Cauchy-Schwarz Inequality

Cauchy-Schwarz Inequality

Keyword : Cauchy–Schwarz inequality Minkowski inequality Young’s inequality Hölder’s inequality

设$f(x)$在区间$[0,1]$上连续，且$1\le f(x)\le 3$ .证明:
$$1\le {\int }_0^{1}f(x)\mathrm{d}x{\int }_0^1\frac{1}{f(x)}\mathrm{d}x\le \frac{4}{3}$$
Proof: According to Cauchy-Schwarz inequality
$${\int }_0^{1}f(x)\mathrm{d}x{\int }_0^1\frac{1}{f(x)}\mathrm{d}x\ge ({\int }_0^1\mathrm{d}x{)}^2=1$$
because $1\le f(x)\le 3$, then
$$\frac{(f(x)-1)(f(x)-3)}{f(x)}\le 0$$
Opening the brakests ,we reduce
$$f(x)+\frac{3}{f(x)}\le 4$$
and also because
$${\int }_0^{1}f(x)\mathrm{d}x{\int }_0^1\frac{3}{f(x)}\mathrm{d}x\le \frac{1}{4}({\int }_0^{1}f(x)\mathrm{d}x+{\int }_0^1\frac{3}{f(x)}\mathrm{d}x{)}^2$$
then we reduce
$${\int }_0^{1}f(x)\mathrm{d}x{\int }_0^1\frac{1}{f(x)}\mathrm{d}x\le \frac{4}{3}$$
so
$$1\le {\int }_0^{1}f(x)\mathrm{d}x{\int }_0^1\frac{1}{f(x)}\mathrm{d}x\le \frac{4}{3}$$

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Cauchy-Schwarz Inequality

https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

The inequality for sums was published by Augustin-Louis Cauchy(1821), while the corresponding inequality for integrals was first proved by Viktor Bun-yakovsky (1859) . Later the integral inequality was rediscovered by Hermann Ama-ndus Schwarz (1888).

In Euclidean space $R^n$ with the standard inner product ,the Cauchy–Schwarz inequality is
$$\sum _{k=1}^n(a_k{b}_k{)}^2\le (\sum _{k=1}^n{a}_k^2\left)\right(\sum _{k=1}^n{b}_k^2)$$
The Cauchy–Schwarz inequality can be proved using only ideas from elemen-tary algebra in this case. Consider the following quadratic polynomial in $t$ ,then
$$H(t)=\sum _{k=1}^n(a_{k}t+b_k{)}^2=(a_{1}t+b_1{)}^2+(a_{2}t+b_2{)}^2+\cdots +(a_{n}t+b_n{)}^2\ge 0$$
which is
$$H(t)=\left(\sum _{k=1}^n{a}_k^2\right)t^2+2\left(\sum _{k=1}^n{a}_k{b}_k\right)t+\sum _{k=1}^n{b}_k^2$$
Discriminant $\Delta$ is Less than or equal to $0$,then organized
$$\sum _{k=1}^n(a_k{b}_k{)}^2\le (\sum _{k=1}^n{a}_k^2\left)\right(\sum _{k=1}^n{b}_k^2)$$

还有积分版the intergral inequality

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Continuous version

If $f$ and $g$ was intergral in $[a,b]$ ,then
$$({\int }_a^{b}f(x)g(x)\mathrm{d}x{)}^2\le {\int }_a^b{f}^2(x)\mathrm{d}x{\int }_a^b{g}^2(x)\mathrm{d}x$$
Proof 1: for all $\lambda \in R$ ,$(f(x)+\lambda g(x){)}^2\ge 0$ ,reduce
$${\int }_a^b\left(f\right(x)+\lambda g(x){)}^2\mathrm{d}x\ge 0$$
Opening the brakests ,we reduce
$${\lambda }^2{\int }_a^b{g}^2(x)\mathrm{d}x+2\lambda {\int }_a^{b}f(x)g(x)\mathrm{d}x+{\int }_a^b{f}^2(x)\mathrm{d}x\ge 0$$
Discriminan $\Delta$ is Less than or equal to $0$ ,then organized
$$({\int }_a^{b}f(x)g(x)\mathrm{d}x{)}^2\le {\int }_a^b{f}^2(x)\mathrm{d}x{\int }_a^b{g}^2(x)\mathrm{d}x$$

