Cauchy-Schwarz inequality for inner product spaces

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

Statement

Let [ilmath]\langle\cdot,\cdot\rangle:X\times X\rightarrow\mathbb{K} [/ilmath] be an inner product so [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] is an inner product space, then[1]:

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

Proof

  • Let [ilmath]x,y\in X[/ilmath] be given
    • We have two cases now, [ilmath]y\eq 0[/ilmath] and [ilmath]y\neq 0[/ilmath]
      1. [ilmath]y\eq 0[/ilmath] case
        • We now have two more cases, [ilmath]x\eq 0[/ilmath] and [ilmath]x\neq 0[/ilmath]
          1. [ilmath]x\eq 0[/ilmath]
            • Now [ilmath]\ip{x,y}\eq 0[/ilmath], [ilmath]\ip{x,x}\eq 0[/ilmath] and [ilmath]\ip{y,y}\eq 0[/ilmath] so:
              • [ilmath]\vert\ip{x,y}\vert\eq 0\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\times 0\eq 0[/ilmath] is the case
                • [ilmath]0\le 0[/ilmath] obviously holds. We're done in this case
          2. [ilmath]x\neq 0[/ilmath]
            • Now [ilmath]\ip{x,y}:\eq\ip{x,0}\eq\overline{\ip{0,x} }\eq \overline{0\ip{z,x} } [/ilmath] for any [ilmath]z\in X[/ilmath]
              [ilmath]\eq 0\ip{x,z}\eq 0[/ilmath]
              • Now [ilmath]\vert\ip{x,y}\vert\eq 0[/ilmath] and [ilmath]\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\sqrt{\ip{x,x} }\eq 0 [/ilmath] - as [ilmath]y\eq 0[/ilmath] [ilmath]\ip{y,y}\eq 0[/ilmath]
                • [ilmath]\vert\ip{x,y}\vert\eq 0\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0[/ilmath] gives us [ilmath]0\le 0[/ilmath] which is obviously true (we have equality so we have [ilmath]\le[/ilmath])
      2. [ilmath]y\neq 0[/ilmath] case
        • Consider [ilmath]\lambda\in\mathbb{K} [/ilmath] then:
          • [ilmath]0\le\ip{x-\lambda y,x-\lambda y} [/ilmath] (as for an inner product [ilmath]\forall z\in X[\ip{z,z}\in\mathbb{R}\wedge\ip{z,z}\ge 0][/ilmath]
            • Then [ilmath]\ip{x-\lambda y,x-\lambda y}\eq \ip{x,x-\lambda y}-\lambda\ip{y,x-\lambda y} [/ilmath]
              [ilmath]\eq\overline{\ip{x-\lambda y,x} }-\lambda\big(\overline{\ip{x-\lambda y,y} }\big)[/ilmath]
              [ilmath]\eq\overline{\ip{x,x}-\lambda\ip{y,x} }-\lambda\big(\overline{\ip{x,y} - \lambda\ip{y,y} }\big) [/ilmath]
              [ilmath]\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \lambda\overline{\lambda}\ip{y,y} [/ilmath]
              [ilmath]\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \vert\lambda\vert^2\ip{y,y} [/ilmath][Note 1]
              [ilmath]\eq\ip{x,x}-\big(\overline{\lambda\ip{y,x} }+\lambda\ip{y,x}\big)+\vert\lambda\vert^2\ip{y,y} [/ilmath][Note 2]
              [ilmath]\eq\ip{x,x}-2\text{Re}(\lambda\ip{y,x})+\vert\lambda\vert^2\ip{y,y} [/ilmath]
              [ilmath]\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} } [/ilmath] - by completing the square
              TODO: I think
               (without considering the [ilmath]-2\text{Re}...[/ilmath])
              • So [ilmath]\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} } [/ilmath]
                • Note that [ilmath]\forall z\in X[\ip{z,z}\ge 0[/ilmath] holds, so:
                  • [ilmath]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} } [/ilmath]
                    • Or just [ilmath]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} } [/ilmath]
                    • Define [ilmath]\theta\in[0,2\pi) [/ilmath] such that [ilmath]\ip{y,x}\eq\vert\ip{y,x}\vert e^{j\theta} [/ilmath] (a form of complex number)
                      • Define [math]\lambda:\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta} [/math] and note that [math]\vert\lambda\vert\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } [/math]
                        • this is fine to do as [ilmath]y\neq 0[/ilmath] so [ilmath]\ip{y,y}>0[/ilmath] so [ilmath]\sqrt{\ip{y,y} }>0[/ilmath] and the division in the fraction is defined
                        • we substitute this into our expression to obtain:
                          • [math]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} } [/math]
                            [math]\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} } [/math]
                            [math]\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} [/math] - but [ilmath]e^0\eq 1[/ilmath][Note 3] so there is no imaginary component of the thing in the [ilmath]\text{Re} [/ilmath]
                            [math]\implies 0\le 2\ip{x,x}-2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } [/math]
                            [math]\implies 2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le 2\ip{x,x} [/math] [math]\implies\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le \ip{x,x} [/math]
                            [ilmath]\implies\vert{\ip{y,x} }\vert\sqrt{\ip{x,x} }\le \ip{x,x}\sqrt{\ip{y,y} } [/ilmath]
                            • [ilmath]\implies\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } [/ilmath] - almost as required
                              • Note that [ilmath]\vert\ip{x,y}\vert\eq\vert\overline{\ip{x,y} }\vert\eq\vert\ip{y,x}\vert [/ilmath][Note 4]
                                • So [ilmath]\vert\ip{x,y}\vert\eq\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } [/ilmath]
                                  • Thus [ilmath]\vert\ip{x,y}\vert\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} } [/ilmath] - as required
  • Since [ilmath]x,y\in X[/ilmath] were arbitrary we have shown the claim holds for all.

