Convex functions and approximate Birkhoff–James orthogonality

Abstract

For the well developed notion of approximate Birkhoff-James orthogonality, in a real or complex normed linear space, we formulate a new characterization. It can be derived from other, already known, characterizations as well as obtained in a more elementary and direct way, on the basis of some simple inequalities for real convex functions.

1 Introduction

The standard orthogonality relation in an inner product space:

$$\begin{aligned} x\bot y\quad \Longleftrightarrow \quad \left\langle x|y\right\rangle =0, \end{aligned}$$

can be well generalized to a normed linear space in terms of the Birkhoff-James orthogonality denoted here by $\bot _{\textrm{B}}$ (cf. the original sources [3, 8,9,10] or recent expositions [1, 2]). From now on, we assume X is a normed linear space with $\dim X\ge 2$ and over the scalar field ${\mathbb {K}}$ of real or complex numbers. Then, for given two vectors $x,y\in X$, we define:

$$\begin{aligned} x\bot _{\textrm{B}}y\quad \Longleftrightarrow \quad \forall \,\lambda \in {\mathbb {K}}:\ \Vert x+\lambda y\Vert \ge \Vert x\Vert . \end{aligned}$$

For a given $\varepsilon \in [0,1)$, the notion of $\varepsilon$-Birkhoff-James orthogonality has been introduced by Chmieliński [4] as follows:

$$\begin{aligned} x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y\quad \Longleftrightarrow \quad \forall \,\lambda \in {\mathbb {K}}:\ \Vert x+\lambda y\Vert ^2\ge \Vert x\Vert ^2-2\varepsilon \Vert x\Vert \,\Vert \lambda y\Vert . \end{aligned}$$

(1.1)

If the norm comes from an inner product, then (1.1) reduces to the natural notion of $\varepsilon$-orthogonality in an inner product space:

$$\begin{aligned} x\bot ^{\hspace{-0.2em}\varepsilon }\hspace{0.1em}y\quad \Longleftrightarrow \quad |\left\langle x|y\right\rangle |\le \varepsilon \,\Vert x\Vert \,\Vert y\Vert . \end{aligned}$$

(1.2)

A slightly different approach was presented by Dragomir [7]:

$$\begin{aligned} x\bot \hspace{-1.5ex}{\ \atop {\small \varepsilon }}\hspace{0.2ex}_{\textrm{B}}y\quad \Longleftrightarrow \quad \forall \,\lambda \in {\mathbb {K}}:\ \Vert x+\lambda y\Vert \ge (1-\varepsilon )\Vert x\Vert \end{aligned}$$

but we are not going to deal with it in the present paper.

Denoting by $X^*$ the dual of X, we define for a vector $x\in X\smallsetminus \{0\}$ the nonempty set of supporting functionals at x:

$$\begin{aligned} J(x)=\{\varphi \in X^*:\ \Vert \varphi \Vert =1,\ \varphi (x)=\Vert x\Vert \,\}. \end{aligned}$$

These functionals can be used for characterizations of the exact and approximate Birkhoff–James orthogonalities. A classical result of James [9, Corollary 2.2] states that:

$$\begin{aligned} x\bot _{\textrm{B}}y\quad \Longleftrightarrow \quad \exists \, \varphi \in J(x):\quad \varphi (y)=0. \end{aligned}$$

(1.3)

1.1 Characterizations of the approximate Birkhoff–James orthogonality

The following theorem collects some known characterizations of the $\varepsilon$-Birkhoff-James orthogonality $\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}$ defined by (1.1).

Theorem 1.1

Let X be a real or complex normed linear space with $\dim X\ge 2$, let $x,y\in X$ and $\varepsilon \in [0,1)$. Then the following conditions are equivalent:

• (A) $x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y$;

• (B) $\exists z\in {\text {Lin}}\{x,y\}:\ x\bot _{\textrm{B}}z\ \text{ and }\ \Vert z-y\Vert \le \varepsilon \Vert y\Vert$;

• (C) $\exists \, \varphi \in J(x):\ |\varphi (y)|\le \varepsilon \Vert y\Vert$.

