Cauchy-Schwarz inequality for inner product spaces

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Flesh out statement

$\newcommand{\ip}[2][]{\langle {#2} \rangle {#1} }$

Statement

Let $\langle\cdot,\cdot\rangle:X\times X\rightarrow\mathbb{K}$ be an inner product so $(X,\langle\cdot,\cdot\rangle)$ is an inner product space, then:

• $\forall x,y\in X\left[\vert\ip{x,y}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\right]$

Proof

• Let $x,y\in X$ be given
  • We have two cases now, $y\eq 0$ and $y\neq 0$
    1. $y\eq 0$ case
       • We now have two more cases, $x\eq 0$ and $x\neq 0$
         1. $x\eq 0$
            • Now $\ip{x,y}\eq 0$, $\ip{x,x}\eq 0$ and $\ip{y,y}\eq 0$ so:
              • $\vert\ip{x,y}\vert\eq 0\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\times 0\eq 0$ is the case
                • $0\le 0$ obviously holds. We're done in this case
         2. $x\neq 0$
            • Now $\ip{x,y}:\eq\ip{x,0}\eq\overline{\ip{0,x} }\eq \overline{0\ip{z,x} }$ for any $z\in X$

              $\eq 0\ip{x,z}\eq 0$

              • Now $\vert\ip{x,y}\vert\eq 0$ and $\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\sqrt{\ip{x,x} }\eq 0$ - as $y\eq 0$ $\ip{y,y}\eq 0$
                • $\vert\ip{x,y}\vert\eq 0\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0$ gives us $0\le 0$ which is obviously true (we have equality so we have $\le$)
    2. $y\neq 0$ case
       • Consider $\lambda\in\mathbb{K}$ then:
         • $0\le\ip{x-\lambda y,x-\lambda y}$ (as for an inner product $\forall z\in X[\ip{z,z}\in\mathbb{R}\wedge\ip{z,z}\ge 0]$
           • Then $\ip{x-\lambda y,x-\lambda y}\eq \ip{x,x-\lambda y}-\lambda\ip{y,x-\lambda y}$

             $\eq\overline{\ip{x-\lambda y,x} }-\lambda\big(\overline{\ip{x-\lambda y,y} }\big)$

             $\eq\overline{\ip{x,x}-\lambda\ip{y,x} }-\lambda\big(\overline{\ip{x,y} - \lambda\ip{y,y} }\big)$

             $\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \lambda\overline{\lambda}\ip{y,y}$

             $\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \vert\lambda\vert^2\ip{y,y}$

             $\eq\ip{x,x}-\big(\overline{\lambda\ip{y,x} }+\lambda\ip{y,x}\big)+\vert\lambda\vert^2\ip{y,y}$

             $\eq\ip{x,x}-2\text{Re}(\lambda\ip{y,x})+\vert\lambda\vert^2\ip{y,y}$

             $\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y } }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$ - by completing the square

             TODO: I think

             (without considering the $-2\text{Re}...$)

             • So $\ip{x-\lambda y,x-\lambda y}\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$
               • Note that $\forall z\in X[\ip{z,z}\ge 0$ holds, so:
                 • $0\le\ip{x-\lambda y,x-\lambda y}\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$
                   • Or just $0\le\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$
                   • Define $\theta\in[0,2\pi)$ such that $\ip{y,x}\eq\vert\ip{y,x}\vert e^{j\theta}$ (a form of complex number)
                     • Define $$\lambda:\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta}$$ and note that $$\vert\lambda\vert\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }$$
                       • this is fine to do as $y\neq 0$ so $\ip{y,y}>0$ so $\sqrt{\ip{y,y} }>0$ and the division in the fraction is defined
                       • we substitute this into our expression to obtain:
                         • $$0\le\left(\sqrt{\ip{x,x} }-\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{y,y} }\right)^2-2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta}\ip{y,x}\right)+2\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$$

                           $$\implies 0\le\underbrace{\left(\sqrt{\ip{x,x} }-\sqrt{\ip{x,x} }\right)^2}_{\eq 0}-2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta}\vert\ip{y,x}\vert e^{j\theta}\right)+2\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$$

                           $$\implies 0\le -2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\vert\ip{y,x}\vert e^0\right)+2\ip{x,x}$$ - but $e^0\eq 1$ so there is no imaginary component of the thing in the $\text{Re}$

                           $$\implies 0\le 2\ip{x,x}-2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }$$

                           $$\implies 2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le 2\ip{x,x}$$ $$\implies\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le \ip{x,x}$$

                           $\implies\vert{\ip{y,x} }\vert\sqrt{\ip{x,x} }\le \ip{x,x}\sqrt{\ip{y,y} }$

                           • $\implies\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$ - almost as required
                             • Note that $\vert\ip{x,y}\vert\eq\vert\overline{\ip{x,y} }\vert\eq\vert\ip{y,x}\vert$
                               • So $\vert\ip{x,y}\vert\eq\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$
                                 • Thus $\vert\ip{x,y}\vert\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} }$ - as required
• Since $x,y\in X$ were arbitrary we have shown the claim holds for all.

This completes the proof.