= Solution
A proper <filter on a set> $X$ is a nonempty family $\mathcal F\subseteq\mathcal P(X)$ that excludes $\varnothing$, is closed under finite <intersections>, and is upward closed under inclusion. In particular $X\in\mathcal F$. In a <topological space>, <convergence of a filter> $\mathcal F\to x$ means that every <neighbourhood> of $x$ belongs to $\mathcal F$. An <ultrafilter> is a proper <filter> maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.
To justify the equivalence, if $A\notin\mathcal U$ and every member of $\mathcal U$ met $A$, adjoining $A$ would generate a larger proper <filter>. Maximality therefore supplies $F\in\mathcal U$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in\mathcal U$. Both complementary sets cannot belong to a proper <filter>. Conversely, a <filter> with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty <intersection>.
The <ultrafilter characterization of compact Hausdorff spaces>, stated without proof, is: \b[a <topological space> is <compact> and <Hausdorff> exactly when every <ultrafilter> on it converges to a unique point.]
For a finite <set partition> $X=A_1\cup\cdots\cup A_n$, if no cell belonged to $\mathcal U$, the <ultrafilter> dichotomy would put all $X\setminus A_j$ in $\mathcal U$. Their finite <intersection> is empty, a contradiction. Thus some $A_j\in\mathcal U$, and at most one cell can belong because distinct cells are disjoint. The converse is: \b[a proper <filter on a set> meeting every finite partition by containing one cell is an <ultrafilter>.] Apply the partition hypothesis to $X=A\cup(X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.
For a <function> $f:X\to Y$, the <image of a filter> is
$$
f_*\mathcal F=\{B\subseteq Y:f^{-1}(B)\in\mathcal F\}.
$$
Since <preimages> preserve inclusions and finite <intersections>, this is a proper <filter on a set> $Y$. Equivalently, it is the <filter> generated by $f(F)$ for $F\in\mathcal F$: if $f^{-1}(B)\in\mathcal F$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an <ultrafilter> $\mathcal U$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_*\mathcal U$ contains exactly one of $B,Y\setminus B$. Therefore \b[the <pushforward filter> of an <ultrafilter> is an <ultrafilter>].
Now let $x\in\ell^\infty$, the <l-infinity sequence space>, and put $a_x(n)=n^{-1}\sum_{j=1}^nx_j$. All these <Cesaro means> lie in the closed disk $K_x=\{z:|z|\leq\|x\|_\infty\}$, a <compact Hausdorff space>. The <pushforward filter> $a_{x*}\mathcal U$ contains $K_x$; restricting it to that disk and applying the <ultrafilter> theorem gives a unique limit
$$
\boxed{l(x)=\lim_{n\to\mathcal U}a_x(n).}
$$
For real sequences the disk can be replaced by a closed interval. This is the <ultrafilter construction of a Banach limit>.
The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal U$, hence so does their <intersection>. On that <intersection> the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha|+|\beta|)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\geq0$, its means lie in $[0,\|x\|_\infty]$, so $l(x)\geq0$. Thus $l$ is a <positive linear functional>. The bound $|l(x)|\leq\|x\|_\infty$ gives $\|l\|\leq1$, while the constant-one sequence satisfies $l(\mathbf1)=1$. Hence $\boxed{\|l\|=1}$.
Because $\mathcal U$ contains no <finite sets>, its dichotomy puts every cofinite set in $\mathcal U$. If $x_n\to L$ ordinarily, the <Cesaro theorem for convergent sequences> follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\varepsilon$ for $j>N$,
$$
|a_x(n)-L|\leq\frac1n\sum_{j=1}^N|x_j-L|+\varepsilon\quad(n>N).
$$
The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal U$, making its <ultralimit> also $L$. Thus $\boxed{x_n\to L\Longrightarrow l(x)=L}$.
Finally, for the <left shift on bounded sequences> $S$,
$$
a_{Sx}(n)-a_x(n)=\frac{x_{n+1}-x_1}{n},\qquad |a_{Sx}(n)-a_x(n)|\leq\frac{2\|x\|_\infty}{n}\to0.
$$
Taking <ultralimits> along the same cofinite-containing <ultrafilter> gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a <Banach limit>.
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