Implications $(B)\Rightarrow (A)$ and $(C)\Rightarrow (A)$ are not difficult to verify. The more demanding reverse implications were first shown for real spaces in [6, Theorem 2.2] and recently Wójcik [12, Theorem 2.2] extended them to the complex case. Note that the characterization $(A)\Leftrightarrow (C)$ extends (1.3).

The considered $\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}$ relation is useful as a tool for describing some geometrical properties of normed spaces. Therefore, it is reasonable to ask for yet more characterizations.

1.2 Another characterization of the approximate orthogonality

Observe first that in the definition (1.1) of $\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}$, it is essential to verify the inequality only for $\lambda$ close to zero. Namely, we have:

Lemma 1.2

For an arbitrary normed space X over ${\mathbb {K}}\in \{{\mathbb {R}},{\mathbb {C}}\}$, vectors $x,y\in X$ and a constant $\varepsilon \in [0,1)$ the following conditions are equivalent:

• (a) $x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y$;

• (b) $\exists \, c>0:\quad (\lambda \in {\mathbb {K}},\ |\lambda |\le c) \Rightarrow \Vert x+\lambda y\Vert ^2\ge \Vert x\Vert ^2-2\varepsilon \Vert x\Vert \,\Vert \lambda y\Vert$.

Proof

Obviously (a) implies (b). Assume now (b) and define

$$\begin{aligned} f(\lambda ):=\Vert x+\lambda y\Vert ^2+2\varepsilon \Vert x\Vert \,\Vert \lambda y\Vert ,\qquad \lambda \in {\mathbb {K}}. \end{aligned}$$

The mapping $f:{\mathbb {K}}\rightarrow [0,+\infty )$ is convex and $f(0)=\Vert x\Vert ^2$. Moreover, it follows from (b) that $f(\lambda )\ge \Vert x\Vert ^2$ whenever $|\lambda |\le c$. The convexity of f yields that the inequality is true for all $\lambda \in {\mathbb {K}}$. Indeed, assume, contrary to our claim, that (a) does not hold and $\Vert x+\lambda _{0}y\Vert ^2<\Vert x\Vert ^2-2\varepsilon \Vert x\Vert \,\Vert \lambda _0 y\Vert$ for some $\lambda _0\in {\mathbb {K}}$. This would mean $f(\lambda _0)<\Vert x\Vert ^2$. Taking $n\in {\mathbb {N}}$ large enough to have $\left| \frac{\lambda _0}{n}\right| \le c$ and using the convexity of f we would then get:

$$\begin{aligned} \Vert x\Vert ^2\le & {} f\left( \frac{\lambda _0}{n}\right) =f\left( \frac{1}{n}\lambda _0+\left( 1-\frac{1}{n}\right) \cdot 0\right) \\\le & {} \frac{1}{n}f(\lambda _0)+\left( 1-\frac{1}{n}\right) f(0)<\frac{1}{n}\Vert x\Vert ^2 +\left( 1-\frac{1}{n}\right) \Vert x\Vert ^2=\Vert x\Vert ^2, \end{aligned}$$

a contradiction. $\square$

Taking $\varepsilon =0$ we get the following characterization of the exact Birkhoff-James orthogonality:

$$\begin{aligned} x\bot _{\textrm{B}}y\quad \Longleftrightarrow \quad \exists \, c>0:\quad (\lambda \in {\mathbb {K}},\ |\lambda |\le c) \Rightarrow \Vert x+\lambda y\Vert \ge \Vert x\Vert . \end{aligned}$$

(1.4)

The following result brings a new characterization of the approximate Birkhoff–James orthogonality, which is given by the condition:

• (D) $\forall \,\lambda \in {\mathbb {K}}:\quad \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert$.

Theorem 1.3

For a real or complex normed space X, vectors $x,y\in X$ and a constant $\varepsilon \in [0,1)$ the following conditions are equivalent:

• (i) $x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y$;

• (ii) $\forall \,\lambda \in {\mathbb {K}}:\quad \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert$;

• (iii) $\exists \, c>0:\quad (\lambda \in {\mathbb {K}},\ |\lambda |\le c) \Rightarrow \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert$.

