Tag Archives: continuity

Some properties of continuous functions : continued

We now turn to the question whether $f(x_{n})$ approximates $f(x)$ when $x_{n}$ approximates x. This is the same as asking: suppose $(x_{n})_{n=1}^{\infty}$ is a sequence of real numbers converging to x, does $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$?

Theorem.

If $f: \Re \rightarrow \Re$ is continuous at $x \in \Re$ if and only if $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$ whenever $(x_{n})_{n=1}^{\infty}$ converges to x, that is,

$\lim_{n \rightarrow \infty}f(x_{n})=f(\lim_{n \rightarrow \infty} x_{n})$.

Proof.

Suppose f is continuous at x and $(x_{n})_{n=1}^{\infty}$ converges to x. By continuity, for every $\varepsilon > 0$, there exists $\delta >0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. Since $(x_{n})_{n=1}^{\infty}$ converges to x, for this $\delta >0$, we can find an $n_{0}$ such that $|x-x_{0}|<\delta$ for $n > n_{0}$.

So $|f(x)-f(x_{n})|<\varepsilon$ for $n > n_{0}$ as $|x_{