Cauchy-Schwarz Inequality

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

设f(x)f(x)f(x)在区间[0,1][0,1][0,1]上连续，且1≤f(x)≤31\leq f(x)\leq 31≤f(x)≤3 .证明:
1≤∫01f(x)dx∫011f(x)dx≤431\leq\int_{0}^{1}f(x)\mathrm{d}x\int_{0}^{1}\frac{1}{f(x)}\mathrm{d}x\leq\frac{4}{3} 1≤∫01f(x)dx∫01f(x)1dx≤34
Proof: According to Cauchy-Schwarz inequality
∫01f(x)dx∫011f(x)dx≥(∫01dx)2=1\int_{0}^{1}f(x)\mathrm{d}x\int_{0}^{1}\frac{1}{f(x)}\mathrm{d}x\geq\big(\int_{0}^{1}\mathrm{d}x\big)^{2}=1 ∫01f(x)dx∫01f(x)1dx≥(∫01dx)2=1
because 1≤f(x)≤31\leq f(x)\leq31≤f(x)≤3, then
(f(x)−1)(f(x)−3)f(x)≤0\frac{\big(f(x)-1\big)\big(f(x)-3\big)}{f(x)}\leq 0 f(x)(f(x)−1)(f(x)−3)≤0
Opening the brakests ,we reduce
f(x)+3f(x)≤4f(x)+\frac{3}{f(x)}\leq4 f(x)+f(x)3≤4
and also because
∫01f(x)dx∫013f(x)dx≤14(∫01f(x)dx+∫013f(x)dx)2\int_{0}^{1}f(x)\mathrm{d}x\int_{0}^{1}\frac{3}{f(x)}\mathrm{d}x\leq \frac{1}{4}\bigg(\int_{0}^{1}f(x)\mathrm{d}x+\int_{0}^{1}\frac{3}{f(x)}\mathrm{d}x\bigg)^{2} ∫01f(x)dx∫01f(x)3dx≤41(∫01f(x)dx+∫01f(x)3dx)2
then we reduce
∫01f(x)dx∫011f(x)dx≤43\int_{0}^{1}f(x)\mathrm{d}x\int_{0}^{1}\frac{1}{f(x)}\mathrm{d}x\leq\frac{4}{3} ∫01f(x)dx∫01f(x)1dx≤34
so
1≤∫01f(x)dx∫011f(x)dx≤431\leq \int_{0}^{1}f(x)\mathrm{d}x\int_{0}^{1}\frac{1}{f(x)}\mathrm{d}x\leq \frac{4}{3} 1≤∫01f(x)dx∫01f(x)1dx≤34

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 RnR^{n}Rn with the standard inner product ,the Cauchy–Schwarz inequality is
∑k=1n(akbk)2≤(∑k=1nak2)(∑k=1nbk2)\sum_{k=1}^{n}\big(a_{k}b_{k}\big)^{2}\leq \big(\sum_{k=1}^{n}a_{k}^{2}\big)\big(\sum_{k=1}^{n}b_{k}^{2}\big) k=1∑n(akbk)2≤(k=1∑nak2)(k=1∑nbk2)
The Cauchy–Schwarz inequality can be proved using only ideas from elemen-tary algebra in this case. Consider the following quadratic polynomial in ttt ,then
H(t)=∑k=1n(akt+bk)2=(a1t+b1)2+(a2t+b2)2+⋯+(ant+bn)2≥0H(t)=\sum_{k=1}^{n}\big(a_{k}t+b_{k}\big)^{2}=(a_{1}t+b_{1})^{2}+(a_{2}t+b_{2})^{2}+\cdots+(a_{n}t+b_{n})^{2}\geq0 H(t)=k=1∑n(akt+bk)2=(a1t+b1)2+(a2t+b2)2+⋯+(ant+bn)2≥0
which is
H(t)=(∑k=1nak2)t2+2(∑k=1nakbk)t+∑k=1nbk2H(t)=\big(\sum_{k=1}^{n}a_{k}^{2}\big)t^{2}+2\big(\sum_{k=1}^{n}a_{k}b_{k}\big)t+\sum_{k=1}^{n}b_{k}^{2} H(t)=(k=1∑nak2)t2+2(k=1∑nakbk)t+k=1∑nbk2
Discriminant Δ\DeltaΔ is Less than or equal to 000,then organized
∑k=1n(akbk)2≤(∑k=1nak2)(∑k=1nbk2)\sum_{k=1}^{n}\big(a_{k}b_{k}\big)^{2}\leq \big(\sum_{k=1}^{n}a_{k}^{2}\big)\big(\sum_{k=1}^{n}b_{k}^{2}\big) k=1∑n(akbk)2≤(k=1∑nak2)(k=1∑nbk2)