Proof

If $x=0$ or $y=0$ all three conditions are clearly satisfied. For $\varepsilon =0$, the equivalence of (i)-(iii) follows from (1.4). From now on, we assume that $x,y\in X\smallsetminus \{0\}$ and $\varepsilon \in (0,1)$.

To prove that (i) implies (ii) we will use the implication $(A)\Rightarrow (B)$ in Theorem 1.1. Assuming (i), there exists $y_0\in X$ such that $x\bot _{\textrm{B}}y_0$ and $\Vert y-y_0\Vert \le \varepsilon \Vert y\Vert$. Thus, for any $\lambda \in {\mathbb {K}}$:

$$\begin{aligned} \Vert x+\lambda y\Vert= & {} \Vert x+\lambda y_0 +\lambda y - \lambda y_0\Vert \ge \Vert x+\lambda y_0\Vert - |\lambda |\,\Vert y-y_0\Vert \\\ge & {} \Vert x\Vert -\varepsilon |\lambda |\,\Vert y\Vert . \end{aligned}$$

The implication (ii)$\Rightarrow$ (iii) is obvious, so we prove now (iii)$\Rightarrow$(i). Assuming that (iii) holds true with some $c>0$ and making use of Lemma 1.2, it is sufficient to show that condition (b) in the lemma is satisfied. Let $d:=\min \left\{ \frac{\Vert x\Vert }{\varepsilon \Vert y\Vert },c\right\}$. Now, if $|\lambda |\le d$, then $\Vert x\Vert -\varepsilon \Vert \lambda y\Vert \ge 0$ whence:

$$\begin{aligned} \Vert x+\lambda y\Vert ^2\ge & {} \left( \Vert x\Vert -\varepsilon \Vert \lambda y\Vert \right) ^2=\Vert x\Vert ^2-2\varepsilon \Vert x\Vert \,\Vert \lambda y\Vert +\varepsilon ^2\Vert \lambda y\Vert ^2\\\ge & {} \Vert x\Vert ^2 -2\varepsilon \Vert x\Vert \,\Vert \lambda y\Vert \end{aligned}$$

and we are done. $\square$

A self-contained proof of the equivalence $(A)\Leftrightarrow (B)\Leftrightarrow (C)\Leftrightarrow (D)$ for the real case was given in [5].

The above proof, seemingly short and easy, depends on the implication $(A)\Rightarrow (B)$ in Theorem 1.1, which is a quite involved result. On the other hand, looking at conditions (A) and (D), one can expect a more elementary verification of its equivalence. That is indeed possible, and actually a yet more general observation can be made. We will show it in Sect. 3, relying on some properties of convex functions, which will be established in the next section.

2 On a property of convex functions

In this section, we describe some property of convex functions defined on the set of nonnegative reals or its initial interval. The property itself is quite elementary but useful, as we will see in the subsequent section. Throughout this section, I stands for an interval of the form [0, c) or [0, c], with $c>0$, or $I=[0,+\infty )$.

The following is a key result.

Theorem 1.4

Let $f,g:I\rightarrow [0,+\infty )$ be convex functions such that $f(0)=1$ and $g(0)=0$. Then, for any $n\in \mathbb {N}$ the following conditions are equivalent:

• (a) $f(t)\ge 1-g(t),\quad t\in I$;

• (b) $f(t)^n\ge 1- ng(t),\quad t\in I$.

