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) $\mu(U)>0$.


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.

  • $\begingroup$ 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 $C([0,1])$? $\endgroup$ – AIM_BLB Nov 13 '19 at 18:32
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    $\begingroup$ Then we just remove the subscript $_0$ from $C_0$ in the third line. $\endgroup$ – Pietro Majer Nov 13 '19 at 18:35
  • $\begingroup$ 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.... $\endgroup$ – AIM_BLB Nov 13 '19 at 18:35
  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Nov 13 '19 at 19:01

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