Key concepts:
Outline:
we consider acts as objects (vectors) in a vector space. We’ll set criteria and inner products on vector spaces of functions, which will permit us to measure”distances” between functions and discuss convergence of sequences of functions.

• Lipschitz works Differentiable works Normed vector spaces
Continuity is a”fine” property for purposes. There are other, similar, properties which functions can have. We briefly discuss three such properties, namely, uniform continuity, Lipschitz continuity, and differentiability. All of these are stricter than continuity; that is, all functions that are uniformly continuous, Lipschitz, or differentiable must also be continuous. Additionally, all Lipschitz functions are uniformly continuous.
This week, we also present the closing foundational topic of the class before beginning the applications. To start with, we’ll show the defining characteristics of a norm and a couple of alternatives to the Euclidean norm that we usually use for vectors at $R^n$.

Definition We say that a function $f$ from $S \subseteq R^n$ to $R^m$ is uniformly continuous if:
for all $\varepsilon>0$, there exists $\delta>0$ such that $|f(\vec{x})-f(\vec{y})|<\varepsilon$ for every $\vec{x}$ and $\vec{y}$ in $S$ satisfying $|\vec{x}-\vec{y}|<\delta$.

## Example

$f(x)=x^{2}$ on the domain $[a, b] \subset \mathbb{R}$ for some real numbers $a<b .$ We show that $f$ is uniformly continuous on $[a, b]$.

Let $\varepsilon>0 .$ For any $x$ and $y$ in $[a, b]$, we have that
\begin{aligned} |f(x)-f(y)|=\left|x^{2}-y^{2}\right|=|(x+y)(x-y)| &=|x+y||x-y| \ & \leq(|x|+|y|)|x-y| \ & \leq 2 \max {|x|,|y|}|x-y| \ & \leq 2 \max {|a|,|b|}|x-y| . \end{aligned}
Since $a0$ and define $\delta=\varepsilon / M$. Note that this $\delta$ does not depend on $x$ or $y .$ We now have that
$$|f(x)-f(y)| \leq M|x-y|<M \cdot \frac{\varepsilon}{M}=\varepsilon \text { whenever }|x-y|<\delta .$$

$f(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$.
Choose $\varepsilon=1$ and consider an arbitrary candidate for $\delta>0$ for the definition of uniform continuity.
For any $x, h>0$, note that $f(x+h)=(x+h)^{2}=x^{2}+2 x h+h^{2}$ so
$$|f(x+h)-f(x)|=f(x+h)-f(x)=2 x h+h^{2}>2 x h$$
Choosing $x=1 / \delta$ and $h=\delta / 2$ and defining $y=x+h$, we satisfy $|y-x|=\delta / 2<\delta$ and $|f(y)-f(x)|>2 \cdot \frac{1}{\delta} \cdot \frac{\delta}{2}=1=\varepsilon$, so $f$ is not uniformly continuous.
$f(x)=1 / x$ on $(0,1]$ is not uniformly continuous. See text book, example 5.5.3.
$f(x)=\sin (1 / x)$ on $(0,1]$ is not uniformly continuous because it oscillates between $-1$ and 1 with higher and higher frequency as $x$ approaches $0 .$ This function does not have a limit as $x$ approaches 0 from the right. Exercise: Justify that this function is not uniformly continuous in a more rigorous manner. See text book, example 5.5.7 for solution.
$f(x)=x \sin (1 / x)$ on $(0,1]$ is uniformly continuous. Even though oscillations become rapid near $x=0$, the amplitude becomes small. Exercise: Justify that this function is uniformly continuous in a more rigorous manner. See text book, example 5.5.8 for solution.

In fact, if you extend this function by defining $f(0)=0$, then you can prove that $f$ is continuous at $x=0$.

In the examples above, you might notice a pattern: uniform continuity can easily be broken by otherwise nice functions when the domain is unbounded or does not include the end points. We can prove the following theorem, which “promotes” continuous functions to uniformly continuous functions on compact domains. This is another example of how compactness leads to nicer properties.

## Lipschitz functions

Definition A function $f$ from $S \subseteq \mathbb{R}^{n}$ to $\mathbb{R}^{m}$ is called a Lipschitz function if there exists $a$ constant $C$ such that
$$|f(\vec{x})-f(\vec{y})| \leq C|\vec{x}-\vec{y}| \text { for all } \vec{x}, \vec{y} \in S$$
Any constant $C$ for which this condition is satisfied is called a Lipschitz constant for $f$. The smallest $C$ for which this condition holds is called the (best) Lipschitz constant.

