The essential norm of multiplication operators on $L_p(\mu )$

Abstract

We show that the formula for the essential norm of a multiplication operator on $L_p$, for $1<p<\infty$, also holds for $p=1$. We also provide a proof for the formula which works simultaneously for all $p\in [1,\infty )$.

1 Introduction

Let $(\Omega ,{\mathcal {A}},\mu )$ be a $\sigma$-finite measure space, and let $1\leqslant p<\infty$. For $u\in L_\infty (\mu )$ let $M_u$ be the bounded multiplication operator on $L_p(\mu )$ defined by

$$\begin{aligned} M_uf:=uf\qquad (f\in L_p(\mu )). \end{aligned}$$

Compactness properties of multiplication operators in various function spaces have been investigated in several papers; see [1,2,3, 5, 6, 10,11,12, 14]. It is only in the recent paper [4] that the essential norm

$$\begin{aligned} \Vert M_u\Vert _\mathrm{e}:=\inf \bigl \{\Vert M_u+K\Vert {;}\;K\in {\mathcal {K}}(L_p(\mu ))\bigr \}, \end{aligned}$$

(1.1)

where ${\mathcal {K}}(L_p(\mu ))$ denotes the space of compact operators on $L_p(\mu )$, has been determined, for $1<p<\infty$. (The essential norm $\Vert M_u\Vert _\mathrm{e}$ is the quotient norm in the Calkin algebra.) In order to describe this result we recall that the measure space can be decomposed as a disjoint union $\Omega =\Omega _d\cup \Omega _a$, where $\Omega _d,\Omega _a\in {\mathcal {A}}$, the restriction $\mu _d$ of $\mu$ to $\Omega _d$ is a diffuse measure, and the restriction $\mu _a$ of $\mu$ to $\Omega _a$ is (purely) atomic. The property of being diffuse means that for every measurable subset A of $\Omega _d$ with $\mu _d(A)>0$ there exists a measurable subset $A'$ of A such that $0<\mu _d(A')<\mu _d(A)$. And the atomic part $\Omega _a$ is the union of a disjoint sequence $(B_n)_{n\in \mathbb {N}}$ of measurable sets, where each $B_n$ is an atom, which means that any measurable subset $B'$ of $B_n$ has measure $\mu _a(B')\in \{0,\mu _a(B_n)\}$. With this notation, the essential norm of $M_u$ is given by

$$\begin{aligned} \Vert M_u\Vert _\mathrm{e}=\max \{\Vert u{|}_{\Omega _d}\Vert _\infty ,\limsup _{n\rightarrow \infty }|u(B_n)|\}. \end{aligned}$$

(1.2)

(By $u(B_n)$ we denote the a.e.-value of u on $B_n$; if $\mu _a(B_n)=0$ we choose $u(B_n):=0$.) The proof of (1.2) given in [4,  Theorem 4.1] does not carry over to the case $p=1$.

In Sect. 2 we show that (1.2) also holds for $p=1$. In Sects. 3 and 4 we provide a second – quite different – proof, which works simultaneously for all $p\in [1,\infty )$.

2 The essential norm of $M_u$

Let $(\Omega ,{\mathcal {A}},\mu )$ be a $\sigma$-finite measure space, and let $\Omega =\Omega _d\cup \Omega _a$ and $\Omega _a=\bigcup _{n\in \mathbb {N}}B_n$ be as described above.

2.1 Theorem

Let $u\in L_\infty (\mu )$, and let $M_u$ be the multiplication operator associated with u on $L_1(\mu )$. Then $\Vert M_u\Vert _\mathrm{e}$ is given by (1.2).

Proof

• (i) For the inequality ‘$\leqslant$’ in (1.2) we refer to [4,  first part of the proof of Theorem 4.1].

