## Some properties of continuous functions

Theorem:

If $g,f : [a,b] \rightarrow \Re$ are continuous functions and c is a constant, then

a) $f+g$ is a continuous functions.

b) $f-g$ is a continuous functions.

c) cf is a continuous function.

d) fg is a continuous function.

Proof:

We shall only prove statement d. Choose and fix any $\varepsilon >0$. Since is continuous at $x_{0}$, we have that for the positive number $\frac{\varepsilon}{(2|g(x_{0})|+1}$ there exists a $\delta_{1}>0$ such that

$|f(x)-f(x_{0}|< \frac{\varepsilon}{2(|g(x_{0}|+1)}$ whenever $|x-x_{0}|<\delta_{1}$

Since $||f(x)|-|f(x_{0})|| \leq |f(x)-f(x_{0})|$, we conclude that

$|f(x)|<|f(x_{0})|+\frac{\varepsilon}{2(|g(x_{0})|+1)}$, whenever $|x-s_{0}|<\delta_{1}$. Let $|f(x_{0})|+\varepsilon 2(|g(x_{0})|+1)=M$. Also, since g is continuous at $x_{0}$, for the positive number $\frac{\varepsilon}{2M}$, there is a $\delta_{2}>0$ such that $|g(x)-g(x_{0})|< \frac{\varepsilon}{2M}$ whenever $|x-x_{0}|<\delta_{2}$. Put $\delta=min(\delta_{1},\delta_{2})$. Then, whenever $|x-x_{0}|<\delta$, we have

$|f(x)g(x)-f(x_{0})g(x_{0})|$ equals

$|f(x)g(x)-f(x)g(x_{0})+f(x)g(x_{0})-f(x_{0})g(x_{0})|$

$\leq |f(x)||g(x)-g(x_{0})|+|g(x_{0})(f(x)-f(x_{0}))|$

which equals

$|f(x)||g(x)-g(x_{0})|+|g(x_{0})||f(x)-f(x_{0})|$

$< M.\frac{\varepsilon}{2M}+|g(x_{0})|.\frac{\varepsilon}{2(|g(x_{0}|+1)}$

which equals $\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

Observe that we have not claimed that the quotient of two continuous functions is continuous. The problem is obvious: $\frac{f(x)}{g(x)}$ cannot have any meaning at x for which $g(x)=0$. So, the question would be, if $g(x) \neq 0$ for every $x \in [a,b]$, is the function $h:[a,b] \rightarrow \Re$, defined by $h(x)=\frac{f(x)}{g(x)}$, continuous? The answer is yes. For a proof, we need a preliminary result.

Lemma.

if $g:[a,b] \rightarrow \Re$ is continuous and $g(x_{0}) \neq 0$, then there is an $m > 0$ and $\delta >0$ such that if $x_{0}-\delta, then $|g(x)|>m$.

Proof.

Let $|g(x_{0})|=2m$. Now, $m>0$. By continuity of g, there is a $\delta>0$ such that

$|g(x)-g(x_{0})| for $x_{0}-\delta

But, $|g(x)-g(x_{0})| \geq ||g(x)|-|g(x_{0})||$ and hence, $-m<|g(x)|-|g(x_{0})|<m$, giving us

$m=|g(x_{0})|-m<|g(x)|$ for $x_{0}-\delta. Hence, the proof.

The lemma says that if a continuous function does not vanish at a point, then there is an interval containing it in which it does not vanish at any point.

Theorem.

If $f,g :[a,b] \rightarrow \Re$ are continuous and $g(x) \neq 0$ for all x, then $h:[a,b] \rightarrow \Re$ defined by $h(x)=\frac{f(x)}{g(x)}$ is continuous.

The proof of the above theorem using the lemma above is left as an exercise.

Examples.

a) $f:\Re \rightarrow \Re$ defined by $f(x)=a_{0}$ for all $x \in \Re$, where $a_{0}$ is continuous.

b) $f:\Re \rightarrow \Re$ defined by $f(x)=x$ is continuous.

c) $g:\Re \rightarrow \Re$ defined by $g(x)=x^{2}$ is a continuous function because $g(x)=f(x)f(x)$, where $f(x)=x$. Since f is continuous by (b), g must be continuous.

d) $h:\Re \rightarrow \Re$ by $h(x)=x^{n}$, n being a positive integer, is continuous by repeated application of the above reasoning.