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Proof 2: we can make
$$F(x)={\int }_a^x{f}^2(t)\mathrm{d}t{\int }_a^x{g}^2(t)\mathrm{d}t-({\int }_a^{x}f(t)g(t)\mathrm{d}t{)}^2$$
and $F(a)=0$ ,derive it
$$\begin{array}{rl}{F}^{{}^{\prime }}(x)& =g^2(x){\int }_a^x{f}^2(t)\mathrm{d}t+f^2(x){\int }_a^x{g}^2(t)\mathrm{d}t-2f(x)g(x){\int }_a^{x}f(t)g(t)\mathrm{d}t\\ & ={\int }_a^x(f(x)g(t)-g(x)f(t){)}^2\mathrm{d}t\ge 0\end{array}$$
so $F(a)\ge F(a)=0$ ,which is
$$({\int }_a^{b}f(x)g(x)\mathrm{d}x{)}^2\le {\int }_a^b{f}^2(x)\mathrm{d}x{\int }_a^b{g}^2(x)\mathrm{d}x$$

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Inference 1: Minkowski inequality

https://en.wikipedia.org/wiki/Minkowski_inequality

In mathematical analysis ,the Minkowski inequality establishes that the Lp spa-ces are normed vector spaces. Let S be a measure space, let $1\le \rho <\infty$ and let $f$and $g$ be elements of $Lp(s)$. Then $f+g$ is in $Lp(s)$ .

If $f$ and $g$ was intergral in $[a,b]$ ,then
$$\sqrt{{\int }_a^b\left(f\right(x)+g(x){)}^2\mathrm{d}x}\le \sqrt{{\int }_a^b{f}^2(x)\mathrm{d}x}+\sqrt{{\int }_a^b{g}^2(x)\mathrm{d}x}$$
Proof : by Cauchy-Schwarz inequality to get
$${\int }_a^b\left(f\right(x)+g(x){)}^2\mathrm{d}x\le {\int }_a^b{f}^2(x)\mathrm{d}x+{\int }_a^b{g}^2(x)\mathrm{d}x+2\sqrt{{\int }_a^b{f}^2(x)\mathrm{d}x{\int }_a^b{g}^2(x)\mathrm{d}x}$$
then we reduce it .

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Inference 2: Hölder’s inequality

https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

Discrete form

Sippose $a_k,b_k>0(k=1,2,\cdots ,n)$ ,$p,q\ge 1$ and $\frac{1}{p}+\frac{1}{q}=1$ ,then
$$\sum _{k=1}^n{a}_k{b}_k\le \left(\sum _{k=1}^n{a}_k^p{)}^{\frac{1}{p}}\right(\sum _{k=1}^n{b}_k^q{)}^{\frac{1}{q}}$$
The inequality hold if $a_k$ and only if is proportional to $b_k$.

Proof : use lemma Young's inequality
$$\begin{array}{rl}\frac{{\sum }_{k=1}^n{a}_k{b}_k}{\left({\sum }_{k=1}^n{a}_k^p{)}^{\frac{1}{p}}\right({\sum }_{k=1}^n{b}_k^q{)}^{\frac{1}{q}}}& =\sum _{k=1}^n\left(\frac{a_k^p}{{\sum }_{k=1}^n{a}_k^p}{)}^{\frac{1}{p}}\right(\frac{b_k^q}{{\sum }_{k=1}^n{b}_k^q}{)}^{\frac{1}{q}}\\ & \le \sum _{k=1}^n\left[\frac{1}{p}\right(\frac{a_k^p}{{\sum }_{k=1}^n{a}_k^p})+(\frac{1}{q}\frac{b_k^q}{{\sum }_{k=1}^n{b}_k^q}\left)\right]\\ & =\frac{1}{p}+\frac{1}{q}\end{array}$$
In mathematical analysis ,Hölder’s inequality ,named after Otto Hölde ,is a fundamental inequality between integrals and an indispensable tool for the study of $Lp$ spaces .

