(0)

\(\boldsymbol{y}=\boldsymbol{0}\)のとき、左辺は\(\left|\left\langle \boldsymbol{x},\boldsymbol{0}\right\rangle \right|=\left|0\right|=0\)となり、右辺は\(\left\Vert \boldsymbol{x}\right\Vert \left\Vert \boldsymbol{0}\right\Vert =\left\Vert \boldsymbol{x}\right\Vert 0=0\)となるので成り立つ。
\(\boldsymbol{y}\ne\boldsymbol{0}\)のときは、
\begin{align*} 0 & \leq\left\langle \boldsymbol{x}-\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\boldsymbol{y},\boldsymbol{x}-\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\boldsymbol{y}\right\rangle \left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \\ & =\left(\left\langle \boldsymbol{x},\boldsymbol{x}\right\rangle -\frac{\overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }}{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle -\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\left\langle \boldsymbol{y},\boldsymbol{x}\right\rangle +\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }}{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \right)\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \\ & =\left(\left\langle \boldsymbol{x},\boldsymbol{x}\right\rangle -\frac{\overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }}{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle -\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }+\frac{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle }}{\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle }\right)\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \\ & =\left(\left\Vert \boldsymbol{x}\right\Vert ^{2}-\frac{\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}}{\left\Vert \boldsymbol{y}\right\Vert ^{2}}\right)\left\Vert \boldsymbol{y}\right\Vert ^{2}\\ & =\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}-\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2} \end{align*} となるので、移項すると、
\[ \left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\leq\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2} \] となり、\(0\leq\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|,0\leq\left\Vert \boldsymbol{x}\right\Vert ,0<\left\Vert \boldsymbol{y}\right\Vert \)なので、
\[ \left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|\leq\left\Vert \boldsymbol{x}\right\Vert \left\Vert \boldsymbol{y}\right\Vert \] となる。
これらより、\(\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|\leq\left\Vert \boldsymbol{x}\right\Vert \left\Vert \boldsymbol{y}\right\Vert \)は\(\boldsymbol{y}=\boldsymbol{0}\lor\boldsymbol{y}\ne\boldsymbol{0}\)で成り立っているので題意は成り立つ。

(0)-2

\(t\)を実数とする。
このとき、
\begin{align*} 0 & \leq\left\Vert \boldsymbol{x}+t\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \boldsymbol{y}\right\Vert ^{2}\\ & =\left\langle \boldsymbol{x}+t\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \boldsymbol{y},\boldsymbol{x}+t\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \boldsymbol{y}\right\rangle \\ & =\left\langle \boldsymbol{x},\boldsymbol{x}\right\rangle +2\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}t+\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle t^{2}\\ & =\left\Vert \boldsymbol{x}\right\Vert ^{2}+2\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}t+\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}t^{2} \end{align*} となり、\(t\)について2次の判別式\(\frac{D}{4}\)が非正でなければいけないので、
\begin{align*} 0 & \geq\frac{D}{4}\\ & =\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{4}-\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\\ & =\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\left(\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}-\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}\right) \end{align*} となるので、
\[ \left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|\leq\left\Vert \boldsymbol{x}\right\Vert \left\Vert \boldsymbol{y}\right\Vert \] となる。
従って題意は成り立つ。

(0)-3

\(G\)をグラム行列とすると、
\begin{align*} \left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}-\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2} & =\det\left(\begin{array}{cc} \left\Vert \boldsymbol{x}\right\Vert ^{2} & \left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \\ \overline{\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle } & \left\Vert \boldsymbol{y}\right\Vert ^{2} \end{array}\right)\\ & =\det\left(\begin{array}{cc} \left\langle \boldsymbol{x},\boldsymbol{x}\right\rangle & \left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \\ \left\langle \boldsymbol{y},\boldsymbol{x}\right\rangle & \left\langle \boldsymbol{y},\boldsymbol{y}\right\rangle \end{array}\right)\\ & =\det\left(G\left(\boldsymbol{x},\boldsymbol{y}\right)\right)\\ & \geq0\cmt{\because G\left(A\right)\geq0} \end{align*} となるので、\(\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2}\leq\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2}\)となり、\(\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|\leq\left\Vert \boldsymbol{x}\right\Vert \left\Vert \boldsymbol{y}\right\Vert \)が成り立つ。

(0)-3

内積がエルミート内積\(\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle =\sum_{k=1}^{n}x_{k}\overline{y}_{k}\)のときで証明をする。
\begin{align*} \sum_{j=1}^{n}\left|x_{j}\right|^{2}\sum_{k=1}^{n}\left|y_{k}\right|^{2} & =\sum_{j=1}^{n}\sum_{k=1}^{n}\left|x_{j}\right|^{2}\left|y_{k}\right|^{2}\\ & =\sum_{j=1}^{n}\sum_{k=1}^{n}\left|x_{j}y_{k}\right|^{2}\\ & =\sum_{k=1}^{n}\left|x_{k}y_{k}\right|^{2}+\sum_{j\ne k}\left|x_{j}y_{k}\right|^{2}\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}-\sum_{j\ne k}\left|x_{j}y_{j}x_{k}y_{k}\right|+\sum_{j\ne k}\left|x_{j}y_{k}\right|^{2}\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}+\sum_{j\ne k}\left(\left|x_{j}y_{k}\right|^{2}-\left|x_{j}y_{j}x_{k}y_{k}\right|\right)\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}+\sum_{j\ne k}\left(\left|x_{j}y_{k}x_{j}y_{k}\right|-\left|x_{j}y_{j}x_{k}y_{k}\right|\right)\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}+\frac{1}{2}\sum_{j\ne k}\left(\left|x_{j}y_{k}x_{j}y_{k}\right|-\left|x_{j}y_{j}x_{k}y_{k}\right|+\left|x_{k}y_{j}x_{k}y_{j}\right|-\left|x_{k}y_{k}x_{j}y_{j}\right|\right)\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}+\frac{1}{2}\sum_{j\ne k}\left(\left|x_{j}y_{k}\right|-\left|x_{j}y_{k}\right|\right)^{2} \end{align*} より、
\begin{align*} \left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2} & =\sum_{j=1}^{n}\left|x_{j}\right|^{2}\sum_{k=1}^{n}\left|y_{k}\right|^{2}\\ & \geq\sum_{j=1}^{n}\left|x_{j}\right|^{2}\sum_{k=1}^{n}\left|y_{k}\right|^{2}-\frac{1}{2}\sum_{j\ne k}\left(\left|x_{j}y_{k}\right|-\left|x_{j}y_{k}\right|\right)^{2}\\ & =\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}\cmt{\because\sum_{j=1}^{n}\left|x_{j}\right|^{2}\sum_{k=1}^{n}\left|y_{k}\right|^{2}=\left|\sum_{k=1}^{n}x_{k}y_{k}\right|^{2}+\frac{1}{2}\sum_{j\ne k}\left(\left|x_{j}y_{k}\right|-\left|x_{j}y_{k}\right|\right)^{2}}\\ & =\left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|^{2} \end{align*} となる。
従って、
\[ \left|\left\langle \boldsymbol{x},\boldsymbol{y}\right\rangle \right|\leq\left\Vert \boldsymbol{x}\right\Vert ^{2}\left\Vert \boldsymbol{y}\right\Vert ^{2} \] となるのでエルミート内積のときは題意が成り立つ。