# 数学代写|数学分析代写Mathematical Analysis代考|Jensen’s Inequality

## 数学代写|数学分析代写Mathematical Analysis代考|Jensen’s Inequality

This section is devoted to the integral Jensen inequality, namely the continuous version of the discrete Jensen inequality proved for convex functions of several variables in Sect. 3.15, Appendix to Chap. 3.

We start by defining the integral of a vector-valued function $\varphi: X \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m$, of components
$$\left(\varphi_1, \varphi_2, \ldots, \varphi_m\right)$$
defined on a measurable and bounded subset $X \subset \mathbb{R}^n$. If the components $\varphi_k, k=$ $1,2, \ldots, m$, are integrable on $X$, we say that $\varphi$ is integrable on $X$. By definition, the integral of $\varphi$ on $X$ is the point in $\mathbb{R}^m$ of components
$$\left(\int_X \varphi_1 d x, \int_X \varphi_2 d x, \ldots, \int_X \varphi_m d x\right),$$
as usual denoted by
$$\int_X \varphi(x) d x$$
(Integral) Jensen Inequality Let $X$ be a compact and measurable set in $\mathbb{R}^n$ and $\varphi: X \subseteq \mathbb{R}^n \rightarrow A \subseteq \mathbb{R}^m$ a continuous function on $X$. Suppose $K$ is a closed convex set in $\mathbb{R}^m$ such that $\varphi(X) \subseteq K$, and $f: K \subseteq \mathbb{R}^m \rightarrow \mathbb{R}$ is a convex function on $K$. Then
$$\begin{gathered} \frac{1}{m(X)} \int_X \varphi(x) d x \in K, \ f\left(\frac{1}{m(X)} \int_X \varphi(x) d x\right) \leq \frac{1}{m(X)} \int_X f(\varphi(x)) d x . \end{gathered}$$
Proof Consider a partition $P=\left{X_1, X_2, \ldots, X_h\right}$ of $X$ made of measurable sets, and let $x_1, x_2, \ldots, x_h$ be $h$ points of $X$ such that $x_i \in X_i$ for any $i=1,2, \ldots, h$. The integral of a continuous function can be computed as a suitable limit (see Sect. 8.1), so the integral of $\varphi$ on $X$ is
$$\lim {\operatorname{diam}\left{X_i\right} \rightarrow 0} \sum{i=1}^h m\left(X_i\right) \varphi\left(x_i\right)=\int_X \varphi(x) d x,$$
when the diameter of $P=\left{X_i\right}$ tends to zero.

## 数学代写|数学分析代写Mathematical Analysis代考|The Gamma Function. The Measure of the Unit Ball in Rna

For any $t>0$ define
$$\Gamma(t)=\int_0^{+\infty} x^{t-1} e^{-x} d x$$
The function $\Gamma:(0,+\infty) \rightarrow(0,+\infty)$ thus defined is called Gamma function. Note that for any $t>0$, the improper integral defining $\Gamma(t)$ is finite.

The above function has interesting applications in Mathematical Analysis. Here we shall illustrate a few properties of $\Gamma$ and show how the use of this function allows to compute the measure of the unit ball $B=\left{x \in \mathbb{R}^n:|x| \leq 1\right}$ in $\mathbb{R}^n$.

The $\Gamma$ function generalises the well-known concept of factorial of a positive integer. For any $t>0$, in fact,
$$\Gamma(t+1)=\int_0^{+\infty} x^t e^{-x} d x=\left[-x^t e^{-x}\right]{x=0}^{x \rightarrow+\infty}+t \int_0^{+\infty} x^{t-1} e^{-x} d x$$ and so $$\Gamma(t+1)=t \Gamma(t), \quad \forall t>0 .$$ Since $$\Gamma(1)=\int_0^{+\infty} e^{-x} d x=1$$ from (8.119) and (8.120) we deduce that $$\Gamma(k+1)=k !, \quad \forall k=0,1,2, \ldots$$ Using $\Gamma$ we may obtain a formula to compute integrals of the type $$I\alpha=\int_{-1}^1\left(1-x^2\right)^\alpha d x, \quad \alpha>-1 .$$
For that, let us prove the following.

# 数学分析代考

## 数学代写|数学分析代写Mathematical Analysis代考|Jensen’s Inequality

$$\left(\varphi_1, \varphi_2, \ldots, \varphi_m\right)$$

$$\left(\int_X \varphi_1 d x, \int_X \varphi_2 d x, \ldots, \int_X \varphi_m d x\right)$$

$$\int_X \varphi(x) d x$$
(积分) Jensen 不等式让 $X$ 是一个紧凑且可测量的集合 $\mathbb{R}^n$ 和 $\varphi: X \subseteq \mathbb{R}^n \rightarrow A \subseteq \mathbb{R}^m$ 上的连续函数 $X$. 认为 $K$ 是闭凸集 $\mathbb{R}^m$ 这样 $\varphi(X) \subseteq K$ ，和 $f: K \subseteq \mathbb{R}^m \rightarrow \mathbb{R}$ 是一个凸 函数 $K$. 然启
$$\frac{1}{m(X)} \int_X \varphi(x) d x \in K, f\left(\frac{1}{m(X)} \int_X \varphi(x) d x\right) \leq \frac{1}{m(X)} \int_X f(\varphi(x)) d x$$

## 数学代写|数学分析代写Mathematical Analysis代考|The Gamma Function. The Measure of the Unit Ball in Rna

$$\Gamma(t)=\int_0^{+\infty} x^{t-1} e^{-x} d x$$

$$\Gamma(t+1)=\int_0^{+\infty} x^t e^{-x} d x=\left[-x^t e^{-x}\right] x=0^{x \rightarrow+\infty}+t \int_0^{+\infty} x^{t-1} e^{-x} d x$$

$$\Gamma(t+1)=t \Gamma(t), \quad \forall t>0$$

$$\Gamma(1)=\int_0^{+\infty} e^{-x} d x=1$$

$$\Gamma(k+1)=k !, \quad \forall k=0,1,2, \ldots$$使用 $\Gamma$ 我们可以获得一个公式来计算类型的积分
$$I \alpha=\int_{-1}^1\left(1-x^2\right)^\alpha d x, \quad \alpha>-1$$

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