Elementary Properties of Cesàro Operator in L^2

An example in elementary calculus

Consider a sequence of real or complex numbers \(\{s_n\}\). If \(s_n \to s\), then \[ \pi_n = \frac{s_1+\cdots+s_n}{n} \to s. \]

Here, \(\pi_n\) is called the Cesàro sum of \(\{s_n\}\). The proof is rather simple. Given \(\varepsilon>0\), there exists some \(N>0\) such that \(|s_n-s|<\varepsilon\) for all \(n > N\). Therefore we can write \[ \begin{aligned} |\pi_n - s| &= \left|\frac{s_1+s_2+\cdots+s_N}{n}+\frac{s_{N+1}+\cdots+s_n}{n}-s\right| \\ &= \left|\frac{(s_1-s)+(s_2-s)+\cdots+(s_N-s)}{n}+\frac{(s_{N+1}-s)+\cdots+(s_n-s)}{n}\right| \\ &\leq \left| \frac{s_1+\cdots+s_N-Ns}{n} \right| + \frac{N}{n}\varepsilon \end{aligned} \] For fixed \(N\), we can pick \(n\) big enough such that \(N/n<1/2\) (i.e. \(n>2N\)) and \[ \left| \frac{s_1+\cdots+s_N-Ns}{n} \right|<\frac{1}{2}\varepsilon. \] Hence \(\pi_n\) converges to \(s\). But the converse is not true in general. For example, if we put \(s_n=(-1)^n\), then it diverges but \(\pi_n \to 0\). If \(\pi_n\) converges, we say \(\{s_n\}\) is Cesàro summable.

If we treat \(\pi_n\) as an integration with respect to the counting measure, things become interesting. Why don't we investigate the operator defined to be \[ C(f)(x)= \frac{1}{x}\int_0^xf(t)dt. \] In this blog post we investigate this operator in Hilbert space \(L^2(0,\infty)\).

The Cesàro operator

Put \(L^2=L^2(0,\infty)\) relative to Lebesgue measure, and the Cèsaro operator \(C\) is defined as follows: \[ \begin{aligned} (Cf)(s) = \frac{1}{s}\int_0^sf(t)dt. \end{aligned} \]

Compactness and boundedness of this operator

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

Here is the proof. Consider a family of functions \(\{\varphi_A\}_{A>0}\) where \[ \varphi_A = \sqrt{A}\chi_{(0,1/A]}. \] (I owe Oliver Diaz for this family of functions.) It's not hard to show that \(\lVert \varphi_A \rVert = 1\). If we apply \(C\) on it we see \[ (C\varphi_A)(x) = \frac{1}{x}\int_0^x\sqrt{A}\chi_{(0,1/A]}dx = \sqrt{A}\left(\chi_{(0,1/A]}(x)+\frac{1}{Ax}\chi_{(1/A,+\infty)}\right) \] Hence \(\lVert C\varphi_A \rVert = \frac{\sqrt{1+A^2}}{A}\). Meanwhile for \(B>A\), we have \[ \begin{aligned} C(\varphi_B-\varphi_A)(x) &=\left(\sqrt{B}-\sqrt{A} \right)\chi_{(0,1/B]}(x)+\left(\frac{1}{\sqrt{B}x}-\sqrt{A}\right)\chi_{(1/B,1/A]}(x) \\ &+\left(\frac{1}{\sqrt{B}} - \frac{1}{\sqrt{A}} \right)\frac{1}{x}\chi_{(1/A,+\infty)}(x) \end{aligned} \] It follows that \[ |C(\varphi_B-\varphi_A)|(x) \geq \left(\frac{1}{\sqrt{A}}-\frac{1}{\sqrt{B}} \right)\frac{1}{x}\chi_{(1/A,\infty)}(x). \] If we compute the norm on the right hand side we get \[ \|C(\varphi_B-\varphi_A)\| \geq \left|1-\sqrt{\frac{A}{B}} \right|. \] As a result, if we pick \(f_n=\varphi_{2^n}\), then for any \(m>n\) we get \[ \|C(f_m-f_n)\| \geq \left|1 - \sqrt{2^{n-m}} \right| \geq 1-\frac{1}{\sqrt{2}}. \] Therefore, we find a sequence \((f_n)\) on the unit ball such that \((Cf_n)\) has no convergent subsequence.

Also we can find its adjoint operator: \[ \begin{aligned} \langle Cf,g \rangle &= \int_0^\infty \left(\frac{1}{s}\int_0^sf(t)dt \right)\overline{g}(s)ds \\ &= \int_0^\infty\left(\int_t^\infty \frac{1}{s}f(t)\overline{g}(s)ds\right)dt \\ &= \int_0^\infty f(t) \left(\int_t^{\infty}\frac{1}{s}\overline{g}(s)ds\right)dt. \end{aligned} \] Hence the adjoint is given by \[ (C^\ast f)(t) = \int_t^{\infty}\frac{1}{s}g(s)ds. \] \(C^\ast\) is not compact as well. Further, another application of Fubini's theorem shows that \[ CC^\ast = C + C^\ast=C^\ast C \implies (I-C)(I-C^\ast)=I=(I-C^\ast)(I-C) \] Hence \(I-C\) is an isometry, and \(C\) is normal.

Bilateral shift, spectrum

In this section we study the spectrum of \(C\) and \(C^\ast\), which will be derived from properties of bilateral shift, which comes from \(\ell^2\) space. For convenience we write \(\mathbb{N}=\mathbb{Z}_{\geq 0}\). This section can also help you understand the connection between \(L^2(0,1)\) and \(L^2(0,\infty)\).

An operator \(U\) on a Hilbert space \(H\) is called a simple unilateral shift if \(H\) has a orthonormal basis \(\{e_n\}\) such that \(U(e_n)=e_{n+1}\) for all \(n \in \mathbb{N}\). 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 \(\mathbb{N}\) and \(\mathbb{Z}\), we have the definition of bilateral shift. An operator \(W\) on \(K\) is called a simple bilateral shift if \(K\) has a orthonormal basis \(\{e_n\}\) such that \(We_{n}=e_{n+1}\) for all \(n \in \mathbb{Z}\). Besides, if we consider the subspace \(H\) which is spanned by \(\{e_n\}\), 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 \((Ue_m,Ue_n)=(e_{m+1},e_{n+1})=\delta_{m+1,n+1}=\delta_{mn}=(e_m,e_n)\). \(\square\)

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

Proof. Note \((We_m,e_n)=(e_{m+1},e_n)=\delta_{m+1,n}=\delta_{m,n-1}=(e_m,W^{-1}e_n)\), from which it follows that \(W^\ast=W^{-1}\). \(\square\)

Now let the Hilbert space \(K\) and its subspace \(H\) (invariant under \(W\)) be given. Consider the 'orthonormal' operator given by \(Re_n=e_{-(n+1)}\). It follows that \(R\) is a unitary involution and \[ Re_0=W^{-1}e_0 \quad RH = H^{\perp} \quad R \circ W = W^{-1} \circ R. \]

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, \[ (Wf)(x) = f(x) - \int_x^\infty \frac{1}{t}f(t)dt, \quad x \in L^2(0,\infty). \] 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 \((W^{k-1}e_0,e_0)=0\), 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 \(W^ke_0 \in L^2(0,\infty)\) 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 (admitting a countable basis, proof), \((W^ke_0)_k\) 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 \(\square\)


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

Elementary Properties of Cesàro Operator in L^2




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