Best approximation theorem
Theorem
Let X be an inner product space with induced norm, and
a non-empty, complete convex subset. Then, for all
, there exists a unique best approximation a0 to x in A.


Proof
Suppose x = 0 (if not the case, consider A − {x} instead) and let
. There exists a sequence (an) in Asuch that

-
.
-
.
-
.
Hence
-
as
-
.
Since
,
. Furthermore


-
as
,
which proves | | a0 | | = d. Existence is thus proved. We now prove uniqueness. Suppose there were two distinct best approximations a0and a0‘ to x (which would imply | | a0 | | = | | a0‘ | | = d). By the parallelogram rule we would have
-
.
Then