taking derivative of each term?
Understanding these basic definition and statement~
- We say that a sequence of functions $\left{f_{n}: D \rightarrow \mathbb{R}\right}$ defined on a subset $D \subseteq \mathbb{R}$ converges pointwise on $D$ if for each $x \in D$ the sequence of numbers $\left{f_{n}(x)\right}$ converge. If $\left{f_{n}\right}$ converges pointwise on $D$, then we define $f: D \rightarrow \mathbb{R}$ with $f(x)=\lim {n \rightarrow \infty} f{n}(x)$ for each $x \in D .$ We denote this symbolically by $f_{n} \rightarrow f$ on $D$.
- We say that a sequence of functions $\left{f_{n}\right}$ defined on a subset $D \subseteq \mathbb{R}$ converges uniformly on $D$ to a function $f$ such that for every $\varepsilon>0$ there is a number $N$ such that
$$
\left|f_{n}(x)-f(x)\right|<\varepsilon \quad \text { for all } x \in D, \quad n \geq N
$$
We denote this type of convergence symbolically by $f_{n} \rightrightarrows f$ on $D$. - If $\left{f_{n}\right}_{0}^{\infty}$ is a sequence of functions defined on $D$, the series $\sum_{n=0}^{\infty} f_{n}$ is said to converge pointwise (respectively, uniformly) on $D$ if and only if the sequence $\left{s_{n}\right}_{n=0}^{\infty}$ of partial sums, given by
$$
s_{n}(x)=\sum_{k=0}^{n} f_{k}(x),
$$
converges pointwise (respectively, uniformly) on $D$. - Weierstrass $M$-Test: Suppose that $\left{f_{n}\right}$ is a sequence of functions defined on $D$ and $\left{M_{n}\right}$ is a sequence of nonnegative numbers such that
- $$
- \left|f_{n}(x)\right| \leq M_{n} \quad \forall x \in D, \quad \forall n \in \mathbb{N}
- $$
- If $\sum_{n=0}^{\infty} M_{n}$ converges, then $\sum_{n=0}^{\infty} f_{n}(x)$ converges uniformly on $D$.
- Assume that $f_{n} \rightrightarrows f$ uniformly on $[a, b]$ and that each $f_{n}$ is integrable. Then, $f$ is integrable and
- $$
- \lim {n \rightarrow \infty} \int{a}^{b} f_{n}=\int_{a}^{b} f
- $$
- As a corollary to this theorem we obtain the following result:
- Termwise Integration: If each function $f_{n}(x)$ is continuous on a closed interval $[a, b]$ and if the series $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly on $[a, b]$, then we have
- $$
- \sum_{n=1}^{\infty} \int_{a}^{b} f_{n}(x) d x=\int_{a}^{b} \sum_{n=1}^{\infty} f_{n}(x) d x
- $$
- Suppose that $\left{f_{n}\right}$ converges to $f$ on the interval $[a, b]$. Suppose also that $f_{n}^{\prime}$ exists and is continuous on $[a, b]$, and the sequence $\left{f_{n}^{\prime}\right}$ converges uniformly on $[a, b]$. Then
- $$
- \lim {n \rightarrow \infty} f{n}^{\prime}(x)=f^{\prime}(x)
- $$
- for each $x \in[a, b] .$ As a corollary to this theorem we obtain the following result:
- Termwise Differentiation: If each function $f_{n}(x)$ has the derivative $f_{n}^{\prime}(x)$ at any point $x \in$ $(a, b)$, if the series $\sum_{n=1}^{\infty} f_{n}(x)$ converges to at least one point $k \in(a, b)$, and if $\sum_{n=1}^{\infty} f_{n}^{\prime}(x)$ converges uniformly on $(a, b)$ to a function $g(x)$, then $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly on $(a, b)$ and is differentiable at any point $x \in(a, b)$, whose derivative is equal to $g(x)$. In other words, termwise differentiation is possible, i.e.,
- $$
- \left(\sum_{n=1}^{\infty} f_{n}(x)\right)^{\prime}=\sum_{n=1}^{\infty} f_{n}^{\prime}(x) .
