 Solution

For every bounded sequence $x$, $x-Sx\in Y$ and $L{|}_Y=0$. Linearity gives $\boxed{L(x)=L(Sx)}$, 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 $x_n\ge 0$ and $a=\Vert x{\Vert }_{\infty }>0$, then $\Vert 1-x/a{\Vert }_{\infty }\le 1$. Since $\Vert L\Vert =L(1)=1$, we have $|1-L(x)/a|\le 1$, whence $L(x)\ge 0$. 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.