4.2 Functional Limits

Delta-Epsilon

Let $f:A\to \mathbb{R}$ and let $c$ be a limit point in the domain of $A$. $\underset{x\to c}{lim}f(x)=L$ if

Topological

Let $f:A\to \mathbb{R}$ and let $c$ be a limit point in the domain of $A$. $\underset{x\to c}{lim}f(x)=L$ if

Sequential Definition

Let $f:A\to \mathbb{R}$ and let $c$ be a limit point in the domain of $A$. $\underset{x\to c}{lim}f(x)=L$ iff for all sequences $(x_n)\subseteq A$ satisfying $x_n\ne c$ and $f(x_n)\to L$

Proof

$⟹$
Assume $\underset{x\to c}{lim}f(x)=L$, $(x_n)\to c$ and $x_n\ne c$. Let $\epsilon >0$, there exists $V_{\delta }(c)$ with the property that all $x\in V_{\delta }(c)$ different from $c$ satisfy $f(x)\in V_{\epsilon }(L)$.

$(x_n)\to c$ so there exists an $x_N$ after which $x_n\in V_{\delta }(c)$. $n\ge N$ implies $f(x_n)\in V_{\epsilon }(L)$

$⟸$
Contrapositive
If $\underset{x\to c}{lim}f(x)\ne L$ then there exists one particular ${\epsilon }_0>0$ for which no matter what $\delta >0$ there will be

If ${\delta }_n=\frac{1}{n}$ for any $n\in \mathbb{N}$ there is a $x_n\in V_{{\delta }_n}(c)$ with $x_n\ne c$ and $f(x_n)\notin V_{{\epsilon }_0}(L)$. Here $(x_n)\to c$ with $x_n\ne c$ where $f(x_n)$ does not converge to $L$.

Algebraic Limit Theorems for Functions

Let $f$ and $g$ be functions defined on $A\subseteq \mathbb{R}$ and assume that $\underset{x\to c}{lim}f(x)=L$ and $\underset{x\to c}{lim}g(x)=M$ for some limit point $c$ of $A$. Then,
(i) $\underset{x\to c}{lim}kf(x)=kL$ for all $k\in \mathbb{R}$
(ii) $\underset{x\to c}{lim}[f(x+g(x)]=L+M$
(iii) $\underset{x\to c}{lim}[f(x)g(x)]=L\cdot M$
(iv) $\underset{x\to c}{lim}\left(\frac{f(x)}{g(x)}\right)=\frac{L}{M}$ provided $M\ne 0$

Divergence Criterion

Let $f$ be defined on $A$ and let $c$ be a limit point of $A$. If there exists two sequences $(x_n)$ and $(y_n)$ in $A$ with $x_n\ne c$ and $y_n\ne c$ but $lim{x}_n=lim{y}_n=c$ but

Then $\underset{x\to c}{lim}f(x)$ does not exist

4.3 Continuous Functions

Continuity

$f:A\to \mathbb{R}$ is continuous at a point $c\in A$ if for all $\epsilon >0$ there exists a $\delta >0$ such that whenever $|x-c|<\delta$ (and $x\in A$), it follows that $|f(x)-f(c)|<\epsilon$

$\epsilon$-Neighbourhood Definition:
$\mathrm{\forall }{V}_{\epsilon }(c),\mathrm{\exists }{V}_{\delta }(c)$ so that $x\in V_{\delta }(c)$ (and $x\in A$) $⟹f(x)\in V_{\epsilon }(f(c))$

Notation:

• If $f$ is continuous on $A$ if $f$ is continuous at each point in $A$
• $f$ is discontinuous on $A$ if $f$ is not continuous at some point of $A$
• isolated points are trivially continuous

Thm: Functional Continuity and Sequences

$f$ is continuous at $c\in A$ iff for all $(x_n)\in A$ that converge to $c$, $f(x_n)\to f(c)$
$⟹$ contrapositive useful for showing not continuous at $c$

Examples:

Polynomials

$\underset{x\to c}{lim}p(x)=p(x)$ so, $p$ is continuous on $\mathbb{R}$.

Rational Polynomials

as long as $q(c)\ne 0$
Generally rationals are not continuous at only a finite number of points.

Drichlet

Discontinuous everywhere

Let $c\in \mathbb{R}\setminus \mathbb{Q}$. For all $\delta >0$ there exists $x\in \mathbb{Q}\cap (c-\delta ,c+\delta ,)$ since $\mathbb{Q}$ is dense. Choose $\epsilon <1$, then $|f(x)-f(c)|=1\nless \epsilon$

Similar if $c\in \mathbb{Q}$

does there exist a function which is discontinuous on a dense set and continuous on a dense set?