还有积分版the intergral inequality

Continuous version

If fff and ggg was intergral in [a,b][a,b][a,b] ,then
(∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx\bigg(\int_{a}^{b}f(x)g(x)\mathrm{d}x\bigg)^{2}\leq\int_{a}^{b}f^{2}(x)\mathrm{d}x\int_{a}^{b}g^{2}(x)\mathrm{d}x (∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx
Proof 1: for all λ∈R\lambda \in Rλ∈R ,(f(x)+λg(x))2≥0\big(f(x) + \lambda g(x)\big)^{2}\geq0(f(x)+λg(x))2≥0 ,reduce
∫ab(f(x)+λg(x))2dx≥0\int_{a}^{b}\bigg(f\big(x\big)+\lambda g\big(x\big)\bigg)^{2}\mathrm{d}x\geq0 ∫ab(f(x)+λg(x))2dx≥0
Opening the brakests ,we reduce
λ2∫abg2(x)dx+2λ∫abf(x)g(x)dx+∫abf2(x)dx≥0\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\geq0 λ2∫abg2(x)dx+2λ∫abf(x)g(x)dx+∫abf2(x)dx≥0
Discriminan Δ\DeltaΔ is Less than or equal to 000 ,then organized
(∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx\bigg(\int_{a}^{b}f(x)g(x)\mathrm{d}x\bigg)^{2}\leq\int_{a}^{b}f^{2}(x)\mathrm{d}x\int_{a}^{b}g^{2}(x)\mathrm{d}x (∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx

Proof 2: we can make
F(x)=∫axf2(t)dt∫axg2(t)dt−(∫axf(t)g(t)dt)2F(x)=\int_{a}^{x}f^{2}(t)\mathrm{d}t\int_{a}^{x}g^{2}(t)\mathrm{d}t-\bigg(\int_{a}^{x}f(t)g(t)\mathrm{d}t\bigg)^{2} F(x)=∫axf2(t)dt∫axg2(t)dt−(∫axf(t)g(t)dt)2
and F(a)=0F(a)=0F(a)=0 ,derive it
F′(x)=g2(x)∫axf2(t)dt+f2(x)∫axg2(t)dt−2f(x)g(x)∫axf(t)g(t)dt=∫ax(f(x)g(t)−g(x)f(t))2dt≥0\begin{aligned} F^{'}(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}\bigg(f(x)g(t)-g(x)f(t)\bigg)^{2}\mathrm{d}t\geq0 \end{aligned} F′(x)=g2(x)∫axf2(t)dt+f2(x)∫axg2(t)dt−2f(x)g(x)∫axf(t)g(t)dt=∫ax(f(x)g(t)−g(x)f(t))2dt≥0
so F(a)≥F(a)=0F(a)\geq F(a)=0F(a)≥F(a)=0 ,which is
(∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx\bigg(\int_{a}^{b}f(x)g(x)\mathrm{d}x\bigg)^{2}\leq\int_{a}^{b}f^{2}(x)\mathrm{d}x\int_{a}^{b}g^{2}(x)\mathrm{d}x (∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx

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≤ρ<∞1\leq \rho <\infty1≤ρ<∞ and let fffand ggg be elements of Lp(s)Lp(s)Lp(s). Then f+gf + gf+g is in Lp(s)Lp(s)Lp(s) .

