Inequality for Integrals

Inequality for Integrals | 积分不等式

1. Jensen’s Inequality | 詹森不等式

Definition | 定义

Jensen’s inequality applies to a convex function $f$ and a random variable $X$. It states that for any convex function $f$ and integrable random variable $X$:

$$f\left(\mathbb{E}[X]\right)\le \mathbb{E}[f(X)]$$

若 $f$ 是凸函数且 $X$ 是可积随机变量，则詹森不等式表明：

$$f\left(\mathbb{E}[X]\right)\le \mathbb{E}[f(X)]$$

Usage | 使用

• Convex Functions: Ensure $f$ is convex, which means $f^{\prime \prime }(x)\ge 0$ for all $x$.

• Expectation of a Random Variable: Useful in probability theory, particularly when working with expectations.

• 凸函数：确保 $f$ 是凸函数，即 $f^{\prime \prime }(x)\ge 0$ 对所有 $x$ 成立

• 随机变量的期望：在概率论中非常有用，特别是在处理期望时

Proof | 证明

1. By definition, a function $f$ is convex if for any $x_1,x_2\in \mathbb{R}$ and $\lambda \in [0,1]$:

$$f(\lambda x_1+(1-\lambda )x_2)\le \lambda f(x_1)+(1-\lambda )f(x_2)$$

根据定义，若函数 $f$ 是凸函数，则对于任意的 $x_1,x_2\in \mathbb{R}$ 和 $\lambda \in [0,1]$，有：

$$f(\lambda x_1+(1-\lambda )x_2)\le \lambda f(x_1)+(1-\lambda )f(x_2)$$

2. Take the expectation over the inequality where $X$ is a random variable, using $\lambda =\frac{1}{n}$ for discrete cases or integral form for continuous cases.

对该不等式取期望，其中 $X$ 是随机变量，在离散情形下使用 $\lambda =\frac{1}{n}$，在连续情形下使用积分形式

3. The inequality leads to:

$$f\left(\mathbb{E}[X]\right)\le \mathbb{E}[f(X)]$$

该不等式导出：

$$f\left(\mathbb{E}[X]\right)\le \mathbb{E}[f(X)]$$

2. Hölder’s Inequality | Hölder 不等式

Definition | 定义

Hölder’s inequality provides a bound on the integral (or sum) of the product of functions. If $p>1$ and $q>1$ satisfy $\frac{1}{p}+\frac{1}{q}=1$, then for integrable functions $f$ and $g$:

$$\int |f(x)g(x)|\mathrm{d}x\le {\left(\int |f(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}{\left(\int |g(x){|}^q\mathrm{d}x\right)}^{\frac{1}{q}}$$

Hölder 不等式为函数积的积分（或和）提供了一个界。若 $p>1$ 且 $q>1$ 满足 $\frac{1}{p}+\frac{1}{q}=1$，则对可积函数 $f$ 和 $g$，有：

$$\int |f(x)g(x)|\mathrm{d}x\le {\left(\int |f(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}{\left(\int |g(x){|}^q\mathrm{d}x\right)}^{\frac{1}{q}}$$

Usage | 使用

• Normed Spaces: Essential in proving the Minkowski inequality and establishing norms.

• L^p Spaces: Often used in functional analysis, particularly in $L^p$ spaces.

• 赋范空间：在证明 Minkowski 不等式和建立范数时至关重要

• $L^p$ 空间：在泛函分析中经常使用，特别是在 $L^p$ 空间中

Proof | 证明

1. Start with the convexity of the exponential function, which is a special case of Young’s inequality:

$$ab\le \frac{a^p}{p}+\frac{b^q}{q}\text{where}\frac{1}{p}+\frac{1}{q}=1$$

从指数函数的凸性开始，这是 Young 不等式的特例：

$$ab\le \frac{a^p}{p}+\frac{b^q}{q}\text{其中}\frac{1}{p}+\frac{1}{q}=1$$

2. Integrate both sides:

$$\int |f(x)g(x)|\mathrm{d}x\le \int \left(\frac{|f(x){|}^p}{p}+\frac{|g(x){|}^q}{q}\right)\mathrm{d}x$$

