \in W,\) we say that \(x\) is orthogonal to \(y\), noted by \(x\perp y,\) if and only if for all \(\alpha, \beta \in \mathbb{C}\) there holds \(\|\alpha y + \beta x \| \geq\|\beta x \|.\)
Definition 2. Let \(W\) be a complex Banach space. If \(P\) and \(Q\) are linear subspaces in \(W\), we say that \(P\) is orthogonal to \(Q\) , denoted by \(P\perp Q\), if \(\|x + y\|\geq \|x\|\) for all \(x \in P\) and all \( y \in Q\). If \(P = {x}\), we simply write \(x\perp Q\).
Remark 1. We note that the orthogonality in the definition above is not symmetric [17] and if \(W\) is a Hilbert space with its inner product then it follows from [18] that \(\langle x, y\rangle= 0\) which means that Birkhoff-James’s orthogonality generalizes the usual sense of orthogonality in a Hilbert space [19].
Definition 3. Let \(F: X \rightarrow Y\) and \(J: Y \rightarrow Z\) be operators between norm-attainable classes. We say \(J\) is orthogonal to \(F\) if \(s\in Ker T\Rightarrow \|s+E(x)\|\geq \|s\|,\, \forall\, x\in X\). Moreover, if \(F=J\), we shall say that \(F\) is orthogonal.
Remark 2. The authors in [10] proved that if \(A\) and \(B\) are normal operators then for all \(X, S \in B (H),\, S \in ker \delta _{A,B} \Rightarrow \|\delta _{A,B} (X) + S\| = \|S\|,\) where the \( ker \delta _{A,B}\) denotes the kernel of \(\delta _{A,B}.\) This means that the kernel of \(\delta _{A,B}\) is orthogonal to its range. This result has been generalized in different directions, to non-normal operators [20], to \(C_{p}(H) \), and to some elementary operators [19].
Proposition 1. Let \(K\) be a norm-attainable subclass of a norm-attainable set \(X\) and \(x \not\in K,\) then \(x \perp K\) if and only if there exists \( \tilde{\varphi} \in D (x)\) such that \( K \subseteq ker \tilde{\varphi}\).
Proof. Since \(K\) is a norm-attainable class, let \(\varphi \in D(x)\) be such that \(\langle \varphi,y \rangle=0,\) for all \(y\in K\). Then \(\varphi(x+y)=\varphi(x)=\varphi\|x\|^{2}\) and \( \|x\| ^{2}= \|\varphi\| \|x + y\| = \|x\| \|x + y\| , \) that is, \(\|x\| = \|x + y\|,\) for all \(y\in K\). Hence, \(x\perp K.\)
Conversely, let \(x \not\in K\) such that \(x \perp K\). Then, for all \(y \in K\), \(x\) and \(y\) are linearly independent vectors. Let \(L\) be the closed norm-attainable subclass spanned by \(K\) and \({x}, L = [K, {x}]\). Define the function \(\varphi\) on \(L\) by \(\varphi (\alpha x + \beta y) = \alpha \|x\|^{2},\) for all \(y \in K\) and \(\alpha\, , \beta \in \mathbb{C}\). Clearly, \(\varphi\) is linear (by the assumption that \(K\) is a linear subset of \(X\) ). To prove the continuity of \(\varphi\), let \(z \in L\), then \(z = \alpha x + \beta y\) and \(\varphi (z) = \alpha \|x\|^{ 2}\). By the definition of \(\varphi\) and the assumption that \(x \perp K,\) we derive that \(\|z\| = \|\alpha x\|.