- $$
- A family of functions $\left{f_{n}\right} \in \mathcal{F}$ mapping a set $A \in \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ is equicontinuous at a point $a \in A$ if for every $\varepsilon>0, \exists \delta>0$ such that
$$
\left|f_{n}(x)-f_{n}(a)\right|<\varepsilon \text { where }|x-a|<\delta \text { and } f_{n} \in \mathcal{F} \text {. }
$$
The family $\mathcal{F}$ is equicontinuous on a set $A$ if it is equicontinuous at every point in $A$. - Arzelà-Ascoli Theorem: Let $A$ be a compact subset of $\mathbb{R}^{n}$ and $\mathcal{C}\left(A, \mathbb{R}^{m}\right)$ the space of continuous functions from $A$ into $\mathbb{R}^{m}$. A subset $\mathcal{B}$ of $\mathcal{C}\left(A, \mathbb{R}^{m}\right)$ is compact if and only if it is closed, bounded, and equicontinuous.
- Weierstrass Approximation Theorem: If $f$ is a continuous function on a closed interval $[a, b]$, then there exists a sequence of polynomials that converge uniformly to $f(x)$ on $[a, b]$.
Try the following problem to test yourself！
a) Show that for any function $f \in C[0,1]$ and any number $\epsilon>0$, there exists a polynomial $p$, all of whose coefficients are rational numbers, such that
$$
|p-f|<\epsilon
$$
b) Show that $C[a, b]$ is separable.
a) Let $\varepsilon>0 .$ By the Weierstrass Approximation Theorem, there exists a polynomial $q(x)$ such that $|q-f|<\varepsilon / 2$. Suppose $q$ has degree $r$, and
$$
q(x)=\sum_{k=0}^{r} a_{k} x^{k}
$$
where some or all of the coefficients $a_{0}, a_{1}, \ldots, a_{r}$ may be irrational. For each coefficient $a_{k}$ we find a rational number $b_{k}$ such that $\left|b_{k}-a_{k}\right|<\varepsilon / 2(r+1) .$ Let $p$ be the polynomial given by
$$
p(x)=\sum_{k=0}^{r} b_{k} x^{k} .
$$
Then, for all $x \in[0,1]$ we have
$$
|p(x)-q(x)|=\left|\sum_{k=0}^{r}\left(b_{k}-a_{k}\right) x^{k}\right| \leq \sum_{k=0}^{r}\left|b_{k}-a_{k} | x\right|^{k} \leq \sum_{k=0}^{r} \frac{\varepsilon}{2(r+1)}=\frac{\varepsilon}{2} \text {. }
$$
Then $|p-q| \leq \varepsilon / 2$, so $|p-f| \leq|p-q|+|q-f|<\varepsilon$.
b) To show $C[a, b]$ separable we must show it contains a countable dense set. Let
$$
A=\left{p(x)=\sum_{i=0}^{n} b_{i} x^{i} \mid b_{i} \in \mathbb{Q}, n \in \mathbb{N}\right} .
$$
Then define
$$
A_{n}=\left{p(x)=\sum_{i=0}^{n} b_{i} x^{i} \mid b_{i} \in \mathbb{Q}\right},
$$
and we have
$$
A=\bigcup_{n \in \mathbb{N}} A_{n}
$$
There is an obvious bijection between $A_{n}$ and $\mathbb{Q}^{n+1}$ which sends each coefficient of a polynomial in $A_{n}$ to one coordinate in $\mathbb{Q}^{n+1}$. Thus $A_{n}$ is countable, and $A$ is countable since it is a countable union of countable sets.
So we must show that $A$ is dense in $C[a, b]$. We follow the same proof as above:
Let $\varepsilon>0 .$ By the Weierstrass Approximation Theorem, there exists a polynomial $q(x)$ such that $|q-f|<\varepsilon / 2$. Suppose $q$ has degree $r$, and $$ q(x)=\sum_{k=0}^{r} a_{k} x^{k} $$ where some or all of the coefficients $a_{0}, a_{1}, \ldots, a_{r}$ may be irrational. Let $c=\max {0 \leq i \leq r}\left{\sup {a \leq x \leq b}\left{|x|^{i}\right}\right}$. Then for each coefficient $a_{k}$ we find a rational number $b_{k}$ such that $\left|b_{k}-a_{k}\right|<\varepsilon / 2 c(r+1)$. Let $p$ be the polynomial given by $$ p(x)=\sum_{k=0}^{r} b_{k} x^{k} . $$ Then, for all $x \in[0,1]$ we have $$ |p(x)-q(x)|=\left|\sum_{k=0}^{r}\left(b_{k}-a_{k}\right) x^{k}\right| \leq \sum_{k=0}^{r}\left|b_{k}-a_{k} | x\right|^{k} \leq \sum_{k=0}^{r} \frac{|x|^{k} \varepsilon}{2 c(r+1)} \leq \sum_{k=0}^{r} \frac{\varepsilon}{2(r+1)}=\frac{\varepsilon}{2} \text {. } $$ Then $|p-q| \leq \varepsilon / 2$, so $|p-f| \leq|p-q|+|q-f|<\varepsilon$. Thus for any $\varepsilon>0$ and $f \in C[a, b]$
we can find a $p \in A$ with $|p-f|<\varepsilon$, so $A$ is dense in $C[a, b]$.
