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Diagonal operator on ell p is compact iff diagonal tends to zero

Example

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let K=R or C, let 1p and let a:NK, nan, be bounded in modulus: there is a real M0 such that anM for every n. Write p(N,K) for the counting-measure space Lp(#;K) on (N,P(N),#), read as the space of scalar sequences x with nxnp< for p< and with supnxn< for p= (Counting measure on an arbitrary set, Counting measure is a measure, p is the Lp space of counting measure, Complex Lp classes and Euclidean test-function conventions), and let

Da:p(N,K)p(N,K),(Dax)n:=anxn,

be the diagonal operator. Then Da is compact (Compact linear operator) if and only if an0, meaning that for every real ε>0 there is NN such that an<ε whenever nN.

Facts & Assumptions

[A1]

On (N,P(N),#) every real function is measurable, for real f one has fpd#=nf(n)p for 0<p< and f=supnf(n), and almost-everywhere equality is equality everywhere (p is the Lp space of counting measure, Counting measure on an arbitrary set); for complex f=u+iv measurability means measurability of u,v, so every complex function on N is measurable, and Lp(#;C) is complete under ACω (Complex Lp classes and Euclidean test-function conventions, Complex completeness, density, and inner product: the consumer interface).

[A2]

Real Lp(#) is complete under countable choice (Riesz-Fischer completeness of Lp for 1p); completeness of the target is what the norm-closure theorem needs for a compact conclusion (Banach space, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[A3]

For λ a scalar and n fixed, the sequence en that is 1 at n and 0 elsewhere lies in p with enp=1 for every 1p (Counting measure on an arbitrary set, p is the Lp space of counting measure, Complex Lp classes and Euclidean test-function conventions).

[A4]

A bounded finite-rank operator is compact (Bounded finite rank operators are compact); under ACω a norm limit of compact operators with Banach target is compact (Norm limit of compact operators is compact); under DC a compact operator sends bounded sequences to sequences with convergent subsequences (Sequential characterization of compact operators, The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, The Axiom of Countable Choice (ACω), Dependent choice implies countable choice).

[A5]

The operator norm bounds TxTx and equals the supremum of Tx over the closed unit ball (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, Open ball, closed ball and sphere in a metric space); operator-norm convergence is metric convergence (Convergence of a sequence in a metric space: xkx iff d(xk,x)0 in R), while scalar convergence has the explicit epsilon meaning in the statement.

Verification

technique · direct

Given: DC, K{R,C}, 1p, a bounded sequence a, the operator Da on p(N,K), and the truncations Da(N) defined by (Da(N)x)n=anxn for n<N and 0 for nN.

1.1

Da is well defined and bounded with Daxpaxp, because anxnaxn for every n; so Daa.

A1A5algebra
1.2

Each truncation Da(N) is bounded, and its range is contained in the linear span of e0,,eN1, so Da(N) has finite rank and is compact by [A4].

A3A4
1.3

If an↛0, then there is a real ε>0 and a strictly increasing n with ankε for all k: this is the negation of convergence, and the indices are chosen by DC (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, A strictly increasing index map satisfies nkk).

A5
2.1

For every N one has DaDa(N)=supnNan: the upper bound is [step 1.1] applied to the tail sequence, and the lower bound follows by testing en for nN, where (DaDa(N))enp=an.

step 1.1A3A5
2.2

Under the hypothesis of [step 1.3], the bounded sequence (enk) has DaenkDaenlpε for all kl (indeed it is at least (εp+εp)1/p for p< and at least ε for p=), so the sequence of images has no convergent subsequence; by the sequential characterization [A4] the operator Da is not compact.

step 1.3A3A4A5
3.1

If an0, then for every real ε>0 there is N with an<ε for all nN, so DaDa(N)ε by [step 2.1]; hence Da is a norm limit of the compact operators Da(N) and is compact by [A4], the target p being Banach by [A1] and [A2].

step 1.2step 2.1A1A2A4
4.1

Steps [step 3.1] and [step 2.2] are the two directions of the equivalence, so the diagonal operator Da on p(N,K) is compact exactly when an0.

step 3.1step 2.2

Depends on

Used by

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Dependency tree · two levels

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Sources