\) If \(\alpha \neq 0\), it is easy to see from known inequalities that \(|\varphi(z)|=|\alpha|\|x\|^{2}.\frac{\|z\|}{\|z\|}\leq \frac{|\alpha|\|x\|^{2}}{\|\alpha x\|}\|z\|=\|x\|\|z\|. \) If \(\alpha =0\), then \(|\varphi(z)|=|\varphi (\beta y)|=0\). Hence, \( |\varphi (z)|\leq \|x\|\|z\|, \) for all \(z\in L\). Therefore, \(\varphi\) is continuous on \(L\) and \(\|\varphi \| = \|x\|.\) By Hahn-Banach theorem there is a continuous linear functional \( \tilde{\varphi}\) on \(X\) such that \( \tilde{\varphi}_{|L}=\varphi\) and \(\| \tilde{\varphi}\|=\|\varphi\|,\) where \(\tilde{\varphi}_{|L}\) is the restriction of \(\varphi\) on \(L\). It follows, by the definition of \(\varphi\) and \(\tilde{\varphi}_{|L}=\varphi\) that \(K \subseteq ker \tilde{\varphi} \) and \(\tilde{\varphi}\in D (x).\)
Theorem 1. Let \(K\) be a norm-attainable subclass of a norm-attainable set \(X\). For all \(x, y \in X\,, \,x \perp y\) if and only if there exists \(\varphi\in D (x)\) such that \(\varphi (y) = 0.\) Moreover, for all \(x \in X \) and for all \( \varphi \in D (x)\), \( x \perp ker \varphi.\)
Proof. From Proposition 1 and an analogous computation from [20], the proof is clear.
Remark 3. We can consider general normed spaces as follows: Let \(K\) be a nonempty subset of a Banach space \(X\) and \(T \in NA(X )\), we denote the duality adjoint of \(T\) by \(T^{\dagger}\) and set \(K^{\perp r}=\{x \in X : x\perp y; \forall y \in K\}.\) It is clear that if \(\{x_{n}\}_{n}\) is a sequence in a subset \(K\) converging to \(y\) and \(x \perp x_{n},\) for all \(n\), then \(x \perp y.\) Hence, \(x \perp K \rightarrow x \perp \overline{K}.\)
Proposition 2. Let \(X\) be a norm-attainable subclass of \(NA(H)\), then
Proof.
Remark 4. Let \(f : X \rightarrow X\) be a map on \(X\), not necessarily linear or additive, and \(F_{f} : X \rightarrow \mathbb{R}^{+}\) be a map defined by \( F_{f}(x)=\|f(x)\|,\,\forall\, x\in X.\) We say that \(F_{f}\) has a global minima at \(a \in X\) if \(\|f (a)\| = \|f (x)\|\), for all \(x \in X \).
As an application of the previous results, the following result gives us a necessary and sufficient conditions in term of Birkhoff-James orthogonality for minimizing the map \(F_{f}.\)Lemma 1. Let \(T\) and \(f : X \rightarrow X\) be norm-attainable maps and \(a \in X.\) Suppose that the relation \(f (x) + T (y) = f (y) + T (x)\) holds for all \(x, y \in X\), if \(f (a)\perp T (x),\, \forall x \in X\) then the map \(F_{f}\) has a global minima at \(a\). Moreover, if we suppose that \(T\) is linear and \(f (x) = T (x) + f (0), \forall x \in X \), then \(F_{f}\) has a global minima at \(a\) if and only if \(f (a)\perp ran T\) if and only if there is \(\varphi \in D (f (a))\) such that \(\varphi \in ker T^{\dagger}\), where \(T^{\dagger}\) is the duality map of \(T.\) Lastly, if \(f(a)\) is a smooth point, then the existence of \(\varphi\) is unique.
