# Understanding these basic definition and statement~

• Let $f: D \rightarrow \mathbb{R}$ and let $c \in D .$ We say that $f$ is continuous at $c$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that $|f(x)-f(c)|<\varepsilon$ whenever $|x-c|<\delta$ and $x \in D .$ If $f$ is continuous at each point of a subset $K \subseteq D$, then $f$ is said to be continuous on $K$. Moreover, if $f$ is continuous on its domain $D$, then we simply say that $f$ is continuous.
• Let $D$ be a nonempty subset of $\mathbb{R}$ and $f: D \rightarrow \mathbb{R}$. We say that $f$ is uniformly continuous on $D$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that $|x-c|<\delta$ and $x, c \in D$ imply $|f(x)-f(c)|<\varepsilon$. Notice that the $\delta$ in this definition depends on $\varepsilon$ and $f$, but not on the point $c$ or $x$. The geometric illustration of uniform continuity is shown in Figure 5.1.
• Let $f: A \subset \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A$ is a closed and bounded subset of $\mathbb{R}$. Then $f$ is uniformly continuous on $A$.
• Intermediate Value Theorem: If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, and if $L$ is a real number satisfying $f(a)L>f(b)$, then there is a point $c \in(a, b)$ where $f(c)=L$
• A function $f$ defined on an interval $I$ is said to be convex if
$$f(\alpha x+(1-\alpha) y) \leq \alpha f(x)+(1-\alpha) f(y)$$
holds true for any $x, y$ in $I$ and $0 \leq \alpha \leq 1$.

We say $f$ is a periodic function on $\mathbb{R}$, if there is a $T \in \mathbb{R}$ such that
$$f(x+T)=f(x)$$
for all $x \in \mathbb{R}$.
For $f: \mathbb{R} \rightarrow \mathbb{R}, c \in \mathbb{R}$, and $\delta>0$, define the modulus of continuity $w_{f}(c, \delta)$ of $f$ by
$$w_{f}(c, \delta)=\sup {|f(x)-f(c)|: x \in \mathbb{R},|x-c|<\delta}$$

# Try the following problem to test yourself！

Problem 1.

Let $f: D \rightarrow \mathbb{R}$ and let $c \in D$. Show that the following conditions are equivalent.
(i) $f$ is continuous at $c$.
(ii) If $\left{x_{n}\right}$ is a sequence in $D$ such that $\left{x_{n}\right}$ converges to $c$, then
$$\lim {n \rightarrow \infty} f\left(x{n}\right)=f(c)$$
(This condition is called sequential continuity).

Proof .

1 Let us first prove (i) $\Rightarrow$ (ii). Suppose that $\left{x_{n}\right}$ is a sequence with $\lim {n \rightarrow \infty} x{n}=c$. We need to show that $\lim {n \rightarrow \infty} f\left(x{n}\right)=f(c) .$ Let $\varepsilon>0$. We must find an integer $N$ so that $n \geq N$ implies $\left|f\left(x_{n}\right)-f(c)\right|<\varepsilon .$ To do this, choose $\delta>0$ so that $|x-c|<\delta$ implies $|f(x)-f(c)|<\varepsilon$. The existence of such a $\delta$ is guaranteed by the continuity of $f .$ Since $\lim {n \rightarrow \infty} x{n}=c$, there exists $N \geq 1$ such that $n \geq N$ implies $\left|x_{n}-c\right|<\delta$. Hence for any $n \geq N$, we have $\left|f\left(x_{n}\right)-f(c)\right|<\varepsilon$ This yields the desired conclusion.
2 Next we show (ii) $\Rightarrow$ (i). Suppose, to the contrary, that $f$ is sequentially continuous but discontinuous at $c$. Then there is a neighborhood $V$ of $f(c)$ such that no neighborhood $U$ of $c$ satisfies $f(U) \subseteq V .$ Set $U_{n}=B(c, 1 / n)={x \in D:|c-x|<1 / n}$, for $n \geq 1 .$ In particular, we have that $f\left(U_{n}\right) \cap V^{c} \neq \emptyset$, for any $n \geq 1$. Hence for $n \geq 1$, choose some $x_{n} \in D$ with $|x-c|<1 / n$ and $f\left(x_{n}\right) \notin V .$ By construction, $\left{x_{n}\right}$ converges to $c$, but $\left{f\left(x_{n}\right)\right}$ does not converge to $f(c)$. This contradicts the sequential continuity of $f$.

Problem 2.

Prove the Dirichlet function $f:(0,1) \rightarrow \mathbb{R}$ defined as
$f(x)= \begin{cases}0 & \text { if } x \in \mathbb{R} \backslash \mathbb{Q} \ \frac{1}{q} & \text { if } x \in \mathbb{Q} \text { and } x=\frac{p}{q} \text { in lowest terms }\end{cases}$
(a) discontinuous at every rational number in $(0,1)$,
(b) continuous at each irrational number in $(0,1)$.