Proof

Notice first that since g is convex and $g(0)=0$, we must have

$$\begin{aligned} \lim _{t\rightarrow 0^+}g(t)\le 0. \end{aligned}$$

But $g\ge 0$, so

$$\begin{aligned} \lim _{t\rightarrow 0^+}g(t)=0. \end{aligned}$$

(2.1)

Similarly, the convexity of f and $f(0)=1$ yields

$$\begin{aligned} \lim _{t\rightarrow 0^+}f(t)\le 1. \end{aligned}$$

(2.2)

Suppose that (a) holds. Because of (2.1), there is a $\delta >0$ such that $1-g(t)>0$ for $t\in [0,\delta ]$. Therefore we have from (a)

$$\begin{aligned} f(t)\ge 1-g(t)>0,\qquad t\in [0,\delta ]. \end{aligned}$$

Now, using the Bernoulli inequality we obtain

$$\begin{aligned} f(t)^n\ge \left( 1-g(t)\right) ^n\ge 1-ng(t),\qquad t\in [0,\delta ], \end{aligned}$$

i.e., (b) holds on $[0,\delta ]$. Suppose that it does not hold for some $t_0>\delta$, that is,

$$\begin{aligned} f(t_0)^n+ng(t_0)<1. \end{aligned}$$

Pick $k\in \mathbb {N}$ such that $\frac{t_0}{k}\le \delta$. We would then have

$$\begin{aligned} 1&\le f\left( \frac{t_0}{k}\right) ^n+ng\left( \frac{t_0}{k}\right) \\&=f\left( \frac{1}{k}t_0+\left( 1-\frac{1}{k}\right) \cdot 0\right) ^n+ng\left( \frac{1}{k}t_0+\left( 1-\frac{1}{k}\right) \cdot 0\right) \\&\le \frac{1}{k}f(t_0)^n+\left( 1-\frac{1}{k}\right) f(0)^n+\frac{1}{k}ng(t_0)+\left( 1-\frac{1}{k}\right) ng(0)\\&= \frac{1}{k}\left( f(t_0)^n+ng(t_0)\right) +\left( 1-\frac{1}{k}\right) (f(0)^n+ng(0))\\&<\frac{1}{k}\cdot 1+\left( 1-\frac{1}{k}\right) (1+0)=1, \end{aligned}$$

a contradiction.

Now we prove the reverse and assume that (b) holds for some $n\in \mathbb {N}$, whence

$$\begin{aligned} \lim _{t\rightarrow 0^+}(f(t)^n-1+ng(t))\ge 0. \end{aligned}$$

Bearing (2.1) and (2.2) in mind, suppose that $\lim _{t\rightarrow 0^+}f(t)=a<1$. We would then have

$$\begin{aligned} \lim _{t\rightarrow 0^+}(f(t)^n-1+ng(t))=a^n-1<0, \end{aligned}$$

a contradiction. Therefore,

$$\begin{aligned} \lim _{t\rightarrow 0^+}f(t)=1. \end{aligned}$$

(2.3)

We consider two cases.

Case 1: For each $t\in I$ we have $f(t)\ge 1$. Then

$$\begin{aligned} 1\le f(t)\le f(t)+g(t) \end{aligned}$$

so (a) holds.

Case 2: There is a $t_1>0$ such that $0\le f(t)\le 1$ for $t\in [0,t_1]$.

Define $F(n,t)=\sum \limits _{k=0}^{n-1}f(t)^k$. Then

$$\begin{aligned} 0\le f(t)^n-1+ng(t)=(f(t)-1)F(n,t)+ng(t) \end{aligned}$$

and thus

$$\begin{aligned} 0\le \frac{F(n,t)}{n}(f(t)-1)+g(t),\qquad t\in I. \end{aligned}$$

(2.4)

Let us fix $\gamma \in (0,1)$. By (2.3), $\lim \limits _{t\rightarrow 0^+}\frac{F(n,t)}{n}=1$ and we can find $\delta >0$ such that for each $t\in [0,\delta ]$,

$$\begin{aligned} 1-\gamma<\frac{F(n,t)}{n}<1+\gamma . \end{aligned}$$

For $t\in [0,t_1]\cap [0,\delta ]$ we have $f(t)-1\le 0$ and from (2.4)

$$\begin{aligned} 0\le \frac{F(n,t)}{n}(f(t)-1)+g(t)\le (1-\gamma )(f(t)-1)+g(t). \end{aligned}$$

Set

$$\begin{aligned} \mu (t)=(1-\gamma )(f(t)-1)+g(t),\qquad t\in I. \end{aligned}$$

