Let $X$ be an inner product space. ($X$ may not be Hilbert nor finite-dimensional.) Let $V$ be a vector subspace of $X$ ($V$ may be infinite-dimensional). Let $x\in X$.
Then we know that the best approximation of $x$ (i.e., a vector $y\in V$ satisfies $\|x-y\|=\inf_{z\in V}\|x-z\|$) may not exist. But suppose it exists, then is such vector unique?
I think the answer is yes, but I have just found a reference saying whether the "uniqueness" is hold is related to some kind of convexity, after he had just shown the existence in a particular circumstance (that is, $V$ is finite-dimensional).
My proof that existence implies uniqueness: Let $y\in V$ satisfies $\|x-y\|=\inf_{z\in V}\|x-z\|$, then one can show that $x-y\in V^\perp$. If there is another $y'$ such that the $\|x-y'\|=\inf_{z\in V}\|x-z\|$, then again $x-y'\in V^\perp$. Since $X=V\oplus V^\perp$, $y=y'$. Is there any potential logical problem in the argument?