对双方积分：

$$\int |f(x)g(x)|\mathrm{d}x\le \int \left(\frac{|f(x){|}^p}{p}+\frac{|g(x){|}^q}{q}\right)\mathrm{d}x$$

3. Apply the definition of $L^p$ and $L^q$ norms to complete the proof.

应用 $L^p$ 和 $L^q$ 范数的定义完成证明

3. Minkowski’s Inequality | Minkowski 不等式

Definition | 定义

Minkowski’s inequality is a generalization of the triangle inequality to integrals. For integrable functions $f$ and $g$, and $p\ge 1$:

$${\left(\int |f(x)+g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}\le {\left(\int |f(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}+{\left(\int |g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}$$

Minkowski 不等式是三角不等式对积分的推广。对可积函数 $f$ 和 $g$，以及 $p\ge 1$：

$${\left(\int |f(x)+g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}\le {\left(\int |f(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}+{\left(\int |g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}$$

Usage | 使用

• $L^p$ Spaces: Fundamental in the study of $L^p$ spaces, particularly in proving that these spaces are normed vector spaces.

• Distance Measurement: Used in defining the distance between functions in $L^p$ spaces.

• $L^p$ 空间：在 $L^p$ 空间研究中具有基础性意义，特别是在证明这些空间是赋范向量空间时

• 距离度量：用于定义 $L^p$ 空间中函数之间的距离

Proof | 证明

1. Apply Hölder’s inequality to the sum of $f$ and $g$ raised to the power $p$.

对 $f$ 和 $g$ 的和取 $p$ 次方后，应用 Hölder 不等式

2. The result follows by combining terms and recognizing the definition of the $L^p$ norm.

通过合并项并识别 $L^p$ 范数的定义，得出结果

$${\left(\int |f(x)+g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}\le {\left(\int |f(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}+{\left(\int |g(x){|}^p\mathrm{d}x\right)}^{\frac{1}{p}}$$

4. Cauchy-Schwarz Inequality | 柯西-施瓦茨不等式

Definition | 定义

The Cauchy-Schwarz inequality is a fundamental result in linear algebra and analysis. It states that for any two integrable functions $f$ and $g$ on a measurable space, the following inequality holds:

$${\left(\int |f(x)g(x)|\mathrm{d}x\right)}^2\le \left(\int |f(x){|}^2\mathrm{d}x\right)\left(\int |g(x){|}^2\mathrm{d}x\right)$$

柯西-施瓦茨不等式是线性代数和分析中的一个基本结果。它表明，对于测度空间上的任意两个可积函数 $f$ 和 $g$，有如下不等式成立：

$${\left(\int |f(x)g(x)|\mathrm{d}x\right)}^2\le \left(\int |f(x){|}^2\mathrm{d}x\right)\left(\int |g(x){|}^2\mathrm{d}x\right)$$

Parametric Form | 参数形式

For real numbers $a$, $b$, and a parameter $t\in [0,1]$, the Cauchy-Schwarz inequality can be expressed as:

$$((1-t)a+tb{)}^2\le (1-t)a^2+t{b}^2$$

对于实数 $a$、$b$ 和参数 $t\in [0,1]$，柯西-施瓦茨不等式可以表示为：

$$((1-t)a+tb{)}^2\le (1-t)a^2+t{b}^2$$

Usage | 使用

• Inner Product Spaces: Provides a bound on the inner product, critical in defining angles and orthogonality.

• Distance Measurement: Used to establish the triangle inequality in normed vector spaces.

• Convexity Proofs: The parametric form is often used in proving the convexity of functionals in variational problems.