Proof. It follows from Theorem 1, that for all \(x \in X , f (a)\perp T (x) \Leftrightarrow \exists \varphi \in X^{\dagger}: \varphi (f (a)) = \|f (a)\|_{2}= \|\varphi\|^{2}\) and \(\varphi (T (x)) = 0\). Then by the relation defined in (i), we get \(\varphi (f (a)) = \varphi (f (a) + T (x)) = \varphi (f (x) + T (a)) =\varphi (f (x))\), so that \(\|\varphi (f (a))\| = \|f (a)\| \|\varphi\| \leq \|\varphi\| \|f (x)\|\). Hence, \(\|f (a)\| \leq \|f (x)\|\), for all \(x \in X \). Since the maps \(f, T\) satisfy the relation cited in (i), the sufficient condition follows from the first part of the proof. By linearity of \(T\), we get \(f (a) + \lambda T (x) = f (a + \lambda x),\) for all \(x \in X , \lambda \in \mathbb{C}\). Hence, \(F_{f}\) has a global minima at a implies \(\|f (a) + \lambda g(x)\| \geq \|f (a + \lambda x)\| = \|f (a)\|\). The other equivalence follows immediately. To complete the proof, if \(f (a)\) is a smooth point, then \(f (a)\) has only one functional support and therefore \(D (f (a))\) has one element.
Lemma 2. If \(\mathcal{J}\) is a separable ideal of norm-attainable operators in \(NA(H)\) equipped with unitary invariant norm, then its dual \(\mathcal{I}\) is isometrically isomorphic to an ideal of compact operator \(\mathcal{Q}\) not necessarily separable, i.e., \[ \phi:\mathcal{Q}\rightarrow \mathcal{J^{\dagger}},\, R\mapsto \phi_{R}(X)=tr (XR).\]
Proof. Following the argument in Lemma 1, the proof is trivial.
Theorem 2. Let \(A \in C_{p}(H), T \in B(C_{p}(H))\) and \(f, F_{f}\) are defined as in Lemma 1, where \(f (A)\) is given by its polar decomposition \(f (A) = u |f (A)|.\) If \(A \in C_{p}(H)\), then \(F_{f}\) has a global minimizer at A if and only if \(f(A)\perp ran T\) if and only if \(|f (A)|^{p-1}u^{*}\in ker T^{\dagger}\). Moreover, if \(A \in C_{1} (H)\) and \(f (A)\) is a smooth point, then \(F_{f}\) has a global minimizer at \(A\) if and only if \(f (A)\perp ran T\) if and only if \(u^{*}\in ker T^{\dagger}\) when \(f (A)\) is injective (or \(u \in kerT^{\dagger}\) when \(f (A)^{*}\)is injective).
Proof. From Lemma 1, we have that \(F_{f}\) has a global minimizer at \(A\) if and only if there exists \(\varphi\in D (f (A))\) such that \(\varphi \in ker T^{\dagger}\). If \(A \in C_{p}(H)\), then by the properties of the isomorphism, it follows that \( \varphi \in C_{p}(H)^{\dagger}\) if and only if there exists \( R\in C_{p}(H)\) such that \(\phi_{R}=\varphi, \) \( \|\varphi\|=\|R\| \) and \(\varphi(X)=tr\, RX,\) for all \(X\in C_{p}(H).\) Hence, the smoothness of \( C_{p}(H),\) \(F_{f}\) has a global minimizer at \(A\) if and only if there is a unique operator \(R\) such that \(\varphi (f (A)) = tr (f (A)R) =\|f (A)\|^{2}_{p}= \|R\|^{2}_{q}\) and \(tr (T^{\dagger}(R) X)= 0\), for all \(X \in C_{p}(H)\). To complete the proof, it is well known that \(C_{1} (H)\) is neither reflexive, nor smooth and its dual \(C_{1} (H)^{\dagger}\) is isometrically isomorphic to \(NA(H).\) This isomorphism is given by \( \phi\in NA(H)\) if and only if \( C_{1} (H)^{\dagger}\) contains \(R\mapsto\phi_{R}\) such that \(\phi_{R}(X)=tr(XR)\) so we have that \( \varphi\in C_{1} (cH)^{\dagger}\Leftrightarrow\exists R \in NA(H):\phi_{R}=\varphi,\) \(\|\varphi\|=\|R\|\) and \(\varphi(X)=tr\, RX\, \forall\, X\in C_{1} (H)\), so that if \(f (A)\) is a smooth point then \(F_{f}\) has a global minimizer at \(A\) if and only if there is a unique operator \(R\) such that \( \varphi (f (A)) = tr (f (A)R) = \|f (A)\|^{2}_{1}= \|R\|^{2}\) and \( tr (T^{\dagger}(R) X)= 0, \forall\, X\in C_{1} (H).\) Since \(f (A)\) is smooth then by [7], either \(f (A)\) or \(f (A)^{*}\) is injective, thus either \(u\) or \(u^{*}\) is an isometry i.e., \(uu^{*}= I\) or \(u^{*}u = I\). So it suffices to take \(R = \|f (A)\|_{1}u^{*} \)or \(R = \|f (A)\|_{1} u\), which is the unique operator required in both cases. So, \(f (A)\) or \(f (A)^{*}\) is injective.
