1 Introduction

For Banach spaces X and Y, let \(\mathcal {B}(X, Y)\) and \(\mathcal {K}(X, Y)\) denote the sets of bounded linear and compact linear operators from X to Y, respectively; \(\mathcal {B}(X) := \mathcal {B}(X, X)\), \(\mathcal {K}(X) := \mathcal {K}(X, X)\); \(I \in \mathcal {B}(X)\) denotes the identity operator. An operator \(P \in \mathcal {B}(X)\) is called a projection if \(P^2 = P\). A closed linear subspace \(X_0 \subset X\) is called 1-complemented (in X) if there exists a projection \(P \in \mathcal {B}(X)\) such that \(P(X) = X_0\) and \(\Vert P\Vert = 1\).

Let \((\Omega , \Sigma , \mu )\) be a nonatomic measure space with \(0< \mu (\Omega ) < \infty \). We will use the following notation:

  • \(\Sigma ^+ :=\{A \in \Sigma : \ \mu (A) > 0\}\),

  • \(\mathbb {I}_A\) is the indicator function of \(A \in \Sigma \), i.e. \(\mathbb {I}_A(\omega ) = 1\) if \(\omega \in A\) and \(\mathbb {I}_A(\omega ) = 0\) if \(\omega \not \in A\),

  • \(\mathbf {1} := \mathbb {I}_\Omega \),

  • \(\mathbf { E}f := \left( \frac{1}{\mu (\Omega )} \int _\Omega f\, d\mu \right) \mathbf {1}\).

We will use the terminology from [6]. A \(\Sigma \)-measurable function g is called a sign if it takes values in the set \(\{-1, 0, 1\}\), and a sign on \(A \in \Sigma \) if it is a sign with the support equal to A, i.e. if \(g^2 = \mathbb {I}_A\). A sign is of mean zero if \(\int _\Omega g\, d\mu = 0\).

An operator \(T \in \mathcal {B}(L^p(\mu ), Y)\), \(1 \le p < \infty \), is called narrow if for every \(A \in \Sigma ^+\) and every \(\varepsilon > 0\), there exists a mean zero sign g on A such that \(\Vert Tg\Vert < \varepsilon \).

Every \(T \in \mathcal {K}(L^p(\mu ), Y)\) is narrow (see [6, Proposition 2.1]), but there are noncompact narrow operators. Indeed, let \(\mathcal {G}\) be a sub-\(\sigma \)-algebra of \(\Sigma \) such that there exists a random variable \(\xi \) on \(\left( \Omega , \Sigma , \frac{1}{\mu (\Omega )}\,\mu \right) \), which is independent of \(\mathcal {G}\) and has a nontrivial Gaussian distribution. Then the corresponding conditional expectation operator \(\mathbf {E}^\mathcal {G} = \mathbf { E}(\cdot | \mathcal {G}) \in \mathcal {B}(L^p(\mu ))\) is narrow (see [6, Corollary 4.25]), but not compact if \(\mathcal {G}\) has infinitely many pairwise disjoint elements of positive measure.

Let

$$\begin{aligned} C_p := \max _{0 \le \alpha \le 1} \left( \alpha ^{p - 1} + (1 - \alpha )^{p - 1}\right) ^{\frac{1}{p}} \left( \alpha ^{\frac{1}{p - 1}} + (1 - \alpha )^{\frac{1}{p - 1}}\right) ^{1 - \frac{1}{p}} \end{aligned}$$
(1.1)

for \( 1< p < \infty \), and \(C_1 := 2\).

In the following theorems, \((\Omega , \Sigma , \mu )= ([0, 1], \mathcal {L}, \lambda )\), where \(\lambda \) is the standard Lebesgue measure on [0, 1] and \(\mathcal {L}\) is the \(\sigma \)-algebra of Lebesgue measurable subsets of [0, 1].

Our starting point is a result due to C. Franchetti.

Theorem 1.1

([3, 4]). Let \(P \in \mathcal {B}(L^p([0, 1])){\setminus }\{0\}\) be a narrow projection operator, \(1 \le p < \infty \). Then

$$\begin{aligned} \Vert I - P\Vert _{L^p \rightarrow L^p} \ge \Vert I - \mathbf {E}\Vert _{L^p \rightarrow L^p} = C_p. \end{aligned}$$
(1.2)

The following theorem was proved in [7], where it was used to show that \(C_p\) is the optimal constant in the bounded compact approximation property of \(L^p([0, 1])\). It implies the inequality in (1.2) in the case when \(P \not = 0\) is a finite-rank projection.