Thomae

Proof:

Let $c=\frac{m}{n}$ and $gcd(m,n)=1$

Let $(x_n)$ be a sequence of irrational numbers tending to $c$.
So $f(x_n)=0$ and $f(x)=\frac{1}{n}$
Hence $f$ is not continuous at $c\in \mathbb{Q}$.

Let $c\in \mathbb{R}\setminus \mathbb{Q}$ let $\epsilon >0$
Suppose $x\in \mathbb{Q}$
Let $N\in \mathbb{N}$ s.t $\frac{1}{N}<\epsilon$

For $(c-1,c+1)$ there are only finitely many $\frac{m}{n}$ with $n<N$

Hence there exists $\delta >0$ s.t $(c-\delta ,c+\delta )$ contains only $\frac{m}{n}$ with $n>N$. Hence, if $\frac{m}{n}\in (c-\delta ,c+\delta )$ then

$f\left(\frac{m}{n}\right)=\frac{1}{n}<\frac{1}{N}$

So $|f\left(\frac{m}{n}\right)-f(c)|<\frac{1}{n}-\frac{1}{N}<\epsilon$

If $x$ is irrational then $f(x)=0$ for for any $\epsilon >0$ $|f(x)-f(c)|=0<\epsilon$

Q: Does there exist a function continuous on $\mathbb{Q}$ but discontinuous on $\mathbb{R}\setminus \mathbb{Q}$ ?

NO

Algebraic Continuity Theorems

Assume: $f:A\to \mathbb{R}$ and $g:A\to \mathbb{R}$ are continuous at $c\in A$ then:

i) $kf(x)$ is continuous at $c$ for all $k\in \mathbb{R}$
ii) $f(x)+g(x)$ is continuous at $c$
iii) $f(x)g(x)$ is continuous at c
iv) $\frac{f(x)}{g(x)}$ is continuous at $c$

Example:

$\sin x,\cos x$ are continuous on $\mathbb{R}$

Tool: $\sin x-\sin y=2\sin (\frac{1}{2}(x-y))\cos (\frac{1}{2}(x+y))$

Given $\epsilon >0$ choose $\delta <\epsilon$ then

Know $\cos x=\sin (x+\frac{\pi }{2})$

Thm: Composition of CTS Functions

Given $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$, assuming that $f(A)\subseteq B$. If $f$ is continuous at $c\in A$ and if $g$ is continuous at $f(c)\in B$ then $g\circ f$ is continuous at $c$

4.4 Continuous Functions on Compact Sets

Thm: Preservation of Compact Sets

$f:A\to \mathbb{R}$ is continuous. If $K\subseteq A$ is compact then $f(K)$ is compact

Proof

WTS that any sequence in $f(K)$ has a convergent subsequence whose limit is in $f(K)$.
Let $(y_n)$ be a sequence such that $\{y_n{\}}_{n=1}^{\mathrm{\infty }}\subseteq f(K)$. Each $y_n=f(x_n)$ for at least one $x_n\in K$. $K$ is compacts so there is subsequence $(x_{n_k})\to l\in K.$ $f(l)\in f(K)$. Since $f$ is continuous $f(x_{n_k})\to f(l)$

Fails if discontinuous
counter

$[-1,1]\to (\mathbb{R}\setminus \{(-1,1)\})\cup \{0\}$

Thm: Extreme Value Theorem (compact)

If $f:K\to \mathbb{R}$ is continuous on a compact set $K\subseteq \mathbb{R}$ then $f$ attains a maximum and minimum value. There exists $x_0,x_1\in K$ such that $f(x_0)\le f(x)\le f(x_1)$ for all $x\in K$

Proof

$f(K)$ is compact, thus it is closed and bounded. It also contains its limit points.
For every $n\in \mathbb{N}$ take

$(x_n)\to supf(K)⟹supf(K)$ is a max and exists in $f(K)$.

Uniform continuity

$f:A\to \mathbb{R}$ is uniformly continuous on $A$ if for every $\epsilon >0$ there exists $\delta >0$ so that for all $x,y\in A$

Example

Set $\delta =\frac{\epsilon }{2}$

Equivalencies

$f:A\to \mathbb{R}$
(i) $f$ is not uniformly continuous
(ii) $\mathrm{\exists }{ϵ}_0>0$ so that for all $\delta >0$ there is an $x,u\in A$ so that $|x-u|<\delta$ but $|f(x)-f(u)|>{\epsilon }_0$

(iii) $\mathrm{\exists }{ϵ}_0>0$ and $(x_n),(y_n)$ so that $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|>{ϵ}_0$

Proof

i) $⟹$ iii)
Negation is $f$ is not uniformly continuous on $A$ if and only if there exists a ${\epsilon }_0>0$ such that for all $\delta >0$ we can find two points $x$ and $y$ satisfying $|x-y|<\delta$ but with $|f(x)-f(y)|\ge {\epsilon }_0$. Thus, if we set ${\delta }_n=\frac{1}{n}$, then there exists point $x_1$ and $y_1$ where $|x_1-y_1|<\delta$ but $|f(x_1)-f(y_1)|\ge {\epsilon }_0$.