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Continous version

If $f$ and $g$ was intergraled in $[a,b]$ ,and $f(X)\ge 0$ ,$g(x)\ge 0$ ,then
$${\int }_a^{b}f(x)g(x)\mathrm{d}x\le \left({\int }_a^b{f}^p\right(x\left)\mathrm{d}x{)}^{1/p}\right({\int }_a^b{g}^q\left(x\right)\mathrm{d}x{)}^{1/q}$$
among $p,q\ge 1$ ,and $\frac{1}{p}+\frac{1}{q}=1$ .

先证一个重要引理Young’s inequality

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lemma : Young’s inequality

https://en.wikipedia.org/wiki/Young%27s_inequality_for_products

In mathematics, Young’s inequality is a mathematical inequality about the pro-duct of two numbers .The inequality is named after William Henry Young and should not be confused with Young's convolution inequation .

Young's inequality can be used to prove Hölder's inequality. It is also wi-dely used to estimate the norm of nonlinear terms in PDE theory ,since it allows one to estimate a product of two terms by a sum of the same terms raised to a power and scaled .

In its standard form, the inequality states that if $a$ and $b$ are nonnegetive real nu-mber and $p$ and $q$ are real numbers greater than $1$ such that $\frac{1}{p}+\frac{1}{q}=1$, then
$$ab\le \frac{a^p}{p}+\frac{b^q}{q}$$
The inequality hold if and only if $a_p=b^q$ .This form of Young's inequality can be proved by Jensen’s inequality and can be used to prove Hölder’s inequality.

Proof : Because $f(x)=e^x$ is an convex function ,that use Jensen's inequality
$$f(\sum _{i=1}^n{\lambda }_i{x}_i)\le \sum _{i=1}^n{\lambda }_{i}f(x_i)$$
among ${\lambda }_i>0$ ,and ${\sum }_i{\lambda }_i=1$

i) while ab≠0
$$\begin{array}{rl}ab& =e^{\ln a}{e}^{\ln b}=\exp \left(\frac{1}{p}\ln a^p\right)\exp \left(\frac{1}{q}\ln b^q\right)\\ & =\exp (\frac{1}{p}\ln a^p+\frac{1}{q}\ln b^q)\le \frac{1}{p}{e}^{\ln a^p}+\frac{1}{q}{e}^{\ln b^q}\\ & =\frac{a^p}{p}+\frac{b^q}{q}\end{array}$$
ii) whlie ab=0

Obviously there is $ab\le \frac{a^p}{p}+\frac{b^q}{q}$

Combining the foregoing i) and ii) .

Now we proof the Integral inequality using Young’s inequality

Proof : we make
$$m=\frac{f^p(x)}{{\int }_a^b{f}^p(x)\mathrm{d}x},n=\frac{g^q(x)}{{\int }_a^b{g}^q(x)\mathrm{d}x}$$
by Young’s inequality ,then we reduce
$$\frac{f^p(x)}{{\int }_a^b{f}^p(x)\mathrm{d}x}\frac{g^q(x)}{{\int }_a^b{g}^q(x)\mathrm{d}x}\le \frac{1}{p}\frac{f^p(x)}{{\int }_a^b{f}^p(x)\mathrm{d}x}+\frac{1}{q}\frac{g^q(x)}{{\int }_a^b{g}^q(x)\mathrm{d}x}$$
now integral for x left and right sides
$${\int }_a^{b}f(x)g(x)\mathrm{d}x\le \left({\int }_a^b{f}^p\right(x\left)\mathrm{d}x{)}^{1/p}\right({\int }_a^b{g}^q\left(x\right)\mathrm{d}x{)}^{1/q}$$
The numbers p and q above are said to be Hölder conjugates of each other. The special case p=q=2 gives a form of the Cauchy-Schwarz inequality.

To be continued … …