- #71
Erland
Science Advisor
- 753
- 140
Advanced Problem 7a.
I refer to post #102 here, and use the notation from that post. In the post, a real linear functional ##F## on ##\mathcal B## such that ##\underline\lim x_n\le F(\{x_n\}_{n=1}^\infty)\le\overline\lim\{x_n\}## for all ##\{x_n\}_{n=1}^\infty\in\mathcal B## is constructed, and Lemma 2 in the post holds for ##F##.
We will give two real linear operators ##G## and ##G'## on ##\mathcal B##, which both satisfy the conditions of Lemma 2, so that, by that Lemma, ##L=F\circ G## and ##L'=F\circ G'## both are generalized limits. We will also show that ##L\neq L'##, and thus we have two unequal generalized limits.
For ##\{x_n\}_{n=1}^\infty##, we put
##G(\{x_n\}_{n=1}^\infty)=\{y_n\}_{n=1}^\infty##, with ##y_n=\frac1{2^{2n-2}}\sum_{2^{2n-2}}^{2^{2n-1}-1}x_k##, and
##G'(\{x_n\}_{n=1}^\infty)=\{y'_n\}_{n=1}^\infty##, with ##y'_n=\frac1{2^{2n-1}}\sum_{2^{2n-1}}^{2^{2n}-1}x_k##, for all ##n\in\mathbb Z_+##.
It is easy to see that ##G## and ##G'## are real linear operators on ##\mathcal B##. Also, with ##\{x_n\}_{n=1}^\infty##, ##\{y_n\}_{n=1}^\infty##, and ##\{y'_n\}_{n=1}^\infty## as above, it is easy to see that for all ##N\in \mathbb Z_+##, ##\sup_{n\ge N} y_n\le \sup_{n\ge N} x_n##. Hence, ##\overline\lim y_n\le \overline \lim x_n##. Applying this to ##-\{x_n\}_{n=1}^\infty## and using linearity and ##\underline\lim x_n=-\overline\lim (-x_n)##, and likewise for ##\{y_n\}_{n=1}^\infty##, we obtain ##\underline\lim x_n\le\underline\lim y_n##, so that
##\underline\lim x_n\le\underline\lim y_n\le\overline\lim y_n\le\overline\lim x_n##, and a similar argument gives
##\underline\lim x_n\le\underline\lim y'_n\le\overline\lim y'_n\le\overline\lim x_n##.
Also, let ##u_n=x_{n+1}## for all ##n\in\mathbb Z_+##, and put ##\{z_n\}_{n=1}^\infty=G(\{u_n\}_{n=1}^\infty)## and ##\{z'_n\}_{n=1}^\infty=G'(\{u_n\}_{n=1}^\infty)##.
Then, for ##n\in\mathbb Z_+##: ##z_n-y_n=\frac1{2^{2n-2}}(\sum_{2^{2n-2}}^{2^{2n-1}-1}x_{k+1}-\sum_{2^{2n-2}}^{2^{2n-1}-1}x_k)=(x_{2^{2n-1}}-x_{2^{2n-2}})/2^{2n-2}\to 0## as ##n\to \infty##, and a similar calculation gives ##z'_n-y'_n=(x_{2^{2n}}-x_{2^{2n-1}})/2^{2n-1}\to 0##, as ##n\to \infty##.
So, ##\lim_{n\to\infty}(z_n-y_n)=0## and ##\lim_{n\to\infty}(z'_n-y'_n)=0##.
All this holds for all ##\{x_n\}_{n=1}^\infty##. This means that both 1) and 2) of Lemma 2 are satisfied for both ##G## and ##G'##. Therefore, Lemma 2 gives that both ##L=F\circ G## and ##L'=F\circ G'## are generalized limits.
We must show that ##L\neq L'##. Put ##x_n=(-1)^{\lfloor\log_2 n\rfloor}##, for all ##n\in \mathbb Z_+##. With ##\{y_n\}_{n=1}^\infty=G(\{x_n\}_{n=1}^\infty)## and ##y'_n=G'(\{x_n\}_{n=1}^\infty)##, we have ##y_n=1## and ##y'_n=-1## for all ##n\in\mathbb Z_+##.
Since ##\{y_n\}_{n=1}^\infty## converges and ##\underline\lim y_n\le F(\{y_n\}_{n=1}^\infty)\le\overline\lim\{y_n\}##, we have ##L(\{x_n\}_{n=1}^\infty)=F(\{y_n\}_{n=1}^\infty)=\lim_{n\to\infty} y_n=1##. Likewise, ##L'(\{x_n\}_{n=1}^\infty)=F(\{y'_n\}_{n=1}^\infty)=\lim_{n\to\infty} y'_n=-1##.
Hence, ##L(\{x_n\}_{n=1}^\infty)\neq L'(\{x_n\}_{n=1}^\infty)##, so ##L\neq L'##.