• (ii) For the proof of ‘$\geqslant$’ we first note that in the infimum of the formula (1.1) (where in the present step we treat the general case $p\in [1,\infty )$) one does not need all compact operators, but it is sufficient to consider operators leaving $L_p(\Omega _d)$ and $L_p(\Omega _a)$ invariant. Indeed, let $P_d$ and $P_a$ denote the canonical projections from $L_p(\mu )$ onto $L_p(\Omega _d,\mu _d)$ and $L_p(\Omega _a,\mu _a)$, respectively. Then for any bounded operator S on $L_p(\mu )$ one has $\Vert (P_d-P_a)S(P_d-P_a)\Vert \leqslant \Vert S\Vert$, and because of

  $$\begin{aligned} P_dSP_d+P_aSP_a =\tfrac{1}{2}\bigl ((P_d+P_a)S(P_d+P_a) + (P_d-P_a)S(P_d-P_a)\bigr ) \end{aligned}$$

  one obtains $\Vert P_dSP_d+P_aSP_a\Vert \leqslant \Vert S\Vert$. In view of $P_dM_uP_d+P_aM_uP_a=M_u$, this yields

  $$\begin{aligned} \Vert M_u+P_dKP_d+P_aKP_a\Vert \leqslant \Vert M_u+K\Vert \end{aligned}$$

  for all compact operators, and $P_dKP_d+P_aKP_a$ is a compact operator leaving $L_p(\Omega _d)$ and $L_p(\Omega _a)$ invariant. As a consequence one also concludes that it is sufficient to prove the inequality ‘$\geqslant$’ separately for diffuse and atomic measure spaces. For the remainder of the proof we now return to the case $p=1$.

• (iii) In this part of the proof we show ‘$\geqslant$’ for the case that $\Omega =\Omega _d$, i.e. that $\mu$ is a diffuse measure. The case $u=0$ being trivial, assume that $\Vert u\Vert _\infty >0$ and let $0<\varepsilon <\Vert u\Vert _\infty$. Then there exists a descending sequence $(A_n)_{n\in \mathbb {N}}$ in ${\mathcal {A}}$ such that $0<\mu (A_n)\rightarrow 0$ as $n\rightarrow \infty$ and $|u|_{A_n}|\geqslant \Vert u\Vert _\infty -\varepsilon$ for all $n\in \mathbb {N}$; without restriction $\mu (A_1)<\infty$. For $n\in \mathbb {N}$ put

  $$\begin{aligned} f_n:=\frac{1}{\mu (A_n)}\mathbf{1 }_{A_n}, \end{aligned}$$

  where $\mathbf{1 }_{A_n}$ denotes the indicator function of the set $A_n$. Let $K\in {\mathcal {K}}(L_1(\mu ))$. Because $(f_n)_{n\in \mathbb {N}}$ is a bounded sequence, the compactness of K implies that there exists a subsequence $(f_{n_j})_{j\in \mathbb {N}}$ such that the sequence $(Kf_{n_j})$ is convergent; by passing to a subsequence, we can assume that $(Kf_n)$ is already convergent. Then there exists $n\in \mathbb {N}$ such that $\Vert Kf_n-Kf_m\Vert \leqslant \varepsilon$ for all $m\geqslant n$. Choose $m\geqslant n$ large enough to obtain additionally $\frac{1}{\mu (A_m)}\geqslant 2\frac{1}{\mu (A_n)}$. Then one has

  $$\begin{aligned} f_n-f_m=\tfrac{1}{\mu (A_n)}\mathbf{1 }_{A_n}-\tfrac{1}{\mu (A_m)}\mathbf{1 }_{A_m} = \tfrac{1}{\mu (A_n)}\mathbf{1 }_{A_n\setminus A_m}-\bigl (\tfrac{1}{\mu (A_m)}-\tfrac{1}{\mu (A_n)}\bigr )\mathbf{1 }_{A_m}, \end{aligned}$$