• 内积空间：为内积提供了一个上界，对定义角度和正交性至关重要。

• 距离度量：用于在赋范向量空间中确立三角不等式。

• 凸性证明：参数形式常用于证明变分问题中泛函的凸性。

Proof | 证明

1. Consider the integral of the square of the linear combination $f(x)+\lambda g(x)$ for a scalar $\lambda$:

   $$\int |f(x)+\lambda g(x){|}^2\mathrm{d}x\ge 0$$

   考虑 $f(x)+\lambda g(x)$ 的线性组合的平方积分，其中 $\lambda$ 为标量：

   $$\int |f(x)+\lambda g(x){|}^2\mathrm{d}x\ge 0$$

2. Expand the square and integrate each term:

   $$\int |f(x){|}^2\mathrm{d}x+2\lambda \int f(x)g(x)\mathrm{d}x+{\lambda }^2\int |g(x){|}^2\mathrm{d}x\ge 0$$

   展开平方并对每一项积分：

   $$\int |f(x){|}^2\mathrm{d}x+2\lambda \int f(x)g(x)\mathrm{d}x+{\lambda }^2\int |g(x){|}^2\mathrm{d}x\ge 0$$

3. The quadratic form in $\lambda$ must be non-negative for all $\lambda$, leading to the discriminant condition:

   $${\left(\int f(x)g(x)\mathrm{d}x\right)}^2\le \left(\int |f(x){|}^2\mathrm{d}x\right)\left(\int |g(x){|}^2\mathrm{d}x\right)$$

   对于所有 $\lambda$，该二次型必须非负，从而得到判别式条件：

   $${\left(\int f(x)g(x)\mathrm{d}x\right)}^2\le \left(\int |f(x){|}^2\mathrm{d}x\right)\left(\int |g(x){|}^2\mathrm{d}x\right)$$

4. This completes the proof of the Cauchy-Schwarz inequality.

   由此完成柯西-施瓦茨不等式的证明。

Proof of Parametric Form | 参数形式的证明

1. Expand the left side of the inequality:

   $$((1-t)a+tb{)}^2=(1-t{)}^2{a}^2+2t(1-t)ab+t^2{b}^2$$

   展开不等式的左边：

   $$((1-t)a+tb{)}^2=(1-t{)}^2{a}^2+2t(1-t)ab+t^2{b}^2$$

2. Subtract the right side:

   $$\begin{array}{rl}& ((1-t)a+tb{)}^2-((1-t)a^2+t{b}^2)\\ & =(1-t{)}^2{a}^2+2t(1-t)ab+t^2{b}^2-(1-t)a^2-t{b}^2\\ & =((1-t{)}^2-(1-t))a^2+2t(1-t)ab+(t^2-t)b^2\end{array}$$

   减去右边：

   $$\begin{array}{rl}& ((1-t)a+tb{)}^2-((1-t)a^2+t{b}^2)\\ & =(1-t{)}^2{a}^2+2t(1-t)ab+t^2{b}^2-(1-t)a^2-t{b}^2\\ & =((1-t{)}^2-(1-t))a^2+2t(1-t)ab+(t^2-t)b^2\end{array}$$

3. Factor out $-t(1-t)$:

   $$\begin{array}{rl}& =-t(1-t)a^2+2t(1-t)ab-t(1-t)b^2\\ & =-t(1-t)(a^2-2ab+b^2)\\ & =-t(1-t)(a-b{)}^2\end{array}$$

   提取公因子 $-t(1-t)$：

   $$\begin{array}{rl}& =-t(1-t)a^2+2t(1-t)ab-t(1-t)b^2\\ & =-t(1-t)(a^2-2ab+b^2)\\ & =-t(1-t)(a-b{)}^2\end{array}$$

4. Since $t\in [0,1]$, we know that $t(1-t)\ge 0$. Therefore:

   $$-t(1-t)(a-b{)}^2\le 0$$

   由于 $t\in [0,1]$，我们知道 $t(1-t)\ge 0$。因此：

   $$-t(1-t)(a-b{)}^2\le 0$$

This proves the parametric form of the Cauchy-Schwarz inequality.

这就证明了柯西-施瓦茨不等式的参数形式。