Some properties of continuous functions


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.


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


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

which equals


< 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.


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<x<x_{0}+\delta, then |g(x)|>m.


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

|g(x)-g(x_{0})|<m for x_{0}-\delta<x<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<x<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.


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.


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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