## 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 $\varepsilon>0$, there exists some $N>0$ such that $|s_n-s|<\varepsilon$ 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 $L^2(0,\infty)$.

## The Cesàro operator

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

### Compactness and boundedness of $C$

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

Consider a family of functions $(\varphi_A)_{A>0}$ where

(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

Hence $\lVert C\varphi_A \rVert = \frac{\sqrt{1+A^2}}{A}$. 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 $f_n=\varphi_{2^n}$, then for any $m>n$ we get

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:

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, $C$ 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 $(e_n)_{n \in \mathbb{N}}$ 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 $(e_n)_{n \in \mathbb{Z}}$ such that $We_{n}=e_{n+1}$ for all $n \in \mathbb{Z}$. Besides, if we consider the subspace $H$ which is spanned by $(e_n)_{n \in \mathbb{N}}$, 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 $(We_m,e_n)=(e_{m+1},e_n)=\delta_{m+1,n}=\delta_{m,n-1}=(e_m,W^{-1}e_n)$, which 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

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,

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

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

meanwhile

Hence $We_0 \perp e_0$. Suppose now we have $(W^ke_0,e_0)=0$, then

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 $(e_n)_{n \geq 0}$.

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 $[1,\infty)$, 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

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 $W^ne_0 \in H$ 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

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 $(e_n)_{n \in \mathbb{Z}}$ 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

Since $(e_0,e_1,\cdots)$ spans $H$, $RH=H^{\perp}$, we see $(Re_0,Re_1,\cdots)$ 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 done $\square$

## References

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