# Invariant ergodic measure Volterra operator on Continuous Functions

This is a follow-up to this question.

Define the Volterra operator $$V$$ on $$C_0([0,1])\triangleq \{g \in C([0,1]):g(0)=0\}$$ by $$f \mapsto \int_0^{\sqrt{\cdot}} f(s)ds.$$ Is there an example of an and locally positive ergodic and $$V$$-invariant Borel probability measure $$\mu$$ on $$C_0([0,1])$$?

Note: Locally positive means that for every non-empty open subset $$U$$ of $$C_0([0,1])$$ (with the usual compact-open topology) $$\mu(U)>0$$.

## 1 Answer

The only $$V$$-invariant probability measure is the delta concentrated on the origin. A quick way to see it is to look at the conjugate of $$V$$ with the multiplication operator $$M$$ by $$e^x$$. Indeed we have for any $$f\in C_0$$ and $$x\in[0,1]$$

$$\big|M^{-1}VMf(x)\big|=\Big|e^{-x}\int_0^{\sqrt x}f(s)e^s ds\Big|\le e^{-x}\int_0^{\sqrt x}e^s ds\|f\|_\infty =$$$$=e^{-x}\big(e^{\sqrt x}-1\big)\|f\|_\infty \le{2\over3}\|f\|_\infty.$$

So $$\|M^{-1}VM\|\le 2/3$$. This implies that $$\|M^{-1}V^nM\|\le (2/3)^n$$ for any $$n$$, and therefore for any $$r>0$$ we have an inclusion $$V^n(B(0,r))\subset V^nM(B(0,r))\subset B(0, ({2/3})^ner)$$, and finally $$B(0,r)\subset V^{-n}B(0, ({2/3})^ner).$$ Being $$r$$ and $$n$$ arbitrary, this implies that any $$V$$-invariant probability measure $$\mu$$ gives the same value to any nbd of $$0$$.

As a general principle, the same conclusion hold for any bounded operator with spectral radius less than $$1$$: up to conjugation it is a norm contraction, and the same argument applies.

• Hi Pietro. Thank you for this very nice and detailed answer (I especially like the comment about the spectral radius). However, what if the question were to be phrased on $C([0,1])$ instead of on $C([0,1])$? – AIM_BLB Nov 13 '19 at 18:32
• Then we just remove the subscript $_0$ from $C_0$ in the third line. – Pietro Majer Nov 13 '19 at 18:35
• I'm noticing this now, as I'm going through the details. So in general, the requirement of local positivity makes there not exist such a measure.... – AIM_BLB Nov 13 '19 at 18:35
• I'd say for the existence of a nontrivial invariant measure for an linear operator T you need the spectrum to have some modulus 1 element. – Pietro Majer Nov 13 '19 at 19:01