Proof .

we showed that $\lim {x \rightarrow a} f(x)=0$ for any $a \in(0,1)$. Therefore (a) if $a \in \mathbb{R} \backslash \mathbb{Q}$, then $f(a)=0$ which implies $\lim {x \rightarrow a} f(x)=f(a)$, i.e., $f(x)$ is continuous at $a$;
(b) if $a \in \mathbb{Q} \cap(0,1)$, then $f(a) \neq 0$ which implies $\lim _{x \rightarrow a} f(x) \neq f(a)$, i.e., $f(x)$ is not continuous at $a .$

Problem 3.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$Problem 5.33 (Banach Contraction Mapping Theorem) Let f be a function defined on all of \mathbb{R}, and assume that there is a constant k such that 0<k<1 and$$
|f(x)-f(y)| \leq k|x-y|
$$for all x, y \in \mathbb{R}. (In this case we call f a contraction mapping.) a) Show that f is continuous on \mathbb{R}. b) Pick some point x_{0} \in \mathbb{R} and construct the sequence x_{n+1}=f\left(x_{n}\right) more precisely$$
\left(x_{0}, f\left(x_{0}\right), f\left(f\left(x_{0}\right)\right), \ldots\right) .
$$Show that the resulting sequence \left(x_{n}\right) is a Cauchy sequence and converges to some point x^{} in \mathbb{R}. c) Show that x^{} is a fixed point of f (i.e., f\left(x^{}\right)=x^{} ). d) Show that f has a unique fixed point. Proof . The equation satisfied by f(x) is usually known as a functional equation. First note that f(n x)= n f(x) for any n \in \mathbb{N}. Indeed, we have f(1 \cdot x)=1 \cdot f(x). Assume that f(n x)=n f(x). Then$$
f((n+1) x)=f(n x+x)=f(n x)+f(x)=n f(x)+f(x)=(n+1) f(x) .
$$By induction we get the desired identity. Let r=\frac{p}{q} \in \mathbb{Q}. Then we have f(q r x)=f(p x)=p f(x), and since f(q r x)=q f(r x), we get f(r x)=\frac{p}{q} f(x)=r f(x) . In particular, we have f(r)=r f(1). Let x \in \mathbb{R}. Then there exists a sequence of rational numbers \left{r_{n}\right} such that \lim r_{n}=x. Since f(x)=f\left(x-r_{n}+r_{n}\right)=f\left(x-r_{n}\right)+f\left(r_{n}\right)=f\left(x-r_{n}\right)+r_{n} f(1), we have f(x)=f\left(x-r_{n}\right)+r_{n} f(1). Now if we take the limit as n \rightarrow \infty, by the existence of \lim _{x \rightarrow 0} f(x)=f(0), we have that f(x)=f(0)+x f(1). Note that f(x)=f(x+0)=f(x)+f(0) for any x \in \mathbb{R}, so f(0)=0. Therefore, we have$$
f

a) Given $c \in \mathbb{R}$ and $\varepsilon>0$, set $\delta=\frac{\varepsilon}{k}$. Then
$|f(x)-f(y)|<\varepsilon$ for all $x \in \mathbb{R}$ such that $|x-c|<\delta$ Hence if $f$ is a contraction, then $f$ is continuous. b) To show that $\left(x_{n}\right)$ is Cauchy, observe: $$\left|x_{n+1}-x_{n}\right|=\left|f\left(x_{n}\right)-f\left(x_{n-1}\right)\right| \leq k\left|x_{n}-x_{n-1}\right|=\left|f\left(x_{n-1}\right)-f\left(x_{n-2}\right)\right| \leq k^{2}\left|\left(x_{n-1}-x_{n-2}\right)\right|$$ we can repeat the above process to obtain $$\left|x_{n+1}-x_{n}\right| \leq k^{n}\left|x_{1}-x_{0}\right| .$$ Thus if $n>m$,
$$\begin{gathered} \left|x_{n}-x_{m}\right| \leq\left|x_{n}-x_{n-1}\right|+\left|x_{n-1}-x_{n-2}\right|+\cdots+\left|x_{m+1}-x_{m}\right| \ \left|x_{n+1}-x_{n}\right| \leq\left(k^{n-1}+k^{n-2}+\cdots+k^{m}\right)\left|x_{1}-x_{0}\right| \leq \frac{k^{m}}{1-k}\left|x_{0}-x_{1}\right| \end{gathered}$$
using the limiting sum of a geometric series, which we may do since $00$ whenever $m$ and $n$ are sufficiently large. Hence $\left(x_{n}\right)$ is a Cauchy sequence. Since $\mathbb{R}$ is a complete space, every Cauchy sequence converges in it, and the existence of $\lim {n \rightarrow \infty} x{n}$ is assured. We set $\lim {n \rightarrow \infty} x{n}=x^{*}$.

c) To show $x^{}$ is a fixed point of $f$, we note that, for any positive integer $n$, $$0 \leq\left|f\left(x^{}\right)-x^{}\right| \leq\left|f\left(x^{}\right)-x_{n}\right|+\left|x_{n}-x^{}\right|=\left|f\left(x^{}\right)-f\left(x_{n-1}\right)\right|+\left|x_{n}-x^{}\right| \leq k\left|x^{}-x_{n-1}\right|+\left|x_{n}-x^{}\right|$$ and so $\left|f\left(x^{}\right)-x^{}\right|=0$ since $\left|x_{n}-x^{}\right| \rightarrow 0\left(\right.$ and $\left.\left|x^{}-x_{n-1}\right| \rightarrow 0\right)$. Thus $f\left(x^{}\right)=x^{}$. d) Suppose there are two fixed points say $x^{}$ and $\tilde{x}$, so also $f(\tilde{x})=\tilde{x}$. Then
$$\left|x^{}-\tilde{x}\right|=\left|f\left(x^{}\right)-f(\tilde{x})\right| \leq k\left|x^{}-\tilde{x}\right|$$ which since $k<1$ can only be true when $\left|x^{}-\tilde{x}\right|=0 ;$ that is, $x^{*}=\tilde{x}$. Hence there is a unique fixed point of $f$.

Fourier analysis代写

## 离散数学代写

Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析