  $\bigl |f_n-f_m\bigr |\geqslant f_n$, $\Vert f_n-f_m\Vert _1\geqslant \Vert f_n\Vert _1=1$; hence

  $$\begin{aligned} \Vert (M_u+K)(f_n-f_m)\Vert _1&\geqslant \Vert M_u(f_n-f_m)\Vert _1-\Vert K(f_n-f_m)\Vert _1\\&\geqslant (\Vert u\Vert _\infty -\varepsilon )\Vert f_n-f_m\Vert _1 - \varepsilon \\&\geqslant (\Vert u\Vert _\infty - 2\varepsilon )\Vert f_n-f_m\Vert _1, \end{aligned}$$

  $\Vert M_u+K\Vert \geqslant \Vert u\Vert _\infty - 2\varepsilon$. As this holds for all $\varepsilon \in (0,\Vert u\Vert _\infty )$, we obtain $\Vert M_u+K\Vert \geqslant \Vert u\Vert _\infty$.

• (iv) It remains to show that ‘$\geqslant$’ holds in the case that $\Omega =\Omega _a$, i.e. that $\mu$ is an atomic measure. If $\mu (B_n)\ne 0$ only for finitely many $n\in \mathbb {N}$, then $\limsup _{n\rightarrow \infty }|u(B_n)|=0$, and the assertion is trivial. Assume that this is not the case, without restriction $\mu (B_n)\ne 0$ for all $n\in \mathbb {N}$. For $n\in \mathbb {N}$ let $P_n$ be the canonical projection from $L_1(\mu )$ onto $L_1(B_n)$, i.e. $P_nf := \mathbf{1 }_{B_n}f$ ($f\in L_1(\mu )$), and put $Q_n:=I-\sum _{j=1}^n P_j$. Iterating the procedure applied in step (ii) above one concludes that for all bounded operators S on $L_1(\mu )$ and all $n\in \mathbb {N}$ one obtains $\Vert \sum _{j=1}^nP_jSP_j+Q_nSQ_n\Vert \leqslant \Vert S\Vert$. Given a compact operator $K\in {\mathcal {K}}(L_1(\mu ))$ we note that $\Vert Q_nK\Vert \rightarrow 0$ as $n\rightarrow \infty$. This holds because for any $g\in L_1(\mu )$ one has $\Vert Q_ng\Vert \rightarrow 0$, and from the relative compactness of $K(B_{L_1(\mu )}[0,1])$ (where $B_{L_1(\mu )}[0,1]$ denotes the closed unit ball of $L_1(\mu )$) together with the equicontinuity of the sequence $(Q_n)$ one concludes that

  $$\begin{aligned} \Vert Q_nK\Vert = \sup _{\Vert f\Vert \leqslant 1}\Vert Q_nKf\Vert = \sup _{g\in K(B_{L_1(\mu )}[0,1])}\Vert Q_ng\Vert \rightarrow 0\qquad (n\rightarrow \infty ). \end{aligned}$$

In particular, we conclude that $\Vert P_nKP_n\Vert \leqslant \Vert (Q_{n-1}-Q_n)K\Vert \rightarrow 0$ ($n\rightarrow \infty$). Note that, for $n\in \mathbb {N}$, there exists $d_n\in \mathbb {K}$ such that $\Vert P_nKP_n\Vert =|d_n|$ and $P_nKP_nf=d_n P_n f$ for all $f\in L_1(\mu )$. Hence the multiplication operator $D_K$, given by $L_1(\mu )\ni f\mapsto \sum _{n\in \mathbb {N}}d_nP_nf\in L_1(\mu )$, is a compact operator. Now we estimate

$$\begin{aligned} \Vert M_u+K\Vert&\geqslant \Vert \sum _{j=1}^nP_j(M_u+K)P_j + Q_n(M_u+K)Q_n\Vert \nonumber \\&=\Vert M_u+D_K+Q_n(K-D_K)Q_n\Vert \nonumber \\&\geqslant \Vert M_u+D_K\Vert -\Vert Q_n(K-D_K)Q_n\Vert . \end{aligned}$$