e) $p: \Re \rightarrow \Re$ defined by $p(x)=a_{0}+a_{1}x+\ldots +a_{n}x^{n}$, where $a_{0}, a_{1}, \ldots , a_{n}$ is also continuous. This is because of the fact that if

$f_{1}, f_{2}, f_{3} \ldots, f_{n}:\Re \rightarrow \Re$ are defined by $f_{1}(x)=x$, $f_{2}=x^{2}$, …, $f_{n}=x^{n}$, then $a_{1}f_{1}, a_{2}f_{2}, \ldots a_{n}f_{n}$ are also continuous functions. Hence,

$a_{0}+a_{1}f_{1}+ \ldots +a_{n}f_{n}=p$ is also a continuous function as the sum of continuous functions is a continuous function. Thus, we have shown that a polynomial is a continuous function.

f) Let p and q be polynomials. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in \Re$ be such that $q(\alpha_{1})=q(\alpha_{2})=\ldots=q(\alpha_{n})$  and $q(\alpha) \neq 0$ for $\alpha \neq \alpha_{1}, \alpha \neq \alpha_{2}, \ldots, \alpha neq \alpha_{n}$.

Now, let $D = \Re - \{ \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}\}$.

Then, $h:D \rightarrow \Re$ defined by $h(x)=\frac{p(x)}{q(x)}$ is a continuous function. What we have said is that a rational function which is defined everywhere except on the finite set of zeroes of the denominator is continuous.

g) $f:\Re \rightarrow \Re$ defined by $f(x)=\sin{x}$ is continuous everywhere. Indeed, $f(x)-f(x_{0})=\sin{x}-\sin{x_{0}}=2\sin{\frac{x-x_{0}}{2}}\cos{\frac{x+x_{0}}{2}}$. Therefore,

$|f(x)-f(x_{0})|=2|\sin{\frac{(x-x_{0})}{2}}| |\cos{\frac{(x+x_{0})}{2}}|\leq |x-x_{0}|$ (because $|\sin{x}| \leq |x|$, where x is measured in radians)

h) $f:\Re \rightarrow \Re$ defined by $f(x)=\cos{x}$ is continuous since

$|f(x)-f(x_{0})|=|\cos{x}-\cos{x_{0}}|=2|\sin{\frac{(x_{0}-x)}{2}}\sin{\frac{x+x_{0}}{2}}| \leq |x-x_{0}|$

i) $f:\Re - \{ (2n+1)\frac{\pi}{2}: n \in \mathbf{Z}\} \rightarrow \Re$ defined by $f(x)=\tan{x}$ is continuous. We had to omit numbers like $\ldots, \frac{-3\pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ from the domain of f as $\tan{x}$ cannot be defined for these values of x.

j) $f:\Re_{+} \rightarrow \Re$ defined by $f(x)=x^{1/n}$ is a continuous function. Indeed,

$f(x)-f(a)=x^{1/n}-a^{1/m}$ which equals

$\frac{(x-a)}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}\frac{1}{a^{n}}+\ldots +a^{\frac{n-1}{n}} }$

Choose $|x-a|<|a/2|$ to start with, so that $|a/2|<|x|<(3/2)|a|$. Thus,

$|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}a^{1/n}+\ldots+a^{\frac{n-1}{n}}|>|a|^{\frac{n-1}{n}} \times ((1/2)^{\frac{n-1}{n}}+(1/2)^{\frac{n-2}{n}}+\ldots+1)$

Given an $\varepsilon >0$, let

$\delta=min\{\frac{|a|}{2}, \varepsilon \times |a|^{\frac{n-1}{n}} \times \left( (1/2)^{\frac{n-1}{n}}+\ldots+1 \right)\}$.

Then, for $|x-a|<\delta$, we have

$|f(x)-f(a)|=\frac{|x-a|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}} \times a^{1/n}+\ldots+a^{\frac{n-1}{n}}|}< \varepsilon$.

It can be shown that f defined by $f(x)=x^{r}$ is also a continuous function for every real $r \in \Re$.

k) Consider the function $f:\Re \rightarrow \Re$ defined by $f(x)=a^{x}$. Is f a continuous function? This is left as an exercise. (Hint: It will suffice to prove continuity at $x=0$. This would follow from $\lim_{m \rightarrow \infty}a^{1/m}$).

k) Suppose $f:\Re \rightarrow \Re$ is defined by $f(x)=1/x$, if $x \neq 0$ and $f(0)=0$. We can see that f is not continuous at 0 as $f(x)$ changes abruptly when x goes over from negative to positive values.

More later,

Nalin Pithwa

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