= Solution
For every <bounded sequence> $x$, $x-Sx\in Y$ and $L|_Y=0$. Linearity gives \b[$\boxed{L(x)=L(Sx)}$], as required. Thus the same $L$ satisfies all three conditions and is a <Banach limit>.
The usual positivity condition for a <Banach limit> follows automatically here. If $x_n\geq0$ and $a=\|x\|_\infty>0$, then $\|\mathbf1-x/a\|_\infty\leq1$. Since $\|L\|=L(\mathbf1)=1$, we have $|1-L(x)/a|\leq1$, whence $L(x)\geq0$. The case $x=0$ is immediate. This explains why the constructed <bounded linear functional> is also a <positive linear functional> extending the ordinary limit while remaining invariant under the <left shift on bounded sequences>.
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