If fff and ggg was intergral in [a,b][a,b][a,b] ,then
∫ab(f(x)+g(x))2dx≤∫abf2(x)dx+∫abg2(x)dx\sqrt{\int_{a}^{b}\bigg(f\big(x\big)+g\big(x\big)\bigg)^{2}\mathrm{d}x}\leq\sqrt{\int_{a}^{b}f^{2}(x)\mathrm{d}x}+\sqrt{\int_{a}^{b}g^{2}(x)\mathrm{d}x} ∫ab(f(x)+g(x))2dx≤∫abf2(x)dx+∫abg2(x)dx
Proof : by Cauchy-Schwarz inequality to get
∫ab(f(x)+g(x))2dx≤∫abf2(x)dx+∫abg2(x)dx+2∫abf2(x)dx∫abg2(x)dx\int_{a}^{b}\bigg(f\big(x\big)+g\big(x\big)\bigg)^{2}\mathrm{d}x\leq \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} ∫ab(f(x)+g(x))2dx≤∫abf2(x)dx+∫abg2(x)dx+2∫abf2(x)dx∫abg2(x)dx
then we reduce it .

Inference 2: Hölder’s inequality

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

Discrete form

Sippose ak,bk>0(k=1,2,⋯,n)a_{k},b_{k}>0(k=1,2,\cdots,n)ak,bk>0(k=1,2,⋯,n) ,p,q≥1p,q\geq1p,q≥1 and 1p+1q=1\frac{1}{p}+\frac{1}{q}=1p1+q1=1 ,then
∑k=1nakbk≤(∑k=1nakp)1p(∑k=1nbkq)1q\sum_{k=1}^{n}a_{k}b_{k}\leq\bigg(\sum_{k=1}^{n}a_{k}^{p}\bigg)^{\frac{1}{p}}\bigg(\sum_{k=1}^{n}b_{k}^{q}\bigg)^{\frac{1}{q}} k=1∑nakbk≤(k=1∑nakp)p1(k=1∑nbkq)q1
The inequality hold if aka_{k}ak and only if is proportional to bkb_{k}bk.

Proof : use lemma Young's inequality
∑k=1nakbk(∑k=1nakp)1p(∑k=1nbkq)1q=∑k=1n(akp∑k=1nakp)1p(bkq∑k=1nbkq)1q≤∑k=1n[1p(akp∑k=1nakp)+(1qbkq∑k=1nbkq)]=1p+1q\begin{aligned} \frac{\sum_{k=1}^{n}a_{k}b_{k}}{\big(\sum_{k=1}^{n}a_{k}^{p}\big)^{\frac{1}{p}}\big( \sum_{k=1}^{n}b_{k}^{q}\big)^{\frac{1}{q}}}&=\sum_{k=1}^{n}\bigg(\frac{a_{k}^{p}}{\sum_{k=1}^{n}a_{k}^{p}} \bigg)^{\frac{1}{p}}\bigg(\frac{b_{k}^{q}}{\sum_{k=1}^{n}b_{k}^{q}} \bigg)^{\frac{1}{q}}\\ &\leq \sum_{k=1}^{n}\bigg[ \frac{1}{p}\bigg(\frac{a_{k}^{p}}{\sum_{k=1}^{n}a_{k}^{p}} \bigg)+\bigg(\frac{1}{q}\frac{b_{k}^{q}}{\sum_{k=1}^{n}b_{k}^{q}} \bigg)\bigg]\\ &=\frac{1}{p}+\frac{1}{q} \end{aligned} (∑k=1nakp)p1(∑k=1nbkq)q1∑k=1nakbk=k=1∑n(∑k=1nakpakp)p1(∑k=1nbkqbkq)q1≤k=1∑n[p1(∑k=1nakpakp)+(q1∑k=1nbkqbkq)]=p1+q1
In mathematical analysis ,Hölder’s inequality ,named after Otto Hölde ,is a fundamental inequality between integrals and an indispensable tool for the study of LpLpLp spaces .