Theorem 1.2

([7]). Let \(1 \le p < \infty \), \(\gamma \in \mathbb {C}\), and let \(T \in \mathcal {K}(L^p([0, 1]))\). Then

$$\begin{aligned} \Vert I - T\Vert _{L^p \rightarrow L^p} + \inf _{\Vert u\Vert _{L^p} = 1} \Vert (\gamma I - T)u\Vert _{L^p} \ge \Vert I - \gamma \mathbf { E}\Vert _{L^p \rightarrow L^p} . \end{aligned}$$
(1.3)

In particular,

$$\begin{aligned} \Vert I - T\Vert _{L^p \rightarrow L^p} + \inf _{\Vert u\Vert _{L^p} = 1} \Vert (I - T)u\Vert _{L^p} \ge C_p . \end{aligned}$$
(1.4)

The following theorem is the main result of the paper. It generalises both Theorems 1.1 and 1.2.

Theorem 1.3

Estimates (1.3) and (1.4) hold for all narrow operators \(T \in \mathcal {B}(L^p([0, 1]))\), \(1 \le p < \infty \).

In the case \(p = 1\), inequality (1.4) (for all narrow operators) follows from a Daugavet-type result due to A.M. Plichko and M.M. Popov ([5, §9, Theorem 8], see also [6, Corollary 6.4]):

$$\begin{aligned} \Vert I - T\Vert = 1 + \Vert T\Vert \quad \text{ for } \text{ every } \text{ narrow } \text{ operator }\quad T \in \mathcal {B}(L^1([0, 1])) . \end{aligned}$$

Indeed,

$$\begin{aligned}&\Vert I - T\Vert _{L^1 \rightarrow L^1} + \inf _{\Vert u\Vert _{L^1} = 1} \Vert (I - T)u\Vert _{L^1} \\&\quad \ge \Vert I - T\Vert _{L^1 \rightarrow L^1} + 1 - \sup _{\Vert u\Vert _{L^1} = 1} \Vert Tu\Vert _{L^1} \\&\quad = 1 + \Vert T\Vert _{L^1 \rightarrow L^1} + 1 - \Vert T\Vert _{L^1 \rightarrow L^1} = 2 = C_1 . \end{aligned}$$

2 Proof of Theorem 1.3

It follows from the definition of a narrow operator that if \(T \in \mathcal {B}(L^p(\mu ))\) is narrow and \(S \in \mathcal {B}(L^p(\mu ))\), then \(ST \in \mathcal {B}(L^p(\mu ))\) is narrow (see [6, Proposition 1.8]). On the other hand, there are \(S, T \in \mathcal {B}(L^p(\mu ))\) such that T is narrow but TS is not (see [6, Proposition 5.1]). The following lemma shows that the latter cannot happen if S is a multiplication operator.

Lemma 2.1

Let \(X = L^p(\mu )\)\(g \in L^\infty (\mu )\),  and \(T \in \mathcal {B}(X, Y)\) be a narrow operator. Then the operator \(TgI \in \mathcal {B}(X, Y)\) is also narrow.

Proof

There is nothing to prove if \(T = 0\). Suppose \(T \not = 0\). Take any \(A \in \Sigma ^+\) and any \(\varepsilon > 0\). There exists a simple function \(g_0 =\sum _{k = 1}^M a_k \mathbb {I}_{B_k} \not \equiv 0\) such that

$$\begin{aligned} \Vert g - g_0\Vert _{L^\infty } < \frac{\varepsilon }{2 \Vert T\Vert (\mu (\Omega ))^{1/p}}\, . \end{aligned}$$

Here \(M \in \mathbb {N}\)\(B_k \in \Sigma \), \(k = 1, \dots , M\), are pairwise disjoint, \(\mu (B_k) > 0\), \(\cup _{k = 1}^M B_k = \Omega \)\(a_k \in \mathbb {C}\).

