Real analysis/Continuity

Definition

Now that we've defined the limit of a function, we're in a position to define what it means for a function to be continuous:

We say a function $f:A\to ℝ$ is continuous at c if $\underset{x\to c}{\lim }f(x)=f(c)$. We say $f$ itself is continuous if this condition holds for all points in A.

Conceptually, this definition means that continuous functions have no sudden jumps or oscillations. We will see several examples of discontinuous functions that illustrate the meaning of the definition.

Algebraic Operations

Since limits are preserved under algebraic operations, we see that if $f(x)$ and $g(x)$ are both continuous at c:

• $af(x)$ is continuous at $c$ for all $a\in ℝ$.
• $f(x)+g(x)$ is continuous at $c$.
• $f(x)g(x)$ is continuous at $c$.
• $f(x)/g(x)$ is continuous at $c$, assuming $g(c)$ is non-zero.

We can use sequential limits to prove that functions are discontinuous as follows:

• $f(x)$ is discontinuous at $c$ if and only if there are two sequences $(x_n)\to c$ and $(y_n)\to c$ such that $\underset{n\to \mathrm{\infty }}{\lim }(f(x_n))=\underset{n\to \mathrm{\infty }}{\lim }(f(y_n))$.

Composition

Another result that will allow us to construct many examples of continuous functions is that any composition of continuous functions is itself continuous:

Theorem

If $f:B\to ℝ$ and $g:A\to B$ are continuous, then the composition $(f\circ g)(x)=f(g(x))$ is continuous on A.

Proof

Let $ϵ>0$. Let $c\in A$.

Since f is continuous, $\mathrm{\exists }{\delta }_1>0:|x-c|<{\delta }_1⟹|f(x)-f(c)|<ϵ$.

Since g is continuous, $\mathrm{\exists }{\delta }_2>0:|x-c|<{\delta }_2⟹|g(x)-g(c)|<{\delta }_1$.

Thus $|x-c|<{\delta }_2⟹|g(x)-g(c)|<{\delta }_1⟹|f(g(x))-f(g(c))|<ϵ$, so $(f\circ g)(x)$ is continuous on A.

The Intermediate Value Theorem

This is the big theorem on continuity. Essentially it says that continuous functions have no sudden jumps or breaks.

Theorem (Intermediate Value Theorem)

Let f(x) be a continuous function. If $a<b$ and $f(a)<m<f(b)$, then $\mathrm{\exists }c\in (a,b):f(c)=m$.

Proof

Let $S=\{x\in (a,b):f(x)<m\}$, and let $c=\sup S$.

Let $ϵ=|f(c)-m|$. By continuity, $\mathrm{\exists }\delta :|x-c|<\delta ⟹|f(x)-f(c)|<ϵ$.

If f(c) < m, then $|f(c+\frac{\delta }{2})-f(c)|<ϵ$, so $f(c+\frac{\delta }{2})<f(c)+ϵ=m$. But then $c+\frac{\delta }{2}\in S$, which implies that c is not an upper bound for S, a contradiction.

If f(c) > m, then since $c=\sup S$, $\mathrm{\exists }x:x\in S,c>x>c-\delta$. But since $|x-c|<\delta$, $|f(x)-f(c)|<ϵ$, so $f(x)>f(c)-ϵ$ = m, which implies that $x\notin S$, a contradiction. $◻$

Construction of Rational Powers

We can now use what we know about continuity to construct rational powers of positive real numbers. We've already defined the integer powers; let's show they're continuous first.

Continuity of x^n

• $f(x)=x^0=1$ is continuous.

Given $ϵ>0$, $|f(x)-f(c)|=|1-1|=0<ϵ$. So, $\mathrm{\forall }\delta >0:|x-c|<\delta ⟹|f(x)-f(c)|<ϵ$.

• $f(x)=x$ is continuous.

Given $ϵ>0$, let $\delta =ϵ$. Then $|x-c|<\delta ⟹|x-c|<ϵ⟹|f(x)-f(c)|<ϵ$ .

• $f(x)=x^n$ is continuous for all $n\in \mathbb{N}$ and all $x\in ℝ$.

We proceed by induction. We have already seen that $f(x)=x^1$ is continuous. Assuming $f(x)=x^{n-1}$ is continuous, we use the that fact that continuity is preserved under algebraic operations to see that $xf(x)=x^n$ is continuous.

• $f(x)=x^{-n}$ is continuous for all $n\in \mathbb{N}$ and all $x\in ℝ\setminus 0$.

Since $x^n$ is continuous and nonzero on the set in question, $\frac{1}{x^n}=x^{-n}$ is continuous since continuity is preserved under division by a nonzero function.

We can now use the continuity of $x^n$ together with the intermediate value theorem to construct positive nth roots. As promised, this is much nicer than the construction of square roots in the first chapter:

Construction of nth roots

Given $c>0$, consider the function $f(x)=x^n-c$(it is clear that 0 has a unique nth root, so we do not consider this case). $f(0)=-c<0$ and since $1+c>1$, $f(1+c)=(1+c{)}^n-c>(1+c{)}^n-(1+c)>0$. By the Intermediate Value Theorem, $\mathrm{\exists }x\in (0,1+c):f(x)=x^n-c=0$. Thus c has a positive nth root.

To prove uniqueness, let x and y be two nth roots of c. If $x>y>0$, then $x^n>y^n>0$. But then it would follow that $c>c$, a contradiction. Similarly we cannot have $x<y$, so it follows that $x=y$.

