# 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*a*_{0}to*x*in*A*.## Proof

Suppose

*x*= 0 (if not the case, consider*A*− {*x*} instead) and let . There exists a sequence (*a*_{n}) in*A*such that- .

- .

Since

*A*is convex, so- .

Hence

- as

- .

Since , . Furthermore

- as ,

which proves | |

*a*_{0}| | =*d*. Existence is thus proved. We now prove uniqueness. Suppose there were two distinct best approximations*a*_{0}and*a*_{0}‘ to*x*(which would imply | |*a*_{0}| | = | |*a*_{0}‘ | | =*d*). By the parallelogram rule we would have- .

Then

which cannot happen since

*A*is convex, and as such , which means , thus completing the proof.