Proposition 3. Let \(K\) be a closed subclass of a norm-attainable class \(X\). If \(X\) is separable and \(K^{\perp r}= {0}\), then \(K = X.\)
Proof. If \(K \neq X\), then there exists \(\varphi \in X^{\dagger} :K\subseteq ker\varphi.\) Since \(D (\varphi)\) is not empty, then there is \(f \in X^{\dagger\dagger}\) with \(f (\varphi) = \|\varphi\|^{2}= \|f \|^{2}\). Let \(J\) be the natural injection between \(X\) and \( X^{\dagger\dagger}\) i.e., \(J : X \rightarrow X^{\dagger\dagger}, \forall x \in X , \forall \psi \in X^{\dagger},\) such that \( J (x)\psi = \psi (x),\; \|J (x)\|= \|x\|.\) So, by the separability of \(X , J\) is a bijection and, then there is \(0 \neq x \in X\) such that \(J (x) = f\). Hence, \(\varphi (x) = \|x\|^{2}= \|\varphi\|^{2}\). Thus, \(\varphi \in D (x)\) and by application of Proposition 1, we get \(0 \neq x \in K^{\perp r}= {0}\), a contradiction.
Proposition 4. Let \(X\) be a separable, smooth and strictly convex norm-attainable class and \(T \in NA(X )\). If \(T^{\dagger}\)is orthogonal, then \(\forall s\in X,\; s\perp ran T\Rightarrow s\in Ker T.\)
Proof. Let \(s \in X\) such that \(\perp ran T\). Then by [7] and the smoothness of \(X\), there is a unique \(\varphi_{s} \in D (s)\) such that \(\varphi_{s}\in ker T^{\dagger}\). Again, by assumptions of the proposition and arguments in [21], there is \(\psi_{\varphi_{s}}\in D (\varphi_{s})\) such that \(\psi_{\varphi_{s}}\in ker T^{\dagger\dagger}\). Let \(J\) be the natural injection between \(X\) and \(X^{*}\) as defined in [22]. We see that \(J (s) \varphi_{s} = \|\varphi_{s}k\|\) and \(\|J (s)\| = \|\varphi_{s}\|,\) which means that \(J (s) \in D (\varphi_{s}).\) By the separability of \(X , J\) is a bijection. Hence, there is \(c \in X\) such that \( \psi_{\varphi_{s}}= J (c)\) and \(\|J (c)\|=\|c\|=\|\varphi\|, J (c) \varphi_{s} =\varphi_{s} (c) = \psi_{\varphi_{s}}(\varphi_{s}) = \|\varphi_{s}\|^{2}\). Then \(\varphi_{s} \in D (s) \bigcap D (c)\) and since \(X\) is strictly convex, we get \(c = s\). Thus, \(J (c) = J (S) \in ker T^{\dagger\dagger}.\) Therefore, it immediately follows that \( (T^{\dagger\dagger} J (s))\varphi =J (s)( T^{\dagger})=(T^{\dagger}\varphi)s = \varphi (T s) = 0,\) for all \(\varphi \in X^{\dagger}\). That is \(s \in ker T\).
