It is exactly $2^{64}$. You need to look for all possible keys for a successful brute-force. One cannot guarantee that the key will be in the half, $2^{63}$, that you are going to search. In, CS there is an adversary argument, that the adversary can always force you to the worst case. Tell him how you the search, he will produce a case that you are not going to find in $2^{63}$ key-space.

Actually, $2^{63}$ is the average case that you will find your keys during the brute-force.

It is exactly $2^{64}$. You need to look for all possible keys for a successful brute-force. One cannot guarantee that the key will be in the half, $2^{63}$, that you are going to search. In, CS there is an adversary argument, that the adversary can always force you to the worst case. Tell him how you the search, he will produce a case that you are not going to find in $2^{63}$ key-space.

Actually, $2^{63}$ is the average case that you will find your keys during the brute-force.

If you have multiple known-plaintext with encrypted different keys then with a multi-target attack you can find some keys faster. The expected cost of finding a key from $t$ target is $2^{64}/t$.