(2.1)

From the argument given above we obtain $\Vert Q_n(K-D_K)Q_n\Vert \leqslant \Vert Q_n(K-D_K)\Vert \rightarrow 0$ ($n\rightarrow \infty$), and from (2.1) we conclude that $\Vert M_u+K\Vert \geqslant \Vert M_u+D_K\Vert$. This shows that

$$\begin{aligned} \Vert M_u\Vert _\mathrm{e}&=\inf \bigl \{\Vert M_u+D\Vert {;}\; D\text { compact multiplication operator}\bigr \}\\&= \inf \bigl \{\sup _{n\in \mathbb {N}}|u(n)+d_n|{;}\;(d_n)_{n\in \mathbb {N}}\text { null sequence}\bigr \}\\&=\limsup _{n\rightarrow \infty }|u(n)|. \square \end{aligned}$$

2.2 Remark

Step (iv) of our proof applies also to $p\in (1,\infty )$ and is an alternative to the last part of [4,  proof of Theorem 4.1]. The idea of our proof is that $\Vert M_u+K\Vert$ can be estimated from below by $\Vert M_u+D\Vert$ for a suitable compact multiplication operator D.

3 The case $\Omega =\Omega _d$, revisited

Let $(\Omega ,{\mathcal {A}},\mu )$ be a diffuse $\sigma$-finite measure space, $p\in [1,\infty )$, and let $u\in L_\infty (\mu )$. In this section we will present a proof of the equality

$$\begin{aligned} \Vert M_u\Vert _\mathrm{e}=\Vert u\Vert _\infty \end{aligned}$$

(3.1)

(i.e. (1.1) for the present special case), which might throw a new light on this property.

We recall that an operator $S\in {\mathcal {L}}(L_p(\mu ))$ (the space of all bounded linear operators) is positive, $S\in {\mathcal {L}}(L_p(\mu ))_+$, if $Sf\geqslant 0$ for all $f\in L_p(\mu )_+$. Then ${\mathcal {L}}^\mathrm {r}(L_p(\mu ))$, defined as the linear hull of ${\mathcal {L}}(L_p(\mu ))_+$, is the space of regular operators. It is a Banach lattice under the lattice operations

$$\begin{aligned} (S\vee T)f&:=\sup \bigl \{Sg+Th{;}\;g,h\geqslant 0,\ g+h=f\bigr \},\\ (S\wedge T)f&:=\inf \bigl \{Sg+Th{;}\;g,h\geqslant 0,\ g+h=f\bigr \} \qquad (f\in L_p(\mu )_+) \end{aligned}$$

(valid for real operators $S,T\in {\mathcal {L}}^\mathrm {r}(L_p(\mu ))$, the absolute value

$$\begin{aligned} |S|f := \sup \bigl \{|Sg|{;}\;|g|\leqslant f\bigr \}\qquad (f\in L_p(\mu )_+), \end{aligned}$$

and with the regular norm $\Vert S\Vert _\mathrm {r}:=\Vert |S|\Vert$. We refer to [13,  Chap. 4], [8,  Sect. 1.3] for more information.

As a preparation to the proof of (3.1) we need the following property, where $q\in (1,\infty ]$ denotes the exponent conjugate to p, $\frac{1}{p}+\frac{1}{q}=1$.

3.1 Lemma

Let $\eta \in L_q(\mu )$ ($=L_p(\mu )'$), $g\in L_p(\mu )$, $\eta , g\geqslant 0$, $K\in {\mathcal {L}}(L_p(\mu ))$ defined by

$$\begin{aligned} Kf:=\Bigl (\int \eta f\,\mathrm {d}\mu \Bigr )\, g\qquad (f\in L_p(\mu )). \end{aligned}$$

(Note that $K\in {\mathcal {L}}(L_p(\mu ))_+\subseteq {\mathcal {L}}^\mathrm {r}(L_p(\mu ))$.) Let $u\in L_\infty (\mu )_+$ be such that $M_u\leqslant K$. Then $u=0$.

