Continuous Functions

if a function $f$ is continuous on $(a,b)$, it’s saying that for any $x_0\in (a,b)$, s.t.:

$$\underset{x\to x_0}{\lim }f(x)=f(x_0)$$

Boundedness Theorem

If a function is continuous on a close-interval $[a,b]$, then its bounded on that interval.

Proof with Bolzano-Weierstrass Theorem.

Suppose that: the function boundless on $[a,b]$

If the function is boundless, loop this process, starts with $I_0=[a,b]$:

1. bisect the interval $I_n$ with $c=\frac{a+b}{2}$
2. select the boundless half $I_{n+1}=$ $[a,c]$ or $[c,b]$
3. select a point $x_n\in I_{n+1}$ making this half boundless to build a sequence $\left\{x_n\right\}$
4. continue this process with $I_{n+1}$

And finally we have a infinite bounded sequence $\left\{x_n\right\}$ and an unbounded sequence $f(x_n)$ s.t.:

1. converge to some number $x_0\in [a,b]$ (Bolzano Weierstrass Theorem)
2. $f(x_n)$ is unbounded, which contradicts to ${\lim }_{x\to x_0}f(x)=f(x_0)$ (Heine Theorem)

So the function is bounded on $[a,b]$

Intermediate Value Theorem

IVT

If a function is continuous on a close-interval $[a,b]$, then for any intermediate value $u\in [f(a),f(b)]$, there’s some value $c\in [a,b]$ making that $f(c)=u$

Proof

Build nested intervals with this process:

1. bisect the interval $[a_n,b_n]$ with $c=\frac{a+b}{2}$
2. if $f(c)=u$, stop.
3. select the half $[a_{n+1},b_{n+1}]$ making $u\in \left[f(a_{n+1}),f(b_{n+1})\right]$
4. continue this process

So we get a nested interval $\left\{a_n,b_n\right\}$, which locate exactly one real number $c$, and $a_n\to c,b_n\to c$ (Nested Interval Theorem). As the function is continuous on $[a,b]$, with Heine’s Theorem:

$$\begin{array}{rl}\underset{n\to \infty }{\lim }f(a_n)& =f(\underset{n\to a_n}{\lim }{a}_n)=f(c)\\ \underset{n\to \infty }{\lim }f(b_n)& =f(\underset{n\to b_n}{\lim }{b}_n)=f(c)\\ ⟹\underset{n\to \infty }{\lim }\left[f(a_n)-f(b_n)\right]& =\underset{n\to \infty }{\lim }f(a_n)-\underset{n\to \infty }{\lim }f(b_n)=0\end{array}$$

So we have another set of nested intervals $\left\{f(a_n),f(b_n)\right\}$ that locate a unique number, which can only be $u$ since $u\in \left[f(a_n),f(b_n)\right],\forall n\in \mathbb{Z}$. (Nested Interval Theorem)

Extreme Value Theorem

Extreme Value Theorem

If a function is continuous on $[a,b]$, then there’s a maxima and minima in $[a,b]$.

As Boundedness Theorem said, the function is bounded, so the function has a supremum and infimum, w.r.t. Dedekind Completeness Theorem. If the supremum is $M$.

Suppose $\forall x\in [a,b],f(x)<M$. $g(x)=\frac{1}{M-f(x)}$ is also a continuous function, thus it’s also bounded. Suppose $K>0$ is an upper bound of $g(x)$ on $[a,b]$:

$$\begin{array}{r}g(x)=\frac{1}{M-f(x)}\le K\\ ⟹f(x)\le M-\frac{1}{K}\end{array}$$

so $M-\frac{1}{K}$ is also an upper bound for $f(x)$ on $[a,b]$, which contradicts to the setting that $M$ is the supremum. So there must be some value $c\in [a,b]$ s.t. $f(c)=M\ge f(x)$.