Let \(A_k := A\cap B_k\). If \(\mu (A_k) > 0\), let \(h_k\) be a mean zero sign on \(A_k\) such that

$$\begin{aligned} \Vert Th_k\Vert < \frac{\varepsilon }{2\sum _{k = 1}^M |a_k|}\, . \end{aligned}$$

Let

$$\begin{aligned} h := \sum _{\{k : \ \mu (A_k) > 0\}} h_k . \end{aligned}$$

It is clear that h is a mean zero sign on A and

$$\begin{aligned} \Vert TgI h\Vert&\le \Vert Tg_0 h\Vert + \Vert T(g - g_0) h\Vert \\&\le \left\| T\sum _{\{k : \ \mu (A_k)> 0\}} a_k h_k\right\| + \Vert T\Vert \Vert g - g_0\Vert _{L^\infty } \Vert h\Vert _{L^p} \\&< \sum _{\{k : \ \mu (A_k)> 0\}} |a_k| \Vert Th_k\Vert + \Vert T\Vert \, \frac{\varepsilon }{2 \Vert T\Vert (\mu (\Omega ))^{1/p}}\, (\mu (\Omega ))^{1/p} \\&\le \frac{\varepsilon }{2\sum _{k = 1}^M |a_k|} \sum _{\{k : \ \mu (A_k) > 0\}} |a_k| + \frac{\varepsilon }{2} \le \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon . \end{aligned}$$

\(\square \)

The above lemma and its proof remain valid if X is a Köthe F-space on \((\Omega , \Sigma , \mu )\) (see [6, Section 1.3]). Similarly, the following lemma and its proof remain valid if X is a rearrangement-invariant Banach space on \(([0, 1], \mathcal {L}, \lambda )\) with absolutely continuous norm. This lemma is a minor modification of [6, Theorem 2.21] and [5, § 8, Proposition 5].

Lemma 2.2

Let \(X = L^p([0, 1])\) and \(T \in \mathcal {B}(X, Y)\) be a narrow operator. Then there exists a 1-complemented subspace \(X_0\) of X isometrically isomorphic to X such that \(\mathbf {1} \in X_0\) and the restriction \(T|_{X_0}\) of T to \(X_0\) is a compact operator.

Proof

Take a mean zero sign \(\overline{g}_1\) on [0, 1] and set \(T_1 := T\overline{g}_1 I\). The operator \(T_1\) is narrow according to Lemma 2.1. The proof of [6, Theorem 2.21] (with \(T_1\) in place of T and with \(\varepsilon \ge 2\Vert T_1 \overline{g}_1\Vert ,\, \varepsilon _1 > \Vert T_1 \overline{g}_1\Vert \)) shows that there exists a 1-complemented (see the proof of [5, §8, Proposition 5]) subspace \(X_1\) of X isometrically isomorphic to X such that \(\overline{g}_1 \in X_1\) and the restriction \(T_1|_{X_1}\) of \(T_1\) to \(X_1\) is a compact operator. Let \(X_0 := \overline{g}_1 X_1\). Since \(\overline{g}_1^2 = \mathbf {1}\), the operator of multiplication \(\overline{g}_1 I\) is an isometric isomorphism of \(X_0\) onto \(X_1\) and of X onto itself. Let \(P_1 \in \mathcal {B}(X)\) be a projection onto \(X_1\) such that \(\Vert P_1\Vert = 1\). Then \(P_0 := \overline{g}_1 P_1 \overline{g}_1 I \in \mathcal {B}(X)\) is a projection onto \(X_0\) such that \(\Vert P_0\Vert = 1\). Hence \(X_0\) is 1-complemented (this follows also from [1, Theorem 4] since \(X_0\) is isometrically isomorphic to \(X = L^p([0, 1])\)). Further, \(\mathbf {1} = \overline{g}_1^2 \in \overline{g}_1 X_1 = X_0\) and \(T|_{X_0} = T_1|_{X_1}\, \overline{g}_1 I|_{X_0}\) is compact.

Proof of Theorem 1.3

Take an arbitrary \(\varepsilon > 0\). Let

$$\begin{aligned} \delta := \inf _{\Vert u\Vert _{L^p} = 1} \Vert (\gamma I - T)u\Vert _{L^p} . \end{aligned}$$
(2.1)