Continous version

If fff and ggg was intergraled in [a,b][a,b][a,b] ,and f(X)≥0f(X)\geq0f(X)≥0 ,g(x)≥0g(x)\geq0g(x)≥0 ,then
∫abf(x)g(x)dx≤(∫abfp(x)dx)1/p(∫abgq(x)dx)1/q\int_{a}^{b}f(x)g(x)\mathrm{d}x\leq\bigg(\int_{a}^{b}f^{p}\big(x\big)\mathrm{d}x\bigg)^{1/p}\bigg(\int_{a}^{b}g^{q}\big(x\big)\mathrm{d}x\bigg)^{1/q} ∫abf(x)g(x)dx≤(∫abfp(x)dx)1/p(∫abgq(x)dx)1/q
among p,q≥1p,q\geq1p,q≥1 ,and 1p+1q=1\frac{1}{p}+\frac{1}{q}=1p1+q1=1 .

先证一个重要引理Young’s inequality

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 aaa and bbb are nonnegetive real nu-mber and ppp and qqq are real numbers greater than 111 such that 1p+1q=1\frac{1}{p}+ \frac{1}{q}=1p1+q1=1, then
ab≤app+bqqab\leq \frac{a^{p}}{p}+\frac{b^{q}}{q} ab≤pap+qbq
The inequality hold if and only if ap=bqa_{p}=b^{q}ap=bq .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)=exf(x)=e^{x}f(x)=ex is an convex function ,that use Jensen's inequality
f(∑i=1nλixi)≤∑i=1nλif(xi)f(\sum_{i=1}^{n}\lambda_{i}x_{i})\leq\sum_{i=1}^{n}\lambda_{i}f(x_{i}) f(i=1∑nλixi)≤i=1∑nλif(xi)
among λi>0\lambda_{i}>0λi>0 ,and ∑iλi=1\sum_{i}\lambda_{i}=1∑iλi=1

i) while ab≠0
ab=eln⁡aeln⁡b=exp⁡(1pln⁡ap)exp⁡(1qln⁡bq)=exp⁡(1pln⁡ap+1qln⁡bq)≤1peln⁡ap+1qeln⁡bq=app+bqq\begin{aligned} ab&=e^{\ln a}e^{\ln b}=\exp\big(\frac{1}{p}\ln a^{p}\big)\exp\big(\frac{1}{q}\ln b^{q}\big)\\ &=\exp\big(\frac{1}{p}\ln a^{p}+\frac{1}{q}\ln b^{q}\big)\leq\frac{1}{p}e^{\ln a^{p}}+\frac{1}{q} e^{\ln b^{q}}\\ &=\frac{a^{p}}{p}+\frac{b^{q}}{q}\end{aligned} ab=elnaelnb=exp(p1lnap)exp(q1lnbq)=exp(p1lnap+q1lnbq)≤p1elnap+q1elnbq=pap+qbq
ii) whlie ab=0

Obviously there is ab≤app+bqqab\leq \frac{a^{p}}{p}+\frac{b^{q}}{q}ab≤pap+qbq

Combining the foregoing i) and ii) .

Now we proof the Integral inequality using Young’s inequality

Proof : we make
m=fp(x)∫abfp(x)dx,n=gq(x)∫abgq(x)dxm=\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} m=∫abfp(x)dxfp(x) , n=∫abgq(x)dxgq(x)
by Young’s inequality ,then we reduce
fp(x)∫abfp(x)dxgq(x)∫abgq(x)dx≤1pfp(x)∫abfp(x)dx+1qgq(x)∫abgq(x)dx\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}\leq\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} ∫abfp(x)dxfp(x)∫abgq(x)dxgq(x)≤p1∫abfp(x)dxfp(x)+q1∫abgq(x)dxgq(x)
now integral for x left and right sides
∫abf(x)g(x)dx≤(∫abfp(x)dx)1/p(∫abgq(x)dx)1/q\int_{a}^{b}f(x)g(x)\mathrm{d}x\leq\bigg(\int_{a}^{b}f^{p}\big(x\big)\mathrm{d}x\bigg)^{1/p}\bigg(\int_{a}^{b}g^{q}\big(x\big)\mathrm{d}x\bigg)^{1/q} ∫abf(x)g(x)dx≤(∫abfp(x)dx)1/p(∫abgq(x)dx)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 … …

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