实分析A Real Analysis A MATH20101

0

这是一份manchester曼切斯特大学 MATH20101作业代写的成功案例

实分析A Real Analysis A MATH20101
问题 1.

$$
W_{f}(\sigma) \geq \int_{a}^{b} f-\int_{a}^{b} f
$$
for every subdivision $\sigma$, therefore
$$
W \geq \int_{a}^{\bar{b}} f-\int_{a}^{b} f .
$$

证明 .

To establish the reverse inequality, choose sequences $\left(\sigma_{n}\right)$ and $\left(\tau_{n}\right)$ of subdivisions of $[a, b]$ such that
$$
s\left(\sigma_{n}\right) \rightarrow \int_{a}^{b} f \text { and } S\left(\tau_{n}\right) \rightarrow \int_{a}^{b} f
$$
(possible by the definitions of upper and lower integrals). For each $n$, let $\rho_{n}$ be a common refinement of $\sigma_{n}$ and $\tau_{n}$;, we have
$$
W \leq W_{f}\left(\rho_{n}\right) \leq S\left(\tau_{n}\right)-s\left(\sigma_{n}\right)
$$
and passage to the limit yields
$$
W \leq \int_{a}^{b} f-\int_{0}^{b} f
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MATH20101 COURSE NOTES :

(Darboux’s theorem ${ }^{2}$ ) For every bounded function $f:[a, b] \rightarrow \mathbb{R}$,
$$
\lim {\mathrm{N}(\sigma) \rightarrow 0} S{f}(\sigma)=\int_{a}^{b} f
$$
in the following sense: for every $\epsilon>0$, there exists a $\delta>0$ such that, for subdivisions $\sigma$ of $[a, b]$,
$$
\mathrm{N}(\sigma) \leq \delta \Rightarrow\left|S_{f}(\sigma)-\int_{a}^{b} f\right| \leq \epsilon .
$$







实分析 Real Analysis MATH20101

0

这是一份manchester帝国理工大学 GG14作业代写的成功案例

实分析 Real Analysis MATH20101
问题 1.

conclusion holds for $k=1$. Let us proceed by induction on $k$. If $C$ has no interior, then by Theorem $6.2 .6$ it is included in a hyperplane $H=f^{-1}{c}$ for some non-zero linear form $f$ and $c \in \mathbb{R}$. Then
$$
H={x: f(x) \geq c} \cap{x: f(x) \leq c},
$$
an intersection of two half-spaces. Also,
$$
H=u+V:={u+v: v \in V}
$$


证明 .

Let $V$ be a real vector space and $C$ a convex set in $V$. A real-valued function $f$ on $C$ is called convex iff
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MATH20101 COURSE NOTES :

$$
t \leq 3 \sum_{i=1}^{q} \lambda\left(J_{i}\right) \leq 6 j \sum_{i=1}^{q} \mu\left(J_{i}\right) \leq 6 j \mu(V) \leq 6 j \varepsilon .
$$
Thus $\lambda\left(P_{j}\right) \leq 6 j \varepsilon$. Letting $\varepsilon \downarrow 0$ gives $\lambda\left(P_{j}\right)=0, j=1,2, \ldots$, and letting $j \rightarrow \infty$ proves the lemma.
Now to prove Theorem 7.2.1, for each rational $r$ let
$$
g_{r}:=\max (g-r, 0), \quad f_{r}(x):=\int_{a}^{x} g_{r}(t) d t .
$$