Proof. Write \(x\) as \(\displaystyle {x=\sum ^N_{n=1}(x,x_n)x_n+\Bigg (x-\sum ^N_{n=1}(x,x_n)x_n\Bigg )}\hspace {0.2cm}\)

Now \(\forall n\leq N\), \begin {align*} \Bigg (x-\sum ^N_{k=1}(x,x_k)x_k,x_n\Bigg ) &=(x,x_n)-\Bigg (\sum ^N_{k=1}(x,x_k)x_k,x_n\Bigg )\\ &=(x,x_n)-\sum ^N_{k=1}(x,x_k)(x_k,x_n)\hspace {0.4cm} \text {at some point}\hspace {0.3cm} k=n\\ &=(x,x_n)-(x,x_n)(x_n,x_n)\\ &=(x,x_n)-(x,x_n)\hspace {0.6cm} \text {since}\hspace {0.3cm} n=m\\ &=0 \end {align*}

This implies that \(\displaystyle {x-\sum ^N_{n=1}(x,x_n)x_n\perp x_n}\) \[\implies x-\sum ^N_{n=1}(x,x_n)x_n\perp \sum ^N_{n=1}(x,x_n)x_n\]
So by Lemma 4.6 \begin {align*} \implies ||x||^2 &=\Bigg |\Bigg |x-\sum ^N_{n=1}(x,x_n)x_n\Bigg |\Bigg |^2+\Bigg |\Bigg |\sum ^N_{n=1}(x,x_n)x_n\Bigg |\Bigg |^2\\ &=\Bigg |\Bigg |x-\sum ^N_{n=1}(x,x_n)x_n\Bigg |\Bigg |^2+\sum ^N_{n=1}|(x,x_n)|^2(x_n,x_n)\\ \end {align*}

Thus \(\displaystyle {||x||^2=\Bigg |\Bigg |x-\sum ^N_{n=1}(x,x_n)x_n\Bigg |\Bigg |^2+\sum ^N_{n=1}|(x,x_n)|^2}\) □

Proof. The proof follows directly from Theorem 5.10 □

Proof. For \(y=0\), the result is trivially true, so assume \(y\neq 0\). Then the vector \(\large {\dfrac {y}{||y||}}\) by itself forms an orthonormal set, and so, it follows from the Bessel’s inequality that for \(x\in V\), \[||x||^2\geq \Bigg |\Bigg (x,\frac {y}{||y||}\Bigg )\Bigg |^2=\frac {|(x,y)|^2}{||y||^2}\] That is \(|(x,y)|^2\leq ||x||^2||y||^2\implies |(x,y)|\leq ||x||||y||.\) □

Proof. We verify only the triangle inequality, since the other properties follow immediately from definition 5.1. 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+2\textbf {Re}(x,y)+||y||^2\\ &\leq ||x||^2+2|(x,y)|+||y||^2\\ &\leq ||x||^2+2||x||||y||+||y||^2\hspace {1cm} \text {by schwarz}\\ &=(||x||+||y||)^2 \end {align*}

i.e \(||x+y||^2\leq (||x||+||y||)^2\)

OR \(||x+y||\leq ||x||+||y||\hspace {0.3cm}\) Proving the triangle inequality. □

Proof. Let \(x,y\in V\) and \(\alpha , \beta \in \mathbb {C}\). Then \begin {align*} \Lambda _z(\alpha x+\beta y) &=(\alpha x+\beta y,z)\\ &=(\alpha x,z)+(\beta y,z)\\ &=\alpha (x,z)+\beta (y,z)\\ &=\alpha \Lambda _z(x)+\beta \Lambda _z(y) \end {align*}

\(\therefore \hspace {0.5cm} \Lambda _z(\alpha x+\beta y)=\alpha \Lambda _z(x)+\beta \Lambda _z(y)\) which proves that \(\Lambda _z\) is linear.

Now \(|\Lambda _z(x)|=|(x,z)|\leq ||x||||z||.\hspace {0.2cm}\) This implies that \[||\Lambda _z||=\sup _{x\in V}\Bigg \{\frac {|\Lambda _z(x)|}{||x||}\Bigg \}\leq ||z||<\infty \] Which proves that \(\Lambda _z\) is bounded.
Therefore \(\Lambda _z\in V^*\). □

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