Proof

Assume on the contrary that $u\ne 0$. Then there exists $\varepsilon >0$ such that $\mu ([u\geqslant \varepsilon ])>0$ (with the notation $[u\geqslant \varepsilon ]:=\bigl \{x\in \Omega {;}\;u(x)\geqslant \varepsilon \bigr \}$). Further there exists $c>0$ such that $\mu ([u\geqslant \varepsilon ]\cap [g\leqslant c])>0$. Let $B\in {\mathcal {A}}$, $B\subseteq [u\geqslant \varepsilon ]\cap [g\leqslant c]$ with $0<\mu (B)<\infty$. Then $M_u\mathbf{1 }_B\geqslant \varepsilon$ and $K\mathbf{1 }_B=\int _B\eta \,\mathrm {d}\mu \, g\leqslant c\int _B\eta \,\mathrm {d}\mu$ on B. There exists B as above and such that $\int _B\eta \,\mathrm {d}\mu <\varepsilon / c$, and this leads to the contradiction $K\mathbf{1 }_B\leqslant c\int \eta \,\mathrm {d}\mu <\varepsilon \leqslant M_u\mathbf{1 }_B$ on B. $\square$

The centre ${\mathcal {Z}}(L_p(\mu ))$ of ${\mathcal {L}}(L_p(\mu ))$ is the linear hull of the order interval

$$\begin{aligned}{}[-I,I] = \bigl \{S\in {\mathcal {L}}(L_p(\mu )){;}\;-f\leqslant Sf\leqslant f\ (f\in L_p(\mu )_+)\bigr \}. \end{aligned}$$

Then ${\mathcal {Z}}(L_p(\mu ))\subseteq {\mathcal {L}}^\mathrm {r}(L_p(\mu ))$ consists of the bounded multiplication operators and is isometrically isomorphic to $L_\infty (\mu )$; see [9,  C-I, Sect. 9].

The centre ${\mathcal {Z}}(L_p(\mu ))$ is a projection band in the Banach lattice ${\mathcal {L}}^\mathrm {r}(L_p(\mu ))$, i.e. for all $S\in {\mathcal {L}}^\mathrm {r}(L_p(\mu ))$ there exists a (unique) decomposition $S=S_1+S_2$, where $S_1\in {\mathcal {Z}}(L_p(\mu ))$ and

$$\begin{aligned} S_2\in {\mathcal {Z}}(L_p(\mu ))^\mathrm d= \bigl \{T\in {\mathcal {L}}^\mathrm {r}(L_p(\mu )){;}\;|T|\wedge R=0\ (R\in {\mathcal {Z}}(L_p(\mu ))_+)\bigr \}; \end{aligned}$$

see [13,  Chap. II, Theorem 2.10] Let ${\mathcal {P}}:{\mathcal {L}}^\mathrm {r}(L_p(\mu ))\rightarrow {\mathcal {Z}}(L_p(\mu ))$, $S\mapsto S_1$ denote the associated band projection. What we have shown in Lemma 3.1 is that ${\mathcal {P}}K=0$ for the special (positive) rank-one operators K. (Indeed, the lemma shows that K belongs to ${\mathcal {Z}}(L_p(\mu ))^\mathrm d$.) It is easy to see that any finite-rank operator $K\in {\mathcal {L}}(L_p(\mu ))$ can be written as a linear combination of rank-one operators as in Lemma 3.1; hence ${\mathcal {P}}K=0$ for all finite-rank operators.

3.2 Theorem

Let $u\in L_\infty (\mu )$. Then

$$\begin{aligned} \Vert M_u+K\Vert \geqslant \Vert M_u\Vert = \Vert u\Vert _\infty \qquad (K\in {\mathcal {K}}(L_p(\mu ))), \end{aligned}$$

(3.2)

and (3.1) holds.