Then $\mu (0)=0\le \mu (t)$ for each $t\in [0,t_1]\cap [0,\delta ]$ and the convexity of $\mu$ implies that $\mu (t)\ge 0$ for each $t\in I$ so we have

$$\begin{aligned} 0\le (1-\gamma )(f(t)-1)+g(t),\qquad t\in I. \end{aligned}$$

Fixing $s\in I$ we have that

$$\begin{aligned} 0\le (1-\gamma )(f(s)-1)+g(s) \end{aligned}$$

holds true for any $\gamma \in (0,1)$. Letting $\gamma \rightarrow 0^+$ we obtain

$$\begin{aligned} 0\le f(s)-1+g(s) \end{aligned}$$

and since s was arbitrary,

$$\begin{aligned} 0\le f(t)-1+g(t),\qquad t\in I, \end{aligned}$$

which is the required condition (a). $\square$

We reformulate the above theorem to omit the assumption $f(0)=1$.

Corollary 1.5

Let $f,g:I\rightarrow [0,+\infty )$ be convex functions such that $g(0)=0$. Then, for $n\in \mathbb {N}$, the following conditions are equivalent:

• (a) $f(t)\ge f(0)-g(t),\quad t\in I$;

• (b) $f(t)^n\ge f(0)^n-nf(0)^{n-1}g(t),\quad t\in I$.

Proof

The case $f(0)=0$ is trivial. Otherwise, we consider convex mappings ${\tilde{f}}=f/f(0)$ and ${\tilde{g}}=g/f(0)$, for which we have ${\tilde{f}}(0)=1$ and ${\tilde{g}}(0)=0$. It is easy to see that functions f, g satisfy conditions (a) or (b) if and only if ${\tilde{f}},{\tilde{g}}$ satisfy, respectively, conditions (a) or (b) in Theorem 2.1. So our assertion follows from that theorem. $\square$

Taking $g(t):=\varepsilon t$, $t\in I$, we get the result which will be directly used in the next section.

Corollary 1.6

Let $f:I\rightarrow [0,+\infty )$ be a convex function such that $f(0)=1$. Then, for any $n\in {\mathbb {N}}$ and $\varepsilon \in [0,1)$, the following conditions are equivalent:

• (a) $f(t)\ge 1-\varepsilon t,\quad t\in I$;

• (b) $f(t)^n\ge 1-n\varepsilon t,\quad t\in I$.

Remark 2.4

In Theorem 2.1, the exponent $n\in {\mathbb {N}}$ can in fact be replaced by an arbitrary real number $\alpha >1$. Namely, condition (b) in Theorem 2.1 can be replaced by

• (b’) $f(t)^{\alpha }\ge 1-\alpha g(t),\quad t\in I$.

Indeed, the Bernoulli inequality holds for real powers as well, hence (a)$\Rightarrow$(b’) follows. For the reverse, we replace F(n, t) by

$$\begin{aligned} F(\alpha ,t):=\frac{f(t)^{\alpha }-1}{f(t)-1} \end{aligned}$$

proving that $\lim _{t\rightarrow 0^{+}}F(\alpha ,t)=\alpha$. Then we use the argument with $\gamma$ (as in the proof of Theorem 2.1) for the function $\frac{F(\alpha ,t)}{\alpha }$ whose limit for $t\rightarrow 0^{+}$ is again 1 and the rest of the proof is the same. We omit the details.

3 New characterization of approximate orthogonality—a direct proof

Let us go back to normed spaces and the Birkhoff–James orthogonality. We are now in a position to give an elementary proof of the characterization (A) $\Leftrightarrow$ (D), which does not require referring to other characterizations.

