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## Created on 2023/03/31

(Refer to the Textbook:Bartle & sherbert, Introduction to Real Analysis, 4th Ed, Wiley)

7.4.11.

If $f$ is a bounded function on $[a, b]$ such that $f(x)=0$ except for $x$ in $\left\{c_1, c_2, \ldots, c_n\right\}$ in $[a, b]$, show that $U(f)=L(f)=0$.

Proof.

We show that $U(f)\leq 0$. Symmetrically we get $L(f)\geq 0$. Then by $L(f)\leq U(f)$ we obtain $U(f)=L(f)=0$.

Let $M=\sup_{x\in[a,b]}f(x).$Let $\mathcal{P}=(x_0,x_1,\cdots,x_m)$ be a partition of $[a,b]$ such that the norm $||\mathcal{P}||\leq \delta$. The upper sum of $f$ corresponding to $\mathcal{P}$ is $$U(f ; \mathcal{P})=\sum_{k=1}^m M_k\left(x_k-x_{k-1}\right) .$$Immediately $M_k=0$ if $[x_{k-1},x_k]\cap \left\{c_1, c_2, \ldots, c_n\right\}=\varnothing$. Suppose $c_i\in [x_{k_i-1},x_{k_i}]$, then

$$U(f ; \mathcal{P})\leq\sum_{i=1}^n M_{k_i}\left(x_{k_i}-x_{k_i-1}\right)\leq \sum_{i=1}^n M\delta\leq nM\delta.$$ By definition, $U(f ; \mathcal{P})\leq U(f ; \mathcal{P})$. Therefore, $U(f ; \mathcal{P})\leq \delta$ holds for any $\delta>0$. We conclude $U(f)\leq 0$.
Q.E.D.
7.4.15.

Let $f$ be defined on $I:=[a, b]$ and assume that $f$ satisfies the Lipschitz condition $|f(x)-f(y)| \leq K|x-y|$ for all $x, y$ in $I$. If $\mathcal{P}_n$ is the partition of $I$ into $n$ equal parts, show that $0 \leq U\left(f ; \mathcal{P}_n\right)-\int_a^b f \leq K(b-a)^2 / n$

Proof.

Let $\mathcal{P}_n=(x_0,x_1,\cdots,x_n)$ such that $x_i-x_{i-1}=\frac{b-a}{n}$. Since $f$ is continuous, we could find $t_i\in[x_{i-1},x_i]$ such that $M_i=f(t_i)$. Then

$$U(f ; \mathcal{P}_n)-\int_a^b f =\sum_{k=1}^n f(t_k)\left(x_k-x_{k-1}\right) -\sum_{k=1}^n\int_{x_{k-1}}^{x_{k}}f(x) \mathrm{d}x=\sum_{k=1}^n\int_{x_{k-1}}^{x_{k}}(f(t_k)-f(x)) \mathrm{d}x.$$ By Lipschitz condition, $f(t_k)-f(x)\leq K|t_k-x|$. And $|t_k-x|\leq |x_k-x_{k-1}|=\frac{b-a}{n}$ if $x\in[x_{k-1},x_k]$. Combine these, we obtain $$U(f ; \mathcal{P}_n)-\int_a^b f\leq \sum_{k=1}^n\int_{x_{k-1}}^{x_{k}}K\frac{b-a}{n} \mathrm{d}x=K\frac{(b-a)^2}{n}.$$ Q.E.D.

8.1.4.

Evaluate $\lim \left(x^n /\left(1+x^n\right)\right)$ for $x \in \mathbb{R}, x \geq 0$.

Solution.

Let $f_n(x)=x^n /\left(1+x^n\right)$. If $x\in[0,1)$, then $\lim f_n(x)=\frac{\lim x^n}{1+\lim x^n}=\frac{0}{1+0}=0$. If $x=1$, then $(f_n(1))=(\frac{1}{2})$ converges to $\frac{1}{2}$. If $x>1$. then then $\lim f_n(x)=\lim \frac{1}{1+x^{-n}}=\frac{1}{1+\lim x^{-n}}=1$.

If we define $$f(x):= \begin{cases}0, & 0< x< 1 \\ \frac{1}{2},& x=1 \\1, & x>1\end{cases}$$ then the sequence $(f_n)$ converges to $f$ for $x\geq 0$.
8.1.6.

Show that $\lim (\operatorname{Arctan} n x)=(\pi / 2) \operatorname{sgn} x$ for $x \in \mathbb{R}$.

Solution.

Let $g_n(x)=\operatorname{Arctan} n x$. If $x=0$, then $(g_n(0))=(0)$ converges to $0$. If $x>0$, then $\lim g_n(x)=\operatorname{Arctan}( \lim n x)=\pi/2$. If $x< 0$, then $\lim g_n(x)=\operatorname{Arctan}( \lim n x)=-\pi/2$. Therefore, $\lim (\operatorname{Arctan} n x)=(\pi / 2) \operatorname{sgn} x$.

8.1.14.

Show that if $0 < b < 1$, then the convergence of the sequence in Exercise 4 is uniform on the interval $[0, b]$, but is not uniform on the interval $[0,1]$.

Proof.

Note that $\frac{u}{1+u}\leq \frac{v}{1+v}$ if $0\leq u\leq v$.

If $0< b< 1$, we have $\lim_n||f_n(x)||_{[0,b]}=\lim_n\operatorname{sup}_{x\in[0,b]}(x^n /\left(1+x^n\right))=\lim_n b^n /\left(1+b^n\right)=0$.

But $||f_n(x)-f(x)||_{[0,1]}\geq b^n /\left(1+b^n\right), \forall b\in [0,1)$. Let $b\to 1$, we get for all $n$, $||f_n(x)-f(x)||_{[0,1]}\geq \frac{1}{2}$.

Q.E.D.
8.1.16

Show that if $a>0$, then the convergence of the sequence in Exercise 5 is uniform on the interval $[a, \infty)$, but is not uniform on the interval $[0, \infty)$.

Proof.

If $a>0$, we have $\lim_n||g_n(x)-\pi/2||_{[a,\infty)}=\lim_n\operatorname{sup}_{x\in[a,\infty)}(\pi/2-\operatorname{Arctan} n x)=\lim_n (\pi/2-\operatorname{Arctan} n a)=0$.

But $||g_n(x)-g(x)||_{[0,\infty)}\geq ||g_n(x)-\pi/2||_{[a,\infty)}= \pi/2-\operatorname{Arctan} n a, \forall a>0$. Let $a\to 0$, we get for all $n$, $||g_n(x)-g(x)||_{[0,\infty)}\geq \pi/2$.

Q.E.D.
Provided by Yunsong Wei