Let be the set of all polynomial functions of degree 4 or less on . If and , we define .
Further, we define .
Will be a complete metric space?
Since is finite dimensional and is a norm it follows that is a Banach space.
If is a Cauchy sequence in , then for there exist such that for . A direct calculation shows that this implies
Taking small enough (the quotient in the last inequality goes to 0 as goes to 0), this shows that is a Cauchy sequence in . Hence there
such that tends to in . But this implies that tends to in , since
.
Thus the answer is yes if you show that is actually a metric.
Showing that is a norm, on , since X is finite dimensional, you obtain that is complete. You can consult any book related to topological vector spaces, in Robertson and Robertson's book "Topological vector spaces" for instance. Perhaps you are not allowed to use this fact?
The completenes of follow from the given argument above, Cauchy in implies Cauchy in , then convergence in and .
I believe that in the same book I have mentioned you can find the clues to show that defines a metric.