If the answer is yes, then you should watch this presentation by Martin Kleppmann:
This proof is interesting, and you have to take into account that this is for a single register. If you're using a reader-writer lock (or the Left-Right technique) and you're only doing reads, then the overhead is that of reading/writing to a couple of variables, which amortized over the entire critical section can become a very low overhead, and therefore make linearizability profitable.
Now you can say, that you can use the same argument for seq-cst, but the problem with such an argument is that there are no seq-cst locks or rw-locks, for the simple reason that seq-cst is not (generically) composable. Therefore, I would argue that, depending on what you're doing, you can have more performance for a linearizable object that you get for a seq-cst object, simply because you can not compose regular reads and writes with seq-cst reads and writes and still expect the result to be seq-cst, but you can for a linearizable object like a mutex, or a rw-lock or a left-right.