Cauchy–Schwarz tengsizligi

Cauchy–Schwarz (o'qilishi: Koshi-Shvarz) tengsizligi (ba'zida Cauchy–Bunyakovsky–Schwarz (o'qilishi: Koshi-Bunyakovskiy-Shvarz) tengsizligi) matematikadagi eng muhim va keng qo'llaniladigan tengsizliklardan biri sifatida qaraladi.

Yig'indilar uchun tengsizlik Augustin-Louis Cauchy tomonidan 1821-yilda nashr etilgan. Integrallar uchun mos keladigan tengsizlik Viktor Bunyakovsky tomonidan 1859-yilda va Hermann Schwarz tomonidan 1888-yilda nashr etilgan. Schwarz integral ko'rinishdagi tengsizlikning zamonaviy isbotini keltirgan.

Tengsizlik bayoni

Cauchy-Schwarz tengsizligi barcha ichki ko'paytma aniqlangan fazodagi barcha $\mathbf {u}$ va $\mathbf {v}$ vektorlar uchunAndoza:NumBlktengsizlikning o'rinli ekanligini ta'kidlaydi, bu yerda $\langle \cdot ,\cdot \rangle$ ichki ko'paytma hisoblanadi. Ichki ko'paytmalarga misollar haqiqiy va kompleks nuqtali ko'paytmalarni o'z ichiga oladi. Ichki ko'paytma mavzusidagi misollarga qarang. Har bir ichki ko'paytma kanonik yoki keltirilgan norma deb ataladigan normani keltirib chiqaradi, bu yerda $\mathbf {u}$ vektorning vektor normasi quyidagicha belgilanadi va aniqlanadi:

$$\|\mathbf {u} \|:={\sqrt {\langle \mathbf {u} ,\mathbf {u} \rangle }}.$$

Bu norma va ichki ko'paytma $\|\mathbf {u} \|^{2}=\langle \mathbf {u} ,\mathbf {u} \rangle$ aniqlovchi shart bilan o'zaro bog'langan bo'lib, bu yerda $\langle \mathbf {u} ,\mathbf {u} \rangle$ har doim nomanfiy haqiqiy son bo'ladi (hattoki ichki ko'paytma kompleks qiymatli bo'lsa ham). Yuqoridagi tengsizlikning har ikki tomonining kvadrat ildizini olib, Cauchy-Schwarz tengsizligini uning koʻproq tanish bo'lgan koʻrinishida yozish mumkin:Andoza:NumBlkBundan tashqari, tengsizlikning ikkala tomoni faqat va faqat $\mathbf {u}$ va $\mathbf {v}$ lar bir-biriga chiziqli bog'liq bo'lsagina bir-biriga teng bo'ladi.

Dalillar

Ixtiyoriy vektor fazolar uchun

Isbot 1

Proof of Andoza:EquationRef

Let $V=\|\mathbf {v} \|^{2}$ and $c=\langle \mathbf {u} ,\mathbf {v} \rangle$ so that ${\overline {c}}c=|c|^{2}=|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}$ and ${\overline {c}}={\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}=\langle \mathbf {v} ,\mathbf {u} \rangle .$ Then $${\begin{alignedat}{4}\left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}&=\|V\mathbf {u} -c\mathbf {v} \|^{2}=\langle V\mathbf {u} -c\mathbf {v} ,V\mathbf {u} -c\mathbf {v} \rangle &&~{\text{ By definition of the norm }}\\&=\langle V\mathbf {u} ,V\mathbf {u} \rangle -\langle V\mathbf {u} ,c\mathbf {v} \rangle -\langle c\mathbf {v} ,V\mathbf {u} \rangle +\langle c\mathbf {v} ,c\mathbf {v} \rangle &&~{\text{ Expand }}\\&=V^{2}\langle \mathbf {u} ,\mathbf {u} \rangle -V{\overline {c}}\langle \mathbf {u} ,\mathbf {v} \rangle -cV\langle \mathbf {v} ,\mathbf {u} \rangle +c{\overline {c}}\langle \mathbf {v} ,\mathbf {v} \rangle &&~{\text{ Pull out scalars (note that }}V:=\|\mathbf {v} \|^{2}{\text{ is real) }}\\&=V^{2}\|\mathbf {u} \|^{2}~~-V{\overline {c}}c~~~~~~~~~-cV{\overline {c}}~~~~~~~~~+c{\overline {c}}V&&~{\text{ Use definitions of }}c:=\langle \mathbf {u} ,\mathbf {v} \rangle {\text{ and }}V\\&=V^{2}\|\mathbf {u} \|^{2}~~-V{\overline {c}}c~=~V\left[V\|\mathbf {u} \|^{2}-{\overline {c}}c\right]&&~{\text{ Simplify }}\\&=\|\mathbf {v} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-\left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|^{2}\right]&&~{\text{ Rewrite in terms of }}\mathbf {u} {\text{ and }}\mathbf {v} .\blacksquare \\\end{alignedat}}$$ This expansion does not require $\mathbf {v}$ to be non-zero; however, $\mathbf {v}$ must be non-zero in order to divide both sides by $\|\mathbf {v} \|^{2}$ and to deduce the Cauchy-Schwarz inequality from it. Swapping $\mathbf {u}$ and $\mathbf {v}$ gives rise to: $$\left\|\|\mathbf {u} \|^{2}\mathbf {v} -{\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\mathbf {u} \right\|^{2}=\|\mathbf {u} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\right]$$ and thus $${\begin{alignedat}{4}\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-\left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|^{2}\right]&=\|\mathbf {u} \|^{2}\left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}\\&=\|\mathbf {v} \|^{2}\left\|\|\mathbf {u} \|^{2}\mathbf {v} -{\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\mathbf {u} \right\|^{2}.\\\end{alignedat}}$$

Manbalar

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