Theorem 1.8

Let X be a normed space over ${\mathbb {K}}\in \{{\mathbb {R}},{\mathbb {C}}\}$, $x,y\in X$ and $\varepsilon \in [0,1)$. For any $n\in {\mathbb {N}}$, the following conditions are equivalent:

$$\begin{aligned}{} & {} \forall \,\lambda \in {\mathbb {K}}:\ \ \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert ; \end{aligned}$$

(3.1)

$$\begin{aligned}{} & {} \forall \,\lambda \in {\mathbb {K}}:\ \ \Vert x+\lambda y\Vert ^n\ge \Vert x\Vert ^n-n\varepsilon \Vert x\Vert ^{n-1}\cdot \Vert \lambda y\Vert . \end{aligned}$$

(3.2)

Proof

Conditions (3.1), (3.2) can be reformulated as:

$$\begin{aligned} \forall \,\alpha \in [0,2\pi )\ \forall \,t\in [0,+\infty ):\Vert x+t e^{i\alpha } y\Vert \ge \Vert x\Vert -\varepsilon t\Vert e^{i\alpha } y\Vert ; \end{aligned}$$

(3.3)

and

$$\begin{aligned} \forall \,\alpha \in [0,2\pi )\ \forall \,t\in [0,+\infty ):\Vert x+ t e^{i\alpha }y\Vert ^n\ge \Vert x\Vert ^n-n\varepsilon \Vert x\Vert ^{n-1}t\Vert e^{i\alpha } y\Vert . \end{aligned}$$

(3.4)

Notice the homogeneity of condition (3.2) in the sense that if it holds for a pair of vectors x, y then it does so for $\alpha x,\beta y$ with any $\alpha ,\beta \in {\mathbb {K}}$. Therefore, we may assume that $\Vert x\Vert =\Vert y\Vert =1$.

Fix $\alpha \in [0,2\pi )$ if $\mathbb {K}=\mathbb {C}$ or $\alpha \in \{0,\pi \}$ if $\mathbb {K}=\mathbb {R}$. Define a function $f:[0,+\infty )\rightarrow (0,+\infty )$ by the formula $f(t):=\Vert x+t e^{i\alpha }y\Vert$. Now the conditions (3.3), (3.4) become

• (a) $f(t)\ge 1-\varepsilon t,\quad t\ge 0$;

• (b) $f(t)^n\ge 1-n\varepsilon t,\quad t\ge 0$,

the equivalence of which is guaranteed by Corollary 2.3. $\square$

Condition (3.2) for $n=2$ means that $x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y$. So we have the characterization (A) $\Leftrightarrow$ (D):

$$\begin{aligned} x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y\ \Longleftrightarrow \forall \,\lambda \in {\mathbb {K}}:\ \ \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert \end{aligned}$$

(3.5)

and actually, for any $n\in {\mathbb {N}}$,

$$\begin{aligned} x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y\ \Longleftrightarrow \forall \,\lambda \in {\mathbb {K}}:\ \ \Vert x+\lambda y\Vert ^n\ge \Vert x\Vert ^n-n\varepsilon \Vert x\Vert ^{n-1}\Vert \lambda y\Vert . \end{aligned}$$

(3.6)

Remark 3.2

Following Remark 2.4, we can replace (3.6) by

$$\begin{aligned} x\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}y\ \Longleftrightarrow \forall \,\lambda \in {\mathbb {K}}:\ \ \Vert x+\lambda y\Vert ^{\alpha }\ge \Vert x\Vert ^{\alpha }-\alpha \varepsilon \Vert x\Vert ^{\alpha -1}\Vert \lambda y\Vert \end{aligned}$$

(3.7)

with an arbitrary $\alpha \ge 1$.

4 Applications in operator theory

The notions of exact and approximate Birkhoff–James orthogonalities have also been applied in the operator theory. Let X and Y be normed linear spaces and let $T,S:X\rightarrow Y$ be linear and continuous operators. In several papers, the authors have adopted the definition (1.1) to define the approximate Birkhoff-James orthogonality of operators T and S:

$$\begin{aligned} T\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}S\quad \Longleftrightarrow \quad \forall \lambda \in {\mathbb {K}}:\ \Vert T+\lambda S\Vert ^2\ge \Vert T\Vert ^2-2\varepsilon \Vert T\Vert \,\Vert \lambda S\Vert \end{aligned}$$

(4.1)

where $\Vert \cdot \Vert$ denotes the operator norm. As it follows from (3.5), the above definition can be, equivalently, formulated as:

$$\begin{aligned} T\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}S\quad \Longleftrightarrow \quad \forall \lambda \in {\mathbb {K}}:\ \Vert T+\lambda S\Vert \ge \Vert T\Vert -\varepsilon \Vert \lambda S\Vert . \end{aligned}$$

(4.2)

We believe that such a simpler form could be helpful in further investigations.