Proof

Clearly, it suffices to show (3.2). There are two ingredients of the proof:

• (i) By the very definition, ${\mathcal {P}}$ is contractive with respect to the regular norm (because band projections are contractive). However, it is shown in [15,  Theorem 1.4] that ${\mathcal {P}}$ is also contractive with respect to the operator norm. This implies that ${\mathcal {P}}$ can be extended by continuity to the closure of ${\mathcal {L}}^\mathrm {r}(L_p(\mu ))$ in ${\mathcal {L}}(L_p(\mu ))$. In particular, for the extension one obtains ${\mathcal {P}}K=0$ for all K in the operator norm closure of the finite rank operators.

• (ii) The space $L_p(\mu )$ enjoys the approximation property, i.e. every compact operator on $L_p(\mu )$ can be approximated in operator norm by finite rank operators. (We refer to [7,  Sects. 3.4 and 4.1] for the approximation property.) This implies that ${\mathcal {P}}K=0$ for all compact operators on $L_p(\mu )$. Putting together these two properties we obtain

  $$\begin{aligned} \Vert M_u\Vert =\Vert {\mathcal {P}}(M_u+K)\Vert \leqslant \Vert M_u+K\Vert \end{aligned}$$

  for all $K\in {\mathcal {K}}(L_p(\mu ))$. $\square$

4 Supplement on the case $\Omega =\Omega _a$

We add that the case $\Omega =\Omega _a$ can be treated analogously to the case $\Omega =\Omega _d$ described in Sect. 3. Then again the centre of ${\mathcal {L}}(L_p(\mu ))$ consists of the bounded multiplication operators. Lemma 3.1 is replaced by the property that multiplication operators are disjoint to positive rank-one operators K of the type

$$\begin{aligned} Kf = \int _{\Omega \setminus B_j}f\eta \,\mathrm {d}\mu \mathbf{1 }_{B_j}=\Bigl (\sum _{k\ne j}f(B_k)\eta (B_k)\mu (B_k)\Bigr )\mathbf{1 }_{B_j}\qquad (f\in L_p(\mu )), \end{aligned}$$

(4.1)

where $j\in \mathbb {N}$ and $\eta \in L_q(\mu )_+$. Indeed, if $u\in L_\infty (\mu )_+$ is such that $M_u\leqslant K$, then clearly $u(B_k)=0$ for all $k\ne j$. But $u(B_j)\mathbf{1 }_{B_j}=M_u\mathbf{1 }_{B_j}\leqslant K\mathbf{1 }_{B_j}=0$; hence also $u(B_j)=0$. (Recall that $u(B_k)=0$ if $\mu (B_k)=0$, by our convention in the Introduction.)

The consequence is that, for a compact operator K, its projection ${\mathcal {P}}K$ onto the centre is the compact operator $D_K$ (described in part (iv) of the proof of Theorem 2.1). This holds because for a compact operator K and $n\in \mathbb {N}$, the finite rank operator $(I-Q_n)K$ (with the notation of the proof of Theorem 2.1, part (iv)) can be decomposed as the multiplication operator $(I-Q_n)D_K$ and a linear combination of rank-one operators of the type (4.1). As $(I-Q_n)K\rightarrow K$ ($n\rightarrow \infty$) in ${\mathcal {L}}(L_p(\mu ))$ and the band projection ${\mathcal {P}}$ onto the centre is contractive with respect to the operator norm, one concludes that ${\mathcal {P}}K = \lim _{n\rightarrow \infty }{\mathcal {P}}(I-Q_n)K = \lim _{n\rightarrow \infty }(I-Q_n)D_K=D_K$.

Hence instead of (2.1) one obtains $\Vert M_u+K\Vert \geqslant \Vert {\mathcal {P}}(M_u+K)\Vert =\Vert M_u+D_K\Vert$, and the proof can be finished as in Sect. 2.