There exists \(u_0 \in L^p([0, 1])\) such that \(\Vert u_0\Vert _{L^p} = 1\) and \(\Vert (\gamma I - T)u_0\Vert _{L^p} < \delta + \epsilon \). Then there exists an approximation \(h :=\sum _{k = 1}^M a_k \mathbb {I}_{A_k}\) of \(u_0\) such that \(A_k\), \(k = 1, \dots , M\), \(M \in \mathbb {N}\), are pairwise disjoint Borel sets of positive measure, \(\cup _{k = 1}^M A_k = [0, 1]\), \(a_k \in \mathbb {C}\), and

$$\begin{aligned} \left\| \gamma h - Th\right\| _{L^p([0, 1])} \le \delta + 2\varepsilon , \quad \Vert h\Vert _{L^p([0, 1])} = 1 . \end{aligned}$$
(2.2)

Partition [0, 1] into subintervals \(I_k\) of length \(\lambda (A_k)\), \(k = 1, \dots , M\). Since \((A_k, \mathcal {L}, \lambda )\) is isomorphic (modulo sets of measure 0) to \((I_k, \mathcal {L}, \lambda )\) (see, e.g., [2, Theorem 9.2.2 and Corollary 6.6.7]), one can easily derive from Lemma 2.2 the existence, for each k, of a 1-complemented subspace \(X_k\) of \(L^p(A_k)\) isometrically isomorphic to \(L^p(I_k)\) such that \(\mathbb {I}_{A_k} \in X_k\) and \(T|_{X_k}\) is a compact operator. Let

$$\begin{aligned} X_0 := \left\{ f \in L^p([0, 1]) : \ f|_{A_k} \in X_k , \, k = 1, \dots , M \right\} . \end{aligned}$$

It is easy to see that \(X_0\) is 1-complemented and isometrically isomorphic to \(L^p([0, 1])\), and that \(T_0 := T|_{X_0}\) is a compact operator. Let \(J : L^p([0, 1]) \rightarrow X_0\) be an isometric isomorphism and \(P_0 \in \mathcal {B}(L^p([0, 1]))\) be a projection onto \(X_0\) such that \(\Vert P_0\Vert = 1\). Then \(T_1 := J^{-1}P_0 T_0 J \in \mathcal {K}(L^p([0, 1]))\) and it follows from Theorem 1.2 that

$$\begin{aligned}&\Vert I - T_0\Vert _{X_0 \rightarrow L^p} + \inf _{f \in X_0, \, \Vert f\Vert _{L^p} = 1} \Vert (\gamma I - T_0)f\Vert _{L^p} \\&\quad \ge \Vert P_0(I - T_0)\Vert _{X_0 \rightarrow X_0} + \inf _{f \in X_0, \, \Vert f\Vert _{L^p} = 1} \Vert P_0(\gamma I - T_0)f\Vert _{L^p} \\&\quad = \Vert J^{-1}P_0(I - T_0)J\Vert _{L^p \rightarrow L^p} + \inf _{\varphi \in L^p, \, \Vert \varphi \Vert _{L^p} = 1} \Vert J^{-1}P_0(\gamma I - T_0)J\varphi \Vert _{L^p} \\&\quad = \Vert I - T_1\Vert _{L^p \rightarrow L^p} + \inf _{\varphi \in L^p, \, \Vert \varphi \Vert _{L^p} = 1} \Vert (\gamma I - T_1)\varphi \Vert _{L^p} \ge \Vert I - \gamma \mathbf { E}\Vert _{L^p \rightarrow L^p} . \end{aligned}$$

Since \(h \in X_0\), it follows from (2.2) that

$$\begin{aligned} \delta + 2\varepsilon \ge \inf _{f \in X_0, \, \Vert f\Vert _{L^p} = 1} \Vert (\gamma I - T_0)f\Vert _{L^p} . \end{aligned}$$

Hence

$$\begin{aligned} \Vert I - T\Vert _{L^p \rightarrow L^p} + \delta + 2\varepsilon&\ge \Vert I - T_0\Vert _{X_0 \rightarrow L^p} + \inf _{f \in X_0, \, \Vert f\Vert _{L^p} = 1} \Vert (\gamma I - T_0)f\Vert _{L^p} \\&\ge \Vert I - \gamma \mathbf { E}\Vert _{L^p \rightarrow L^p} \end{aligned}$$

and

$$\begin{aligned} \Vert I - T\Vert _{L^p \rightarrow L^p} + \inf _{\Vert u\Vert _{L^p} = 1} \Vert (\gamma I - T)u\Vert _{L^p} + 2\varepsilon \ge \Vert I - \gamma \mathbf { E}\Vert _{L^p \rightarrow L^p} \end{aligned}$$

for all \(\varepsilon > 0\) (see (2.1)). \(\square \)