Proof. Let \(x,y\in V\). Then \begin {align*} ||x+y||^2 &=(x+y,x+y)\\ &=(x,x)+(x,y)+(y,x)+(y,y)\\ &=||x||^2+0+0+||y||^2\\ &=||x||^2+||y||^2 \end {align*}

\(\therefore \hspace {0.4cm} ||x+y||^2=||x||^2+||y||^2\). □

Proof.

1.

Let \(y_1,y_2\in \{x\}^{\perp }\) and \(\alpha ,\beta \in \mathbb {F}\). Then \begin {align*} (\alpha y_1+\beta y_2,x) &=(\alpha y_1,x)+(\beta y_2,x)\\ &=\alpha (y_1,x)+\beta (y_2,x)\\ &=\alpha .0+\beta .0\\ &=0 \end {align*}

\(\implies \alpha y_1+\beta y_2\in \{x\}^{\perp }\) \(\implies \{x\}^{\perp }\) is a subspace of \(V\).

Now, if \(\{y_n\}^{\infty }_{n=1}\) is a sequence in \(\{x\}^{\perp }\) with \(\lim \limits _{n\rightarrow \infty }y_n=y\) then \[(y,x)=\Big (\lim _{n\rightarrow \infty }y_n,x\Big )=\lim _{n\rightarrow \infty }(y_n,x)=0.\] \(\implies y\in \{x\}^{\perp }\) \(\implies \{x\}^{\perp }\) is closed.

2.

Let \(y_1,y_2\in A^{\perp }\), \(z\in A\), and \(\alpha , \beta \in \mathbb {F}\). Let \(z\) play the role of \(x\) in (1).

3.

Let \(x\in A\) and \(y\in A^{\perp }\). Then \[(x,y)=0\implies x\in (A^{\perp })^{\perp }=A^{\perp \perp }.\] \(\implies A\subset A^{\perp \perp }\).

4.

Let \(x\in B^{\perp }\) and \(y\in A\). If \(A\subset B\) then \(y\in B\), \((x,y)=0\). Thus \(x\in A^{\perp }\implies B^{\perp }\subset A^{\perp }\).

5.

Let \(x\in A\) and \(y\in A^{\perp }\) then \((x,y)=0\).
\(\implies x\in A^{\perp \perp }\) \(\implies y\in A^{\perp \perp \perp }\)

So \(A^{\perp }\subset A^{\perp \perp \perp }\) (1) Conversely, let \(x\in A^{\perp \perp \perp }\) and \(y\in A^{\perp \perp }\). Then \((x,y)=0\implies x\in A^{\perp }\).

\(\implies A^{\perp \perp \perp }\subset A^{\perp }\) (2) Combining (1) and (2) gives \(A^{\perp }=A^{\perp \perp \perp }\).