Loosely, this means that the function $f$ cannot change too rapidly as you move a point $\vec{x}$ away from $\vec{y}$

## Di ffrentiable functions

Definition Let $S \subseteq \mathbb{R}$ be a set and let $f: U \rightarrow \mathbb{R}$ be a function. We say that $f$ is differentiable at a point $x_{0} \in \operatorname{int}(S)$ if $\lim {h \rightarrow 0} \frac{f\left(x{0}+h\right)-f\left(x_{0}\right)}{h}$ exists.
If $f$ is differentiable at $x_{0}$, then we define
$$f^{\prime}\left(x_{0}\right)=\lim {h \rightarrow 0} \frac{f\left(x{0}+h\right)-f\left(x_{0}\right)}{h}$$
to be the derivative of $f$ at $x_{0}$.
We say that $f$ is differentiable on a set $X \subseteq S$ if it is differentiable at every point in $X$.
Remark 7.3.2. We can expand the definition of derivative for functions on intervals. Consider a function defined on a closed interval, i.e., $f:[a, b] \rightarrow \mathbb{R}$. The definition above applies for $x_{0} \in(a, b) .$ We say that $f$ is differentiable at a if $\lim _{h \rightarrow 0+} \frac{f(a+h)-f(a)}{h}$ exists. Differentiability at $b$ is defined analogously, in terms of a limit from the left.

Proposition Suppose $f:(a, b) \rightarrow \mathbb{R}$ is differentiable at $x_{0} \in(a, b) .$ Then, $f$ is continuous at $x_{0}$

## Normed vector spaces

Recall that vector spaces are defined abstractly as a set of objects, called vectors, together with vector addition and scalar multiplication operations. Certain rules must be followed. (Refer to text book, Sec. $1.2$ for how vector spaces are defined.)

We now consider vectors that are not necessarily from $\mathbb{R}^{n} .$ We will write these vectors as $x, y$, etc. rather than $\vec{x}, \vec{y}$ as we did for points in $\mathbb{R}^{n}$. We will continue to write the zero vector as $\overrightarrow{0}$ to make a distinction from the real number, $0 .$

Definition Let $V$ be a vector space over $\mathbb{R}($ i.e., a real vector space). A norm is a function $|\cdot|: V \rightarrow \mathbb{R}$ that satisfies the following properties for all $x, y \in V$ and for all $\alpha \in \mathbb{R}:$
N1. $|x| \geq 0$ with equality if and only if $x=\overrightarrow{0}$ (positive definite),
N2. $|\alpha x|=|\alpha||x|$ (homogeneous), and
N3. $|x+y| \leq|x|+|y|$ (triangle inequality).

Example (Norms on $\left.R^{n}\right) .$ We will present a few norms and evaluate them for the vector $\vec{v}=(1,2,-3) \in \mathbb{R}^{3} .$ Notice that they have different numerical values.

1. The Euclidean norm, which we worked with up to now, is one example of a norm. We use a subscript 2 to identify this particular norm (see the next example to understand why):
$|\vec{x}|_{2}=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{2}\right)^{1 / 2}$
$|\vec{x}|_{2}=\sqrt{1+2^{2}+|-3|^{2}}=\sqrt{14} \approx 3.7$
2. For real numbers $p \geq 1$, the $p$ -norm, $|\vec{x}|_{p}=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{p}\right)^{1 / p}$ is a generalization of the $E u$ clidean norm (which is the case $p=2$ ).
For $p=3$, we calculate $|\vec{x}|_{3}=\left(1^{3}+2^{3}+|-3|^{3}\right)^{1 / 3}=36^{1 / 3} \approx 3.3$.
3. The case $p=1$ is known as the taxicab norm, $|\vec{x}|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|$.
$|\vec{x}|_{1}=1+2+|-3|=6$.
4. As $p \rightarrow \infty$, we have the infinity or maximum norm, $|\vec{x}|_{\infty}=\max {i}\left|x{i}\right|$.
$|\vec{x}|_{\max }=\max {1,2,|-3|}=3 .$