Mean Value Theorems

Fermat’s Lemma

Fermat's Lemma

derivative on maxima / minima must be zero

Proof:

Suppose $x_0$ is a maxima.

$$f^{\prime }(x_0)=\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}$$

It said that:

$$\begin{array}{r}\forall ϵ>0,\exists \delta >0,\forall x\in (x_0-\delta ,x_0+\delta ),\left|\frac{f(x)-f(x_0)}{x-x_0}-A\right|<ϵ\end{array}$$

Rolle’s MVT

Rolle's MVT

if function $f$ is continuous on $[a,b]$ and $f(a)=f(b)$, then: $\exists \xi \in [a,b],f^{\prime }(\xi )=0$

A function is continuous on $[a,b]$, then with Extreme Value Theorem, there must be a maxima $\xi \in [a,b]$, and then with Fermat’s Lemma, $f^{\prime }(\xi )=0$

Lagrange MVT

Lagrange MVT

If function $f$ is continuous on $[a,b]$, then:

$$\exists \xi \in [a,b],f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$$

, which saids the derivative on some intermediate point concludes the change of endpoint.

Proof:

build a flat function $F$ containing $f(x)$ to apply Rolle’s MVT:

$$F(x)=(f(x)-f(a))-\frac{f(b)-f(a)}{b-a}(x-a)$$

So that $F(b)=F(a)=0$

So with Rolle’s MVT, $\exists \xi \in [a,b],F^{\prime }(\xi )=0$

$$f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$$

Cauchy MVT

Cauchy MVT

if function $f,g$ are both continuous on $[a,b]$, then:

$$\exists \xi \in [a,b],\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}=\frac{f(b)-f(a)}{g(b)-g(a)}$$

construct a function whose derivative is in the form that fit the theorem:

$$\begin{array}{rl}h(x)& =f(x)\left[g(b)-g(a)\right]-g(x)\left[f(b)-f(a)\right]\\ h(a)& =f(a)g(b)-f(b)g(a)\\ h(b)& =f(a)g(b)-f(b)g(a)\end{array}$$

so by Rolle’s MVT:

$$\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}=\frac{f(b)-f(a)}{g(b)-g(a)}$$

MVT for Integrals

MVT for Integrals

If function $f$ is continuous on $[a,b]$, then the mean integral is an intermediate value of $f(x)$

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx=f^{\prime }(\xi )$$

$f(x)$ is continuous on $[a,b]$, so with Boundedness Theorem, $f(x)$ is bounded:

$$\begin{array}{r}{\int }_a^{b}mdx\le {\int }_a^{b}f(x)dx\le {\int }_a^{b}Mdx\\ \frac{1}{b-a}{\int }_a^{b}f(x)dx\in [m,M]\end{array}$$

then with IVT:

$$f(c)=\frac{1}{b-a}{\int }_a^{b}f(x)dt$$

Fundamental Theorem of Calculus

if $f$ is continuous at $x=a$

accumulation function is one of the antiderivative

$${\left({\int }_a^{x}f(t)dt\right)}^{\prime }=f(x)$$

Integration is difference:

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

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Proof:

$$\frac{{\int }_a^{x+\Delta x}f(t)dt-{\int }_a^{x}f(t)dt}{\Delta x}=\underset{\Delta x\to 0}{\lim }f(\xi )=f(\underset{\Delta x\to 0}{\lim }\xi )=f(a)$$

So any original function of $f(x)$, denoted by $F(x)$, can be represented by adding some constant $C$ to ${\int }_a^{x}f(t)dt$:

$$F(x)={\int }_a^{x}f(t)dt+C$$

substituting $x$ with $a$, we know the constant is $F(a)$

$$F(a)={\int }_a^{a}f(t)dt+C=C$$

substituting $x$ with $b$, then the integration of $f(t)$ from $a$ to $b$ is the difference of the value of the endpoints of any antiderivative $F$:

$${\int }_a^{x}f(t)dt=F(x)-F(a)⟹{\int }_a^{b}f(t)dt=F(b)-F(a)$$