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May
11 |
awarded | Supporter |
May
6 |
accepted | Multiple ergodic averages with varying number of terms |
May
6 |
comment |
Multiple ergodic averages with varying number of terms
Overcomplicating indeed. Thank you. |
May
5 |
comment |
Multiple ergodic averages with varying number of terms
You're perfectly right, weak-mixing is the good working hypothesis. As for more expedient ways to prove that $f=0$ a.e.: For integers $n$ and $p$, $f_{n+p} \leq f_n S^n (f_p)$ so that when $p \rightarrow \infty$ and considering that $f_n \leq f_1$, the inequality $f \leq f_1 S^n (f)$ holds for all $n$. Taking averages and integrals, one finds that $\int f \leq 1/n \sum_{k=0}^n \int f_1 S^k (f) \rightarrow \int f_1 \int f$ so that $\int f \leq \int f_1 \int f$ which proves that $f=0$ almost everywhere. But this does not help for the convergence of $\sum f_n$ unless I am missing something? |
May
5 |
awarded | Editor |
May
5 |
revised |
Multiple ergodic averages with varying number of terms
added 18 characters in body; edited title |
May
5 |
asked | Multiple ergodic averages with varying number of terms |
Apr
11 |
awarded | Scholar |
Apr
11 |
accepted | Uniqueness of analytic continuation on a domain of C^n. |
Apr
11 |
awarded | Student |
Apr
11 |
asked | Uniqueness of analytic continuation on a domain of C^n. |
Apr
6 |
comment |
Convexity of a cone in CxC
Indeed. thank you. |
Apr
6 |
asked | Convexity of a cone in CxC |