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

Here, ${\pi }_n$ is called the Cesàro sum of $\{s_n\}$. The proof is rather simple. Given $\epsilon >0$, there exists some $N>0$ such that $|s_n-s|<\epsilon$ for all $n>N$. Therefore we can write

For fixed $N$, we can pick $n$ big enough such that $N/n<1/2$ (i.e. $n>2N$) and

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

In this blog post we investigate this operator in Hilbert space $L^2(0,\mathrm{\infty })$.

The Cesàro operator

Put $L^2=L^2(0,\mathrm{\infty })$ relative to Lebesgue measure, and the Cèsaro operator $C$ is defined as follows:

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 $\Vert C\Vert =2$. I have organised several proofs of this. But $C$ is not compact.

Here is the proof. Consider a family of functions $\{{\phi }_A{\}}_{A>0}$ where

(I owe Oliver Diaz for this family of functions.) It’s not hard to show that $\Vert {\phi }_A\Vert =1$. If we apply $C$ on it we see

Hence $\Vert C{\phi }_A\Vert =\frac{\sqrt{1+A^2}}{A}$. Meanwhile for $B>A$, we have

It follows that

If we compute the norm on the right hand side we get

As a result, if we pick $f_n={\phi }_{2^n}$, then for any $m>n$ we get

Therefore, we find a sequence $(f_n)$ on the unit ball such that $(C{f}_n)$ has no convergent subsequence.

Also we can find its adjoint operator:

Hence the adjoint is given by

$C^{\ast }$ is not compact as well. Further, another application of Fubini’s theorem shows that

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 $\mathbb{N}={\mathbb{Z}}_{\ge 0}$. This section can also help you understand the connection between $L^2(0,1)$ and $L^2(0,\mathrm{\infty })$.

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 $W{e}_n=e_{n+1}$ 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 $(U{e}_m,U{e}_n)=(e_{m+1},e_{n+1})={\delta }_{m+1,n+1}={\delta }_{mn}=(e_m,e_n)$. $◻$

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

Proof. Note $(W{e}_m,e_n)=(e_{m+1},e_n)={\delta }_{m+1,n}={\delta }_{m,n-1}=(e_m,W^{-1}{e}_n)$, from which it follows that $W^{\ast }=W^{-1}$. $◻$

Now let the Hilbert space $K$ and its subspace $H$ (invariant under $W$) be given. Consider the ‘orthonormal’ operator given by $R{e}_n=e_{-(n+1)}$. It follows that $R$ is a unitary involution and

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 $L^2(0,\mathrm{\infty })$ 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 $L^2(0,\mathrm{\infty })$.

Explicitly,

Also we claim the basis to be generated by $e_0={\chi }_{(0,1)}$. First of all we show that $(W{e}_n{)}_{n\ge 0}$ is orthonormal. Note as we have proved, $W^{\ast }W=(I-C)(I-C^{\ast })=I$. Without loss of generality we assume that $m\ge n$ and therefore

If $m=n$, then $(e_m,e_n)=(e_0,e_0)=1$. Hence it is reduced to prove that $(W^k{e}_0,e_0)=0$ for all $k>0$. First of all we have

meanwhile

Hence $W{e}_0\perp e_0$. Suppose now we have $(W^{k-1}{e}_0,e_0)=0$, then

Note $W^k{e}_0\in L^2(0,\mathrm{\infty })$ always vanishes when $x\ge 1$: when we are doing inner product, $[1,\mathrm{\infty })$ is automatically excluded. With these being said, $(W^n{e}_0{)}_{n\ge 0}$ 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), $(W^k{e}_0{)}_k$ 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

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 $[1,\mathrm{\infty })$, 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 $e_0\left(\frac{1}{x}\right)={\chi }_{[1,\mathrm{\infty })}$ almost everywhere, we shall put $g(x)=-\frac{1}{x}$. It is then clear that $R{e}_0=W^{-1}{e}_0$ and $RH=H^{\perp }$. For the third condition, we need to show that

Note

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 $W^n{e}_0\in H$ for all $n\ge 0$ and $\{e_n=W^n{e}_0{\}}_{n\ge 0}$ forms a orthonormal basis of $H$. $R$ is a unitary involution on $K$ such that

$$R{e}_0=W^{-1}{e}_{0}RH=H^{\perp }R\circ W=W^{-1}\circ R$$

then $W$ is a simple bilateral shift.

Indeed, objects mentioned in step 1 fit in this lemma. To begin with, we write $e_n=W^n{e}_0$ 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 $m+j,n+j\ge 0$. Therefore

Since $(e_0,e_1,\cdots )$ spans $H$, $RH=H^{\perp }$, we see $(R{e}_0,R{e}_1,\cdots )$ spans $H^{\perp }$. But

hence $\{e_{-1},e_{-2},\cdots \}$ spans $H^{\perp }$. By definition of $W$, it is indeed a bilateral shift. And our proof is now complete. $◻$

References

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