Cesaro means and Banach limits - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:42:52Z http://mathoverflow.net/feeds/question/93516 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93516/cesaro-means-and-banach-limits Cesaro means and Banach limits kap44 2012-04-08T19:58:49Z 2013-01-12T00:57:40Z <p>Consider the class of bounded sequences to which every Banach limit (non-negative shift-invariant continuous functional on $l^\infty$ taking convergent sequences in the usual sense to their limits) assigns the same limit value.</p> <p><b> Does a sequence belong to this class if its Cesaro means have a limit? </b></p> <p>Also, is the converse true?</p> http://mathoverflow.net/questions/93516/cesaro-means-and-banach-limits/93547#93547 Answer by Aaron Tikuisis for Cesaro means and Banach limits Aaron Tikuisis 2012-04-09T06:10:16Z 2012-04-09T10:38:05Z <p>We can characterize Banach limits as continuous functionals on $\ell^\infty$ which vanish on <code>$$X := \{(x_n - x_{n+1}): (x_n) \in \ell^\infty\}$$</code> and which send the constant sequence $(1,1,\dots)$ to $1$.</p> <p>Note that $X$ is a subspace. The Hahn-Banach Theorem tells us that we are asking: if $(y_n) \in \ell^\infty$ has Cesaro mean $0$, is it in the closure of $X$? (And the converse question is: does every element of $X$ have Cesaro mean $0$? Yes; since the $n^\text{th}$ Cesaro mean of $(x_n-x_{n+1})$ is $(x_1-x_{n+1})/n$, which converges to $0$ since $(x_n)$ is uniformly bounded.)</p> <p>The answer is no. Consider the sequence $(y_n)$ that has $1$ once, followed by $-1$ three times, then $1$ five times, and so on. One can compute the Cesaro mean, and see that it approaches $0$ in the limit. But $(y_n)$ is not in the closure of $X$.</p> <p>Surely, if it were, then let $(x_n) \in \ell^\infty$ be such that $$\|(y_n) - (x_n-x_{n+1})\|_\infty &lt; 1/2.$$ Let $M$ be a natural number, $M \geq \|(x_n)\|$. Let $n$ be an index such that $$y_n = \cdots = y_{n+4M} = 1.$$ Then for $i=1,\dots,4M$, $$x_{n+i} &lt; x_{n + i-1} - y_{n + i - 1} + 1/2 = x_{n + i - 1} - 1/2,$$ and summing these up, we find $$x_{n+4M} &lt; x_n - 4M/2.$$ This contradicts the assumption that $\|(x_n)\| \leq M$.</p> http://mathoverflow.net/questions/93516/cesaro-means-and-banach-limits/118595#118595 Answer by Daniel Mansfield for Cesaro means and Banach limits Daniel Mansfield 2013-01-11T01:03:19Z 2013-01-12T00:57:40Z <p>If $(x_n) \in \ell^\infty$. According to Lorenz the Banch limit is unique (also known as <em>almost convergent</em>) iff $$\lim_{p\mapsto\infty} \frac{ x_n + x_{n+1} + \cdots + x_{n+p}}{p} = L \quad (*)$$ uniformly in $n$. Setting $n=0$ yields Cesaro summability.</p> <p>As Aaron says, the converse is false. If each $x_n$ is chosen uniformly at random from $\{0,1\}$ then this sequence almost never has property $(*)$ (see Connor's appropriately named article <em>Almost none of the sequences of 0's and 1's are almost convergent</em>)</p> <p>However the Cesaro limit of this random sequence $(x_n)$ is almost always $1/2$ by the law of large numbers.</p>