We have proved that there exists two unequal generalized limits.
I refer to post #102 here, and use the notation from that post. In the post, a real linear functional ##F## on ##\mathcal B## such that ##\underline\lim x_n\le F(\{x_n\}_{n=1}^\infty)\le\overline\lim\{x_n\}## for all ##\{x_n\}_{n=1}^\infty\in\mathcal B## is constructed, and Lemma 2 in the post holds for ##F##.
We will give two real linear operators ##G## and ##G'## on ##\mathcal B##, which both satisfy the conditions of Lemma 2, so that, by that Lemma, ##L=F\circ G## and ##L'=F\circ G'## both are generalized limits. We will also show that ##L\neq L'##, and thus we have two unequal generalized limits.
For ##\{x_n\}_{n=1}^\infty##, we put
##G(\{x_n\}_{n=1}^\infty)=\{y_n\}_{n=1}^\infty##, with ##y_n=\frac1{2^{2n-2}}\sum_{2^{2n-2}}^{2^{2n-1}-1}x_k##, and
##G'(\{x_n\}_{n=1}^\infty)=\{y'_n\}_{n=1}^\infty##, with ##y'_n=\frac1{2^{2n-1}}\sum_{2^{2n-1}}^{2^{2n}-1}x_k##, for all ##n\in\mathbb Z_+##.
It is easy to see that ##G## and ##G'## are real linear operators on ##\mathcal B##. Also, with ##\{x_n\}_{n=1}^\infty##, ##\{y_n\}_{n=1}^\infty##, and ##\{y'_n\}_{n=1}^\infty## as above, it is easy to see that for all ##N\in \mathbb Z_+##, ##\sup_{n\ge N} y_n\le \sup_{n\ge N} x_n##. Hence, ##\overline\lim y_n\le \overline \lim x_n##. Applying this to ##-\{x_n\}_{n=1}^\infty## and using linearity and ##\underline\lim x_n=-\overline\lim (-x_n)##, and likewise for ##\{y_n\}_{n=1}^\infty##, we obtain ##\underline\lim x_n\le\underline\lim y_n##, so that
##\underline\lim x_n\le\underline\lim y_n\le\overline\lim y_n\le\overline\lim x_n##, and a similar argument gives
##\underline\lim x_n\le\underline\lim y'_n\le\overline\lim y'_n\le\overline\lim x_n##.
Also, let ##u_n=x_{n+1}## for all ##n\in\mathbb Z_+##, and put ##\{z_n\}_{n=1}^\infty=G(\{u_n\}_{n=1}^\infty)## and ##\{z'_n\}_{n=1}^\infty=G'(\{u_n\}_{n=1}^\infty)##.
Then, for ##n\in\mathbb Z_+##: ##z_n-y_n=\frac1{2^{2n-2}}(\sum_{2^{2n-2}}^{2^{2n-1}-1}x_{k+1}-\sum_{2^{2n-2}}^{2^{2n-1}-1}x_k)=(x_{2^{2n-1}}-x_{2^{2n-2}})/2^{2n-2}\to 0## as ##n\to \infty##, and a similar calculation gives ##z'_n-y'_n=(x_{2^{2n}}-x_{2^{2n-1}})/2^{2n-1}\to 0##, as ##n\to \infty##.
So, ##\lim_{n\to\infty}(z_n-y_n)=0## and ##\lim_{n\to\infty}(z'_n-y'_n)=0##.
All this holds for all ##\{x_n\}_{n=1}^\infty##. This means that both 1) and 2) of Lemma 2 are satisfied for both ##G## and ##G'##. Therefore, Lemma 2 gives that both ##L=F\circ G## and ##L'=F\circ G'## are generalized limits.
We must show that ##L\neq L'##. Put ##x_n=(-1)^{\lfloor\log_2 n\rfloor}##, for all ##n\in \mathbb Z_+##. With ##\{y_n\}_{n=1}^\infty=G(\{x_n\}_{n=1}^\infty)## and ##y'_n=G'(\{x_n\}_{n=1}^\infty)##, we have ##y_n=1## and ##y'_n=-1## for all ##n\in\mathbb Z_+##.
Since ##\{y_n\}_{n=1}^\infty## converges and ##\underline\lim y_n\le F(\{y_n\}_{n=1}^\infty)\le\overline\lim\{y_n\}##, we have ##L(\{x_n\}_{n=1}^\infty)=F(\{y_n\}_{n=1}^\infty)=\lim_{n\to\infty} y_n=1##. Likewise, ##L'(\{x_n\}_{n=1}^\infty)=F(\{y'_n\}_{n=1}^\infty)=\lim_{n\to\infty} y'_n=-1##.
Hence, ##L(\{x_n\}_{n=1}^\infty)\neq L'(\{x_n\}_{n=1}^\infty)##, so ##L\neq L'##.
We have proved that there exists two unequal generalized limits.