This completes the proof.

References

  1. ↑ Warwick 2014 Lecture Notes - Functional Analysis - Richard Sharp

Notes

  1. ↑ Let [ilmath]a+bj\in\mathbb{C} [/ilmath], then:
    • [ilmath](a+bj)\overline{(a+bj)}\eq(a+bj)(a-bj) [/ilmath]
      [ilmath]\eq a^2-j^2b^j +j(ab-ab)[/ilmath]
      [ilmath]\eq a^2+b^2\ (\ +0j\ )[/ilmath]
      [ilmath]\eq\vert a+bj\vert^2[/ilmath] - as required
    • If [ilmath]a+bj[/ilmath] is the complex representation of [ilmath]\lambda[/ilmath] then we see [ilmath]\lambda\overline{\lambda}\eq\vert\lambda\vert^2[/ilmath]
  2. ↑ Let [ilmath]a+bj\in\mathbb{C} [/ilmath], then:
    • [ilmath](a+bj)+\overline{(a+bj)}\eq (a+bj)+(a-bj)\eq 2a[/ilmath]
    • Thus if [ilmath]a+bj[/ilmath] is the complex representation of [ilmath]\lambda\ip{y,x} [/ilmath] then
      • [ilmath]\overline{\lambda\ip{y,x} }+\lambda\ip{y,x}\eq 2\text{Re}(\lambda\ip{y,x})[/ilmath]
  3. ↑ From [ilmath]e^{-j\theta}e^{j\theta}\eq e^{j\theta-j\theta}\eq e^0[/ilmath]
  4. ↑ As for [ilmath]a+bj\in\mathbb{C} [/ilmath] we see:
    • [ilmath]\vert a+bj\vert:\eq \sqrt{a^2+b^2} [/ilmath] and
    • [ilmath]\vert \overline{a+bj}\vert\eq\vert a-bj\vert\eq\vert a+(-b)j\vert:\eq \sqrt{a^2+(-b)^2}\eq\sqrt{a^2+b^2}\eq\vert a+bj\vert[/ilmath] as required
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