(Bernstein Polynomials) The $n$th Bernstein polynomial of a continuous function $f:[0,1] \rightarrow \mathbb{R}$ defined by
$$
B_{n}(f)(x)=\sum_{k=0}^{n} f\left(\frac{k}{n}\right)\left(\begin{array}{l}
n \
k
\end{array}\right) x^{k}(1-x)^{n-k} .
$$
a) Show that $B_{n}$ is linear, monotone map, and $B_{n} 1=1$ and $B_{n} x=x$.
b) Show that $n$th Bernstein polynomial for $f(x)=e^{x}$ is $B_{n}(x)=\left[1+\left(e^{\frac{1}{n}}-1\right) x\right]^{n} .$
c) Show that $B_{n}\left(e^{x}\right)$ converges uniformly to $e^{x}$ on $[0,1]$.
a) It is clear that the map $B_{n}$ is linear, since $B_{n}(f+g)=B_{n} f+B_{n} g$ and $B_{n}(\alpha f)=\alpha B_{n} f$. Notice that when $f \geq 0$, then $B_{n} f$ is also positive. In particular, $|f| \leq g$ means $-g \leq f \leq g$ and hence $-B_{n} g \leq B_{n} f \leq B_{n} g$. This also proves that $\left|B_{n} f\right|<B_{n} g$ if $|f|<g$ (since it is straightforward to show that $f \geq g \Rightarrow B_{n} f \geq B_{n} g$ ).
b) $B_{n}(1)=\sum_{k=0}^{n} 1\left(\begin{array}{c}n \ k\end{array}\right) x^{k}(1-x)^{n-k}=1$ (From the Binomial Theorem). Next notice that
$$
\frac{k}{n}\left(\begin{array}{l}
n \
k
\end{array}\right)=\frac{k}{n} \frac{n !}{k !(n-k) !}=\frac{(n-1) !}{(k-1) !(n-k) !}=\left(\begin{array}{l}
n-1 \
k-1
\end{array}\right) .
$$
Using the Binomial Theorem again, we have
$$
\begin{aligned}
&B_{n}(x)=\sum_{k=0}^{n} \frac{k}{n}\left(\begin{array}{l}
n \
k
\end{array}\right) x^{k}(1-x)^{n-k} \
&=x \sum_{k=0}^{n}\left(\begin{array}{l}
n-1 \
k-1
\end{array}\right) x^{k-1}(1-x)^{n-k} \
&=x(x+(1-x))^{n-1}=x .
\end{aligned}
$$
c)
$$
\begin{gathered}
B_{1}(x)=f(0)(1-x)+f(1) x=(1-x)+e x=1+(e-1) x . \
B_{2}(x)=f(0)(1-x)^{2}+2 f\left(\frac{1}{2}\right) x(1-x)+f(1) x^{2}=(10 x)^{2}+2 e^{1 / 2} x(1-x) e x^{2} \
=\left((1-x)+e^{1 / 2} x\right)^{2}=\left(1+\left(e^{1 / 2}-1\right) x\right)^{2} .
\end{gathered}
$$
More generally, we have
$$
B_{n}(x)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right)\left(e^{1 / n} x\right)^{k}(1-x)^{n-k}=\left(e^{1 / n} x+(1-x)\right)^{n}=\left(1+\left(e^{1 / n}-1\right) x\right)^{n} .
$$
d) It can be shown that $B_{n}\left(e^{x}\right)$ may be written as $\left(1+\frac{x}{n}+\frac{c_{n}}{n^{2}}\right)$ where $0 \leq c_{n} \leq 1$ and hence $B_{n}\left(e^{x}\right)$ converges uniformly to $e^{x}$.
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