Continuous Functions

Definition

if a function $f$ is continuous on $(a,b)$, it’s saying that for any $x\_0\in (a,b)$, s.t.:

$$\lim \_x\to x\_0f(x)=f(x\_0)$$

Boundedness Theorem

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

Proof with Bolzano-Weierstrass Theorem.

Suppose that: the function boundless on $\text{[}a,b]$

If the function is boundless, loop this process, starts with $I\_0=\text{[}a,b]$:

1. bisect the interval $I\_n$ with $c=\frac{a+b}{2}$
2. select the boundless half $I\_n+1=$ $\text{[}a,c]$ or $\text{[}c,b]$
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 $x\_0\in \text{[}a,b]$ (Bolzano Weierstrass Theorem)
2. $f(x\_n)$ is unbounded, which contradicts to $\lim \_x\to x\_0f(x)=f(x\_0)$ (Heine Theorem)

So the function is bounded on $\text{[}a,b]$

Intermediate Value Theorem

IVT

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

Proof

Build nested intervals with this process:

1. bisect the interval $\text{[}a\_n,b\_n]$ with $c=\frac{a+b}{2}$
2. if $f(c)=u$, stop.
3. select the half $\text{[}a\_n+1,b\_n+1]$ 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 $\text{[}a,b]$, with Heine’s Theorem:

\begin{align} \lim\_{ n \to \infty } f(a\_{n}) & = f(\lim\_{ n \to a\_{n} } a\_{n}) = f(c) \\ \lim\_{ n \to \infty } f(b\_{n}) & = f(\lim\_{ n \to b\_{n} } b\_{n}) = f(c) \\ \implies \lim\_{ n \to \infty } \left\[ f(a\_{n}) - f(b\_{n}) \right] & = \lim\_{ n \to \infty } f(a\_{n}) - \lim\_{ n \to \infty } f(b\_{n}) = 0 \end{align}

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 $\text{[}a,b]$, then there’s a maxima and minima in $\text{[}a,b]$.

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

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

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

$$f^{\prime }(x\_0)=\lim \_x\to x\_0\frac{f(x)-f(x\_0)}{x-x\_0}$$

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 $\text{[}a,b]$ and $f(a)=f(b)$, then: $\exists \xi \in \text{[}a,b],f^{\prime }(\xi )=0$

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

Lagrange MVT

Lagrange MVT

If function $f$ is continuous on $\text{[}a,b]$, then:

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

, 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, $\exists \xi \in \text{[}a,b],F^{\prime }(\xi )=0$

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

Cauchy MVT

Cauchy MVT

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

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

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

\begin{align} 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{align}

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 $\text{[}a,b]$, 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 $\text{[}a,b]$, so with Boundedness Theorem, $f(x)$ is bounded:

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

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:

Prove it by MVT for Integrals:

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

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 $F$:

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

Prerequisite of differentiable

it must be continuous to be derivable, or equivalently, differentiable.

if $f$ is differentiable, then $\Delta y$ is also an infinitesimal — which implies that $f$ is continuous.

$$\lim \_\Delta x\to 0\Delta y=\lim \_\Delta x\to 0\left(f^{\prime }(x)\Delta x+o(\Delta x)\right)=0$$