Other direction is obvious.

Thm: Compact and Uniform

let $f:K\to \mathbb{R}$ be cts then $f$ is uniform cts on $[a,b]$

Proof

Proof by contrapositive:
Suppose $f$ is not uniformly continuous

Then $\mathrm{\exists }\epsilon >0$ so that ${\delta }_n=\frac{1}{n}$ there exists $x_n,y_n\in [a,b]$ so that $|x_n-y_n|<{\delta }_n$ but $|f(x_n)-f(y_n)|>\epsilon$

$x_n$ is bounded $\stackrel{⟹}{BW}$ $\mathrm{\exists }(x_{n_k})\to z\in K$

So $(x_{n_k}),(y_{n_k})\to z\in [a,b]$ but $|f(x_{n_k})-f(y_{n_k})|>\epsilon$ so not cts at $z$

Thm: Continuous Extension

$f:(a,b)\to \mathbb{R}$ is uniform continuous then there exists a cts extensions $\tilde{f}$ on $[a,b]$

Lipschitz

$f:A\to \mathbb{R}$ is Lipschitz on $A$ if there exists $k>0$ so that

for all $x,y\in A$

Thm: Lipschitz then Uniformly CTS

4.4.9

4.5 Intermediate Value Theorem

Thm: Intermediate Value Theorem

If we have a continuous function $f:[a,b]\to \mathbb{R}$. If $L\in \mathbb{R}$, $f(a)<L<f(b)$ or $f(a)>L>f(b),$ then there exists a point $c\in (a,b)$ where $f(c)=L$

Proof:

Lemma:
$f:[a,b]\to \mathbb{R}$ is cts. Suppose $f(a)<0$ and $f(b)>0$ then there exist $x\in [a,b]$ with $f(x)=0$.
2.1 Bisection Method

Set $a_1=a$ and $b_1=b$, $x_n=\frac{b_n-a_n}{2}$

1. $f(x_n)<0⟹a_{n+1}=a$ and $b_{n+1}=x_n$
2. $f(x_n)>0⟹b_{n+1}=b$ and $a_{n+1}=x_n$
3. $f(x_n)=0⟹a_{n+1}=b_{n+1}=(x_n)$

$a_n,b_n,x_n\in [a_N,b_N]$ for all $n\ge N$

so $|a_n-a_m|,|b_n-b_m|,|x_n-x_m|\le |a_N-b_N|$
$⟹(a_n),(b_n),(x_n)$ is Cauchy $⟹$ converges

Converges to 0 by squeeze theorem.

This completes the proof of IVT using the Nested Interval Property.

Now take $f(a)<L<f(b)$
Take $g(x)=f(x)-L$
$⟹g(a)<0$ and $g(b)>0$

Better: Use NIP Similar idea:
$f(a)<0<f(b)$. Let $I_0=[a,b]$ and:

If $f(z)\ge 0$ then set $a_1=a$ and $b_1=z$. If $f(z)<0$ then $a_1=z$ and $b_1=b$. Continuing like above. $I_n\subseteq I_{n-1}$, $I_n\ne \mathrm{\varnothing }$ and let $c\in \bigcap _{n=1}^{\mathrm{\infty }}{I}_n.$

Suppose for contradiction that $f(c)<L$; then there must be some $ϵ>0$ so that $Vϵ(f(c))<L$. Since f is continuous, there must also be some $\delta$ so that $f(V\delta (c))<L$ - but this contradicts our construction in that one endpoint of In is mapped to a number greater than $L$, no matter how large n gets and how small $I_n$ is as a result. A similar argument shows $f(c)$ cannot be larger than $L$, and therefore $f(c)=L$.

Example

$f(x)=x{e}^x-2$
$f$ is cts on $[0,1]$ and $f(0)=-2$ , $f(1)=e-2>0$ so there exists a root

Example

If $c^2-\epsilon >0$ then $|x^2-c^2|<\epsilon$ iff

$|x-c_1|<\delta ⟹|f(x)-f(c_1)|<\epsilon$ iff $\delta \le \sqrt{1+\epsilon }-1\approx 0.2$
$|x-c_2|<\delta ⟹|f(x)-f(c_2)|<\epsilon$ iff $\delta \le \sqrt{9+\epsilon }-3\approx 0.08$