Hint: A generic simple polygon is monotone with respect to the $x$-axis if and only if it has exactly one vertex whose $x$-coordinate is smaller than its neighbors. This observation immediately suggests an $O(n)$-time algorithm, at least if no edge of your polygon is vertical.
Spoilers ahoy:
IsMonotone(X[0..n-1], Y[0..n-1])
local_mins ← 0
for i ← 0 to n-1
if (X[i] < X[i+1 mod n]) and (X[i] < X[i-1 mod n])
local_mins ← local_mins + 1
return (local_mins = 1)
If you're worried that your polygon might have vertical edges, use the following subroutine in place of the comparison X[i] < X[j]
to consistently break ties:
IsLess(X, i, j):
return ((X[i] < X[j]) or (X[i] = X[j] and i < j))
Finally, if $L$ is some other line of the form $ax+by=c$, modify IsLess
as follows:
IsLess(X, Y, i, j):
Di ← a·X[i] + b·Y[i]
Dj ← a·X[j] + b·Y[j]
return ((Dj < Dj) or (Di = Dj and i < j))