Definition and Properties of Rational Powers

Given $n\in \mathbb{N}$ we define $x^{\frac{1}{n}}=\sqrt[n]{x}$ to be the unique nonnegative nth root of x. We then define all rational powers as follows:

If $r=\frac{m}{n}$ is in lowest terms(i.e. p and q have no common factors and $q>0$), we define $x^r=(\sqrt[q]{x}{)}^p$.

Our definition would work just as well if $\frac{p}{q}$ were not in lowest terms, as we'll see in a minute. First we must prove some basic facts:

• $\sqrt[m]{\sqrt[n]{x}}=\sqrt[mn]{x}$

Note that $(\sqrt[m]{\sqrt[n]{x}}{)}^{mn}=((\sqrt[m]{\sqrt[n]{x}}{)}^m{)}^n={\sqrt[n]{x}}^n=x$. Thus $\sqrt[m]{\sqrt[n]{x}}$ is an mn-th root of x. The result follows immediately from uniqueness of positive roots.

• $x^{\frac{p}{q}}=\sqrt[q]{x^p}={\sqrt[q]{x}}^p$

Using what we know about integer powers we see that $(x^{\frac{1}{q}}{)}^p)=\sqrt[q]{({\sqrt[q]{x}}^p{)}^q)}=\sqrt[q]{({\sqrt[q]{x}}^q{)}^p)}=\sqrt[q]{x^p}$

As promised, our definition does not depend on the fraction representing r:

• If $\frac{p}{q}=\frac{m}{n}$, then $x^{\frac{p}{q}}=\sqrt[n]{x^m}$.

If $\frac{p}{q}=\frac{m}{n}$, then $m=cp$ and $n=cq$ for some $c\in \mathbb{Z}$. Thus $\sqrt[n]{x^m}=({\sqrt[c]{\sqrt[q]{x}}}^c{)}^p=\sqrt[q]{x^p}=x^{\frac{p}{q}}$.

Now we'll prove the standard algebraic facts about rational powers:

• If $r,s\in \mathbb{Q}$ and $x>0$, then $x^r{x}^s=x^{r+s}$ and $(x^r{)}^s=x^{rs}$

Proof: Let $r=\frac{a}{b}$ and $s=\frac{c}{d}$. Then $x^r{x}^s=x^{\frac{a}{b}}{x}^{\frac{c}{d}}=x^{\frac{ad}{bd}}{x}^{\frac{bc}{bd}}=(x^{\frac{1}{bd}}{)}^{ad}(x^{\frac{1}{bd}}{)}^{bd}=(x^{\frac{1}{bd}}{)}^{ad+bc}=x^{\frac{ad+bc}{bd}}=x^{(r+s)}$

Also, $x^{rs}$ $=x^{\frac{a}{b}\frac{c}{d}}$ $=x^{\frac{ac}{bd}}$ $=(x^{ac}{)}^{\frac{1}{bd}}$ = $(((x^a{)}^c{)}^{\frac{1}{b}}{)}^{\frac{1}{d}}$ $=(((x^a{)}^{\frac{1}{b}}{)}^c{)}^{\frac{1}{d}}$$=(x^{\frac{a}{b}}{)}^{\frac{c}{d}}$

• If $r=\frac{a}{b}>0\in \mathbb{Q}$ and $y>x>0$, then $y^r>x^r>0$.

Proof: If $y^{\frac{1}{b}}\le x^{\frac{1}{b}}$, then $(y^{\frac{1}{b}}{)}^b\le (x^{\frac{1}{b}}{)}^b$, and $y\le x$, contradicting the assumption $y>x>0$. So $y^{\frac{1}{b}}>x^{\frac{1}{b}}>0$. Since a > 0, $y^{\frac{a}{b}}>x^{\frac{a}{b}}>0$. Thus $y^r>x^r>0$

Continuity of rational powers

Now we'll use the preceding algebraic properties to prove continuity of all rational powers:

• $f(x)=x^{\frac{1}{n}}$ is continuous for all $n\in \mathbb{N}$and $x\ge 0$.

Proof: Given $ϵ>0$, let $\delta =|c^{\frac{n-1}{n}}|ϵ$. Then $|x-c|<\delta ⟹$

$|x^{\frac{1}{n}}-c^{\frac{1}{n}}|=\frac{|x^{\frac{1}{n}}-c^{\frac{1}{n}}||x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}{c}^{\frac{1}{n}}+\cdots +x^{\frac{1}{n}}{c}^{\frac{n-2}{n}}+c^{\frac{n-1}{n}}|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}{c}^{\frac{1}{n}}+\cdots +x^{\frac{1}{n}}{c}^{\frac{n-2}{n}}+c^{\frac{n-1}{n}}|}$$=\frac{|x-c|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}{c}^{\frac{1}{n}}+\cdots +x^{\frac{1}{n}}{c}^{\frac{n-2}{n}}+c^{\frac{n-1}{n}}|}$$\le \frac{|x-c|}{|c^{\frac{n-1}{n}}|}<\frac{|c^{\frac{n-1}{n}}|ϵ}{|c^{\frac{n-1}{n}}|}=ϵ$.

The preceding argument works for $c=0$. If $c=0$, then let $\delta ={ϵ}^n$. Then:

$|x-0|<\delta ⟹|x|<{ϵ}^n⟹|x{|}^{\frac{1}{n}}<ϵ⟹|x^{\frac{1}{n}}-0|<ϵ$

So, $x^{\frac{1}{n}}$ is continuous for all $x\ge 0$.

• $x^q$ is continuous for all $q\in \mathbb{Q}$

Proof: If $q=\frac{a}{b}$, where a and b are integers and $b>0$, then $x^q=(x^a{)}^{\frac{1}{b}}$. Thus $x^q$ is the composition of continuous functions, and therefore is continuous itself.