In the next result we consider general Banach spaces. If \(X\) is a reflexive separable Banach space and \(T^{\dagger}\) is orthogonal then the implication of orthogonality holds with respect to a suitable norm in \(X\). Indeed, if \(X\) is separable, then there is an equivalent norm which is smooth and strictly convex in \(X\). This can be seen in the next theorem.Theorem 3. Let \(X\) be a reflexive, smooth and strictly convex Banach space and \(T \in B (X ).\) If \(T\) and \(^{\dagger}\) are orthogonal. Then \( \forall s \in X : s\perp ran T \Leftrightarrow s \in ker T ,\) where \( X = ker T \oplus \overline{ran (T )}. \)
Proof. If \(T\) is orthogonal then, by Definition 2, it follows that \( \forall s \in X \) such that \( s\perp ran T\), it implies that \( s \in kerT \) and the reverse implication follows by Proposition 1. Let us prove the decomposition. Let \( y \in X\) such that \(y \in (ker T \oplus ran (T ))^{\perp r}\), then there is \(\varphi _{y} \in D (y)\) such that \(\varphi _{y}(s \oplus T X ) = 0\), for all \(s \in ker T\) and all \(x \in X\). For \(s = 0\), it follows, by [12], that \(Y \perp ran T \), and by [13], \(ran T \subseteq ker T\). So, we can choose \(x = 0\) and \(s = y\), such that this yields \(\varphi_{y}(y) = 0\). This means \(y\perp y,\) and hence \(y = 0\). Finally, the decomposition follows immediately.
Proposition 5. Let \(H, K\) be Hilbert spaces, \(A \in NA(H),\, B \in NA(K)\) and \(E \in B (NA(K, H))\) such that \(E (X) = AXB+BXA\). If \(A \) and \(B^{*}\) are injective operators then \(E\) is injective.
Proof. If either \(AXB = 0\) or \(BXA = 0\) with \(A\) injective, then we have that \(XB = 0 = B^{*}X= 0\) implies \(X^{*}= 0 =X\) since \(B^{*}\) is injective. Thus, \(E\) is injective.
Proposition 6. Let \(A = (A_{1}, A_{2}, ..A_{n})\) and \(B = (B_{1}, B_{2}, ..B_{n})\) with \(A_{i} , B_{i}\) be operators in \(B (H)\) such that \( \sum\limits_{i=1}^{n}A_{i}A_{i}^{*}\leq 1, \sum\limits_{i=1}^{n}A_{i}A_{i}^{*}\leq 1, \sum\limits_{i=1}^{n}B_{i}B_{i}^{*}\leq 1,\,\text{and}\ \ \sum\limits_{i=1}^{n}B_{i}B_{i}^{*}\leq 1. \) If \(E\) is the elementary operator defined on \(C_{p:1\leq p< \infty}\) by \( E(X)=\sum_{i=1}^{n}A_{i}XB_{i}-X \) then \(ker E = ker \tilde{E} \Rightarrow ker E^{\dagger}= ker \tilde{E^{\dagger}}\). Moreover, if \( E (S) = 0 =\tilde{E} (S)\) for some compact operator \(S\), then \([|S| , B_{i}] = 0 \) \(\forall 1 \leq i < n\).
Proof. We have that \(E^{\dagger}(S)=0\Leftrightarrow \tilde{E}S^{*}=0.\) Then from the equality, \(Ker E=Ker \tilde{E}. \) It follows that \(E^{\dagger}(S)=0\Leftrightarrow E(S^{*})=0\Leftrightarrow \tilde{E^{\dagger}}(S)=0\). The rest is trivial.
At this point we give certain necessary conditions and characterization of the operators in \(C_{p}\)-classes whose kernels are orthogonal to the ranges of certain kinds of elementary operators, in particular, we consider the Jordan elementary operator.Proposition 7. Let \(E\) be an elementary operator defined on \(C_{p}\) then \(\forall S, X \in C_{p}(H),\; \|E (X) + S\|_{p}\geq \|S\|_{p}\Rightarrow S\in ker E\) then \(E (X) = AXB\) with \(A^{*}\) and \(B\) injective operators \(E (X) = AXB – CXD,\) where \(A, B\) normal operators, \(D, C^{*}\) hyponormal operators with \([A, C] = [B, D] = 0\) and \(ker A^{*}\bigcap ker C^{*}= {0} = ker B \bigcap ker D\).
