By Theorem 3.3-2 of Erwin Kreyszig's Introductory Functional Analysis with Applications
$H$ be a Hilbert Space, and $Y$ be a closed subspace of it then $$H=Y \oplus Y^\perp$$
Can we extend this Lemma to any Inner-Product Space $X$ and its complete subspace $Y$ i.e.
$X$ be any Inner-Product Space, and Y be a complete subspace of it then $$X = Y \oplus Y^\perp$$
My thinking was
Given any $x \in X$ we can find unique $y \in Y$ such that $\delta = inf_{\tilde y \in Y} ||x-\tilde y|| = ||x-y||$(by Theorem 3.3-1) and we have $z=x-y \perp Y$(by Theorem 3.3-2). Hence $x=y+z$ is a representation of $x$ where $y \in Y$ and $z \in Y^\perp$. Now we need to show that the decomposition of $x$ is unique. For that we assume it is not correct i.e. $x$ has a different decomposition. Let it be $y_1$ and $z_1$. We would have $x=y_1+z_1=y+z \implies y_1-y=z-z_1$. We can prove that for any given subset $Y$ of $X$, $Y^\perp$ will be a closed subspace of $X$. Using the fact $Y$ and $Y^\perp$ both are vector subspaces of $X$ we have $y_1-y \in Y$ and $z-z_1 \in Y^\perp$. This makes $y_1-y \perp z-z_1$ and both of them are linearly independent which contradicts $y_1-y = z-z_1$.
Is this correct?