Does Reverse Polish Notation have an LL grammar?
Yes,

$
G = (\{E,S,R\}, \{n,o\}, P, E)\text{, with productions}\\
E \rightarrow n\ |\ SR \\
S \rightarrow nEo \\
R \rightarrow EoR\ |\ \epsilon \\
$

Proof:


Observation $(*)$: Each word of L represents an arithmetic expression, the result of calculating such an expression is essentially again a number. So whenever we have $w \in L$, we can treat $w$ as a number.
Theorems:


$L(G) \subset L$
$L(S) \subset L$
$w \in L(R) \implies w=\epsilon \lor (w=w_1 o, w_1 \in L)$


Inductive proof of above theorems on derivation step count $n$:


Basis $n=1$


$E \Rightarrow n\\
            \land \\
            n \in L
            $
$S\text{ can't generate any words in one step. }\\
            \land \\
            \emptyset \subset L
            $
$R \Rightarrow \epsilon, w=\epsilon$

Basis $n=2$

In all 3 cases nothing can be generated in 2 steps, and the theorems hold trivially.
Inductive step: $n=k&gt;1,$ induction hypothesis (IH): assume theorems hold for $n&lt;k$


$E \Rightarrow SR \Rightarrow^{k-1} w_1 w_2\, \\
           \text{where }w_1 \in L\text{ because of IH of theorem 2,} \\
           \text{and } w_2 = \epsilon\text{ or }w_2 = w_3 o, w_3 \in L\text{ because of IH of theorem 3.}$


First case: $w_2 = \epsilon$, then $w_1 w_2 = w_1 \in L$.
Second case: $w_1 w_2 = w_1 w_3 o =^{(*)} nno$, which is a valid RPN expression, thus $w_1 w_3 o \in L$

$S \Rightarrow nEo \Rightarrow^{k-1} nwo$, where $w \in L$ because of IH of theorem 1. $nwo =^{(*)} nno$, which is a valid RPN expression, and thus $n w o \in L$
$R \Rightarrow EoR \Rightarrow^{k-2} w_1 o R$, where $w_1 \in L$ because of IH of theorem 1. The last step can only be $w_1 o R \Rightarrow w_1 o \epsilon$, which satisfies theorem 3.



$\square$
Theorem: $L \subset L(G)$

Proof: induction on length of $w \in L$, show that $E \Rightarrow^* w$

(I got lazy from this point on, I might fill this in later)

$\square$
$(1) \land (2) \implies L(G) = L$
Theorem: G is LL(k) grammar, $k \in \mathbb{N}$

Proof: Pick a k large enough, then show that for each non-terminal you can decide which production rule to use by looking at the k next terminals.

$\square$


$\square$