Proof. The duality adjoint \(E^{\dagger}\) is defined by \(E^{\dagger}(X) = BXA\) and using a result of [19], we get \(E^{\dagger}\) is injective and hence orthogonal. So, the result follows by Proposition 1. Next, we have \(E^{\dagger}(X) = BXA – DXC\) and applying the result of [20], we get \(E^{\dagger}\) is orthogonal, and so by [16], the proof is complete.
Theorem 4. Let \(A,B\in NA(H)\) be hyponormal operators, such that \(AB=BA\), and let \(\mathcal{U}(X)=AXB+BXA.\) Furthermore, suppose that \(A^{*}A+B^{*}B>0.\) If \(S\in Ker\mathcal{U},\) then \(|\|\mathcal{U}(X)+S\||\geq|\|S\||.\)
Proof. Follows trivially from the proof of the sum of two basic elementary operators as shown in [5] and from the fact that \(|\|.|\|\) is a unitarily invariant norm.
We extend Theorem 4 to distinct hyponormal operators \(A,B,C,D\in NA(H)\) in the theorem below:Theorem 5. Consider \(A,B,C,D\in NA(H)\) as hyponormal operators, such that \(AC=CA\), \(BD=DB,\) \(AA^{*}\leq CC^{*}\) , \(B^{*}B\leq D^{*}D\). Let the Jordan elementary operator be given as \(\mathcal{U}(X)=AXB+CXD\) and \(S\in NA(H)\) satisfying \(ASB=CSD,\) then \(\|\mathcal{U}(X)+S\|\geq \|S\|,\) for all \(X\in NA(H).\)
Proof. Since \(\mathcal{U}(X)=AXB+CXD\) and the fact that \(AC=CA\), \(BD=DB,\) \(AA^{*}\leq CC^{*}\) , \(B^{*}B\leq D^{*}D\), then it is easy to see that the operator is injective. So, with \(S\in NA(H)\) satisfying \(ASB=CSD,\) then \(\|\mathcal{U}(X)+S\|\geq \|S\|\) follows analogously from the proof of the sum of two basic elementary operators as shown in [8].
Theorem 6. Consider \(A,B,C,D\in NA(H)\) as hyponormal operators, such that \(AC=CA\), \(BD=DB,\) \(AA^{*}\leq CC^{*}\) , \(B^{*}B\leq D^{*}D\). Let the Jordan elementary operator be give as \(\mathcal{U}(X)=AXB-CXD\) and \(S\in NA(H)\) satisfying \(ASB=CSD,\) then \(\|\mathcal{U}(X)+S\|\geq \|S\|,\) for all \(X\in NA(H).\)
Proof. From \(AA^{*}\leq CC^{*}\) and \(B^{*}B\leq D^{*}D,\) let \(A=CU,\) and \(B=VD,\) where \(U,V\) are unitaries. So we have \(AXB-CXD=CUXVD-CXD=C(UXV-X)D.\) Assume \(C\) and \(D^{*}\) are injective, \(ASB=CSD\) if and only if \(USV=S.\) Moreover, \(C\) and \(U \) commute. Indeed, from \(A=CU\) we obtain \(AC=CUC.\) Therefore, \(C(A-UC)=0.\) Thus since \(C\) is injective \(A=CU.\) Similarly, \(D\) and \(V \) commute. So, \[ \|\mathcal{U}(X)+S\| = \|[AXB-CXD]+S\| = \|[U(CXD)V-CXD] +S\| \geq \|S\|, \forall X\in NA(H). \] The rest is clear from the analogous assertions in [10] for hyponormal operators.