Proof. Let \(d=\inf _{y\in A}||x-y||\). Choose a sequence \(\{y_n\}\), \(y_n\in A\) so that \(||x-y_n||\rightarrow d\) (*). Then \begin {align*} ||y_n-y_m||^2 &=||(y_n-x)-(y_m-x)||^2\\ &=2||y_n-x||^2+2||y_m-x||^2-||-2x+y_n+y_m||^2\\ &=2||y_n-x||^2+2||y_m-x||^2-4||x-1/2(y_n+y_m)||^2\\ &\leq 2||y_n-x||^2+2||y_m-x||^2-4d^2\\ &\rightarrow 2d^2+2d^2-4d^2=0 \end {align*}

Where the second equality follows from the parallelogram law, and the inequality follows from the fact that \(\frac {1}{2}(y_n+y_m)\in A\).

Therefore, \(\{y_n\}\) is Cauchy in \(A\), and since \(A\) is closed, \(y_n\rightarrow z\in A\) it then follows from (*) that \(||x-z||=d\).

Now suppose that \(z_0\in A\) is another element closest to \(x\), then \(||x-z_0||=d=||x-z||\) so that \(z_0=z\). □

Proof. Let \(x\in H\). Then by the Lemma 5.22 there exist a unique \(z\in A\) closest to \(x\).
Define \(w=x-z\), then \(x=z+w\). Since uniqueness follows directly from uniqueness of \(z\), it suffices to show that \(w\in A^{\perp }\).
Let \(y\in A\) and \(\alpha \in \mathbb {R}\). If \(d=||x-z||\), then \[d^2\leq ||x-(z+\alpha y)||^2=||w-\alpha y||^2=d^2-2\alpha \textbf {Re}(w,y)+\alpha ^2||y||^2.\] Thus, for all real \(\alpha \), \(-2\alpha \textbf {Re}(w,y)+\alpha ^2||y||^2\geq 0,\hspace {0.2cm}\) which implies that \(\textbf {Re}(w,y)=0\), since \(\alpha \) is arbitrary.
A similarly argument using \(i\alpha \) in place of \(\alpha \) shows that \(\textbf {Im}(w,y)=0\). Therefore,
\((w,y)=\textbf {Re}(w,y)+\textbf {iIm}(w,y)=0.\) So that \(w\in A^{\perp }\).

The projection theorem sets up a natural isomorphism between \(A\oplus A^{\perp }\) and \(H\) given by \((z,w)\longmapsto z+w\).

However, we often suppress the isomorphism and simply write \(H=A\oplus A^{\perp }\) □