Just as an example, we apply it to a characterization from [11]. Here by $M_T$ we denote the norm attaining set for the operator T, i.e.,

$$\begin{aligned} M_T=\{x\in X:\ \Vert x\Vert =1,\ \Vert Tx\Vert =\Vert T\Vert \}. \end{aligned}$$

Theorem 1.10

( [11, Theorem 3.3]) Let X be a real reflexive Banach space and let Y be a real normed space. Let T and S be linear and compact operators. Then $T\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}S$ if and only if there exist $x,y\in M_T$ such that

$$\begin{aligned} \Vert Tx+\lambda Sx\Vert ^2\ge \Vert T\Vert ^2-2\varepsilon \Vert T\Vert \,\Vert \lambda S\Vert \ \text{ for } \text{ all }\ \lambda \ge 0 \end{aligned}$$

(4.3)

and

$$\begin{aligned} \Vert Ty+\lambda Sy\Vert ^2\ge \Vert T\Vert ^2-2\varepsilon \Vert T\Vert \,\Vert \lambda S\Vert \ \text{ for } \text{ all }\ \lambda \le 0. \end{aligned}$$

(4.4)

Using Theorem 2.1 (or actually Corollary 2.2) we can modify the above result. For a fixed $x\in M_T$ let us define

$$\begin{aligned} f(\lambda )=\Vert Tx+\lambda Sx\Vert ,\qquad \lambda \ge 0 \end{aligned}$$

and

$$\begin{aligned} g(\lambda )=\varepsilon \lambda \Vert S\Vert ,\qquad \lambda \ge 0. \end{aligned}$$

Then f and g are convex functions on $[0,+\infty )$, $f(0)=\Vert Tx\Vert =\Vert T\Vert$ and $g(0)=0$. It follows easily that (4.3) is equivalent to

$$\begin{aligned} f(\lambda )^2\ge f(0)^2-2f(0)g(\lambda ),\qquad \lambda \ge 0. \end{aligned}$$

On the other hand, the condition

$$\begin{aligned} f(\lambda )\ge f(0)-g(\lambda ),\qquad \lambda \ge 0 \end{aligned}$$

is equivalent to

$$\begin{aligned} \Vert Tx+\lambda Sx\Vert \ge \Vert T\Vert -\varepsilon \Vert \lambda S\Vert ,\qquad \lambda \ge 0. \end{aligned}$$

(4.5)

Therefore, Corollary 2.2 yields that (4.3) is equivalent to (4.5). Similarly, taking $y\in M_T$ and

$$\begin{aligned} f(\lambda )=\Vert Ty-\lambda Sy\Vert ,\qquad \lambda \ge 0 \end{aligned}$$

we obtain that (4.4) is equivalent to

$$\begin{aligned} \Vert Ty+\lambda Sy\Vert \ge \Vert T\Vert -\varepsilon \Vert \lambda S\Vert ,\qquad \lambda \le 0. \end{aligned}$$

(4.6)

In consequence, we can reformulate Theorem 4.1:

Theorem 1.11

Let X be a real reflexive Banach space and let Y be a real normed space. For linear and compact operators $T,S:X\rightarrow Y$, $T\bot ^{\hspace{-0.2em}\varepsilon }_{\textrm{B}}S$ if and only if there exist $x,y\in M_T$ such that

$$\begin{aligned} \Vert Tx+\lambda Sx\Vert \ge \Vert T\Vert -\varepsilon \Vert \lambda S\Vert ,\qquad \lambda \ge 0 \end{aligned}$$

and

$$\begin{aligned} \Vert Ty+\lambda Sy\Vert \ge \Vert T\Vert -\varepsilon \Vert \lambda S\Vert ,\qquad \lambda \le 0. \end{aligned}$$