Use map k (u:us) = k u : map k us, list-map fusion, and the left fold equation
foldl f e (x:ys) = foldl f (f e x) ys
For the empty list, the prefix definition immediately gives scanl f e [] = [e]. For a nonempty list, calculate:
scanl f e (x:xs)
  = map (foldl f e) ([] : map (x:) (inits xs))
  = e : map ((foldl f e) . (x:)) (inits xs)
  = e : map (foldl f (f e x)) (inits xs)
  = e : scanl f (f e x) xs
The second-to-third line uses the displayed fold equation as an equality of functions of ys. Hence the efficient left scan is:
scanl :: (b -> a -> b) -> b -> [a] -> [b]
scanl f e [] = [e]
scanl f e (x:xs) = e : scanl f (f e x) xs
On a finite input of length , fully demanding the original prefix-fold results performs accumulator updates. The recurrence performs . Thus the update count improves from to , assuming constant-cost updates. Both return results. This does not assert constant space for arbitrary accumulators or imply that lazy accumulators are evaluated strictly.

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