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 $\square$

References

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