Elementary Properties of Cesàro Operator in L^2

An example in elementary calculus

Consider a sequence of real or complex numbers {sn}. If sn→s, then πn=s1+⋯+snn→s.

Here, πn is called the Cesàro sum of {sn}. The proof is rather simple. Given ε>0, there exists some N>0 such that |sn−s|<ε for all n>N. Therefore we can write |πn−s|=|s1+s2+⋯+sNn+sN+1+⋯+snn−s|=|(s1−s)+(s2−s)+⋯+(sN−s)n+(sN+1−s)+⋯+(sn−s)n|≤|s1+⋯+sN−Nsn|+Nnε For fixed N, we can pick n big enough such that N/n<1/2 (i.e. n>2N) and |s1+⋯+sN−Nsn|<12ε. Hence πn converges to s. But the converse is not true in general. For example, if we put sn=(−1)n, then it diverges but πn→0. If πn converges, we say {sn} is Cesàro summable.

If we treat πn as an integration with respect to the counting measure, things become interesting. Why don't we investigate the operator defined to be C(f)(x)=1x∫x0f(t)dt. In this blog post we investigate this operator in Hilbert space L2(0,∞).

The Cesàro operator

Put L2=L2(0,∞) relative to Lebesgue measure, and the Cèsaro operator C is defined as follows: (Cf)(s)=1s∫s0f(t)dt.

Compactness and boundedness of this operator

From the example above, we shouldn't expect C to be too normal or well-behaved. But fortunately it is at the very least continuous: due to Hardy's inequality, we have ‖C‖=2. I organised several proofs of this. But C is not compact.

Consider a family of functions {φA}A>0 where φA=√Aχ(0,1/A]. (I owe Oliver Diaz for this family of functions.) It's not hard to show that ‖φA‖=1. If we apply C on it we see (CφA)(x)=1x∫x0√Aχ(0,1/A]dx=√A(χ(0,1/A](x)+1Axχ(1/A,+∞)) Hence ‖CφA‖=√1+A2A. Meanwhile for B>A, we have C(φB−φA)(x)=(√B−√A)χ(0,1/B](x)+(1√Bx−√A)χ(1/B,1/A](x)+(1√B−1√A)1xχ(1/A,+∞)(x) It follows that |C(φB−φA)|(x)≥(1√A−1√B)1xχ(1/A,∞)(x). If we compute the norm on the right hand side we get ‖C(φB−φA)‖≥|1−√AB|. As a result, if we pick fn=φ2n, then for any m>n we get ‖C(fm−fn)‖≥|1−√2n−m|≥1−1√2. Therefore, we find a sequence (fn) on the unit ball such that (Cfn) has no convergent subsequence.

Also we can find its adjoint operator: ⟨Cf,g⟩=∫∞0(1s∫s0f(t)dt)¯g(s)ds=∫∞0(∫∞t1sf(t)¯g(s)ds)dt=∫∞0f(t)(∫∞t1s¯g(s)ds)dt. Hence the adjoint is given by (C∗f)(t)=∫∞t1sg(s)ds. C∗ is not compact as well. Further, another application of Fubini's theorem shows that CC∗=C+C∗=C∗C⟹(I−C)(I−C∗)=I=(I−C∗)(I−C) Hence I−C is an isometry, C is normal.

Bilateral shift, spectrum

In this section we study the spectrum of C and C∗, which will be derived from properties of bilateral shift, which comes from ℓ2 space. For convenience we write N=Z≥0. This section can also help you understand the connection between L2(0,1) and L2(0,∞).

An operator U on a Hilbert space H is called a simple unilateral shift if H has a orthonormal basis {en} such that U(en)=en+1 for all n∈N. This is nothing but right-shift operator in the sense of basis. Besides, we call U a unilateral shift of multiplicity m if U is a direct sum of m simple unilateral shifts (note: m can be any cardinal number, finite or infinite).

If we consider the difference between N and Z, we have the definition of bilateral shift. An operator W on K is called a simple bilateral shift if K has a orthonormal basis {en} such that Wen=en+1 for all n∈Z. Besides, if we consider the subspace H which is spanned by {en}, we see W|H is simply a unilateral shift. Before we begin, we investigate some elementary properties of uni/bilateral shifts.

(Proposition 1) A simple unilateral shift U is an isometry.

Proof. Note (Uem,Uen)=(em+1,en+1)=δm+1,n+1=δmn=(em,en). ◻

(Proposition 2) A simple bilateral shift W is unitary, hence is also an isometry.

Proof. Note (Wem,en)=(em+1,en)=δm+1,n=δm,n−1=(em,W−1en), which follows that W∗=W−1. ◻

Now let the Hilbert space K and its subspace H (invariant under W) be given. Consider the 'orthonormal' operator given by Ren=e−(n+1). It follows that R is a unitary involution and Re0=W−1e0RH=H⊥R∘W=W−1∘R.

With these tools, we are ready for the most important theorems.

W=I-C^\ast is a simple bilateral shift on K=L^2.

Step 1 - Obtaining missing subspace, operator and basis

Here we put H=L^2(0,1), which can be canonically embedded into L^2(0,\infty) in the obvious way (consider all L^2 functions vanish outside (0,1)). It is natural to put this, as there are many similarities between L^2(0,1) and L^2(0,\infty).

Explicitly, (Wf)(x) = f(x) - \int_x^\infty \frac{1}{t}f(t)dt. Also we claim the basis to be generated by e_0= \chi_{(0,1)}. First of all we show that (We_n)_{n \geq 0} is orthonormal. Note as we have proved, W^\ast W = (I-C)(I-C^\ast)=I. Without loss of generality we assume that m \geq n and therefore (e_m,e_n)=(W^me_0,W^ne_0)=((W^\ast)^nW^me_0,e_0)=((W^\ast W)^nW^{m-n}e_0,e_0)=(W^{m-n}e_0,e_0). If m=n, then (e_m,e_n)=(e_0,e_0)=1. Hence it is reduced to prove that (W^ke_0,e_0)=0 for all k>0. First of all we have (We_0,e_0)=(e_0,e_0)-(C^\ast e_0,e_0)=1-(C^\ast e_0,e_0) meanwhile \begin{aligned} (C^\ast e_0,e_0) &= \int_0^1 \left(\int_x^1 \frac{1}{t}dt \right)dx \\ &= \int_0^1(-\ln{x})dx \\ &= (-x\ln{x}+x)|_0^1 = 1 \end{aligned} Hence We_0 \perp e_0. Suppose now we have (W^ke_0,e_0)=0, then $$ \begin{aligned} (W^ke_0,e_0)&=(WW^{k-1}e_0,e_0) \\ &=((I-C^\ast)W^{k-1}e_0,e_0) \\ &= (W^{k-1}e_0,e_0)-(C^\ast W^{k-1}e_0,e_0) \\ &= -(W^{k-1}e_0,C e_0) \\ &= -\int_0^1W^{k-1}e_0(x)\frac{1}{x}\left(\int_0^xdt\right)dx \\ &= -\int_0^1 W^{k-1}e_0(x)\frac{1}{x} \cdot x dx \\ &= -(W^{k-1}e_0,e_0) \\ &= 0. \end{aligned}

$$ Note W^ke_0 always vanishes when x \geq 1: when we are doing inner product, [1,\infty) is automatically excluded. With these being said, (W^ne_0)_{n \geq 0} forms a orthonormal set. By The Hausdorff Maximality Theorem, it is contained in a maximal orthonormal set. But since H=L^2(0,1) is separable (if and only if it admits a countable basis) (proof), (W^ke_0) forms a basis of H. From now on we write \{e_n\}.

To find the involution R, note first W=I-C^\ast is already unitary (also, if it is not unitary, then it cannot be a bilateral shift, we have nothing to prove), whose inverse or adjoint is W^\ast=I-C as we have proved earlier. Hence we have Re_0=e_{-1}=(I-C)e_0=\chi_{(0,1)}-\frac{1}{x}\int_0^xdt = -\frac{1}{x}\chi_{[1,\infty)} But we have no idea what R is exactly. We need to find it manually (or we have to guess). First of all it shall be guaranteed that RH=H^\perp. Since H contains all L^2 functions vanish on [1,\infty), functions in RH should vanish on (0,1). It is natural to put R(f)(x)=g(x)f\left( \frac{1}{x}\right) for the time being. g should be determined by e_{-1}. Note e_0\left(\frac{1}{x}\right)=\chi_{[1,\infty)} almost everywhere, we shall put g(x)=-\frac{1}{x}. It is then clear that Re_0=W^{-1}e_0 and RH=H^\perp. For the third condition, we need to show that W \circ R \circ W = R. Note \begin{aligned} W \circ R \circ W(f) &= W \circ R \left(f(x)-\int_x^\infty\frac{1}{t}f(t)dt\right) \\ &= W \left(-\frac{1}{x}f\left(\frac{1}{x}\right)+\frac{1}{x}\int_{1/x}^{\infty}f(t)dt \right) \\ &= -\frac{1}{x}f\left(\frac{1}{x}\right)+\underbrace{\frac{1}{x}\int_{1/x}^{\infty}f(t)dt + \int_x^\infty \frac{1}{t^2}f\left(\frac{1}{t}\right)dt + \int_x^\infty \frac{1}{t^2}\int_{1/t}^{\infty}f(u)du}_{=0 \text{ by Fubini's theorem, similar to proving }CC^\ast=C+C^\ast.} \\ &= R(f). \end{aligned} Step 2 - With these, W in step 1 has to be a simple bilateral shift

This is independent to the spaces chosen. To finish the proof, we need a lemma:

Suppose K is a Hilbert space, H is a subspace and e_0 \in H. W is a unitary operator such that W^ne_0 \in H for all n \geq 0 and \{e_n=W^ne_0\}_{n \geq 0} forms a orthonormal basis of H. R is a unitary involution on K such that Re_0 = W^{-1}e_0 \quad RH=H^\perp \quad R \circ W = W^{-1} \circ R then W is a simple bilateral shift.

Indeed, objects mentioned in step 1 fit in this lemma. To begin with, we write e_n=W^ne_0 for all n \in \mathbb{Z}. Then \{e_n\} is an orthonormal set because for arbitrary m,n \in \mathbb{Z}, there is a j \in \mathbb{Z} such that m+j,n+j \geq 0. Therefore (e_m,e_n)=(W^je_m,W^je_n)=(W^{m+j}e_0,W^{n+j}e_0)=(e_{m+j},e_{n+j})=\delta_{m+j,n+j}=\delta_{m,n}. Since (e_0,e_1,\cdots) spans H, RH=H^{\perp}, we see (Re_0,Re_1,\cdots) spans H^{\perp}. But Re_n=RW^ne_0=W^{-n}Re_0=W^{-n-1}e_0=e_{-n-1}, hence \{e_{-1},e_{-2},\cdots\} spans H^\perp. By definition of W, it is indeed a bilateral shift. And our proof is done \square

References

• Walter Rudin, Functional Analysis.
• Arlen Brown, P. R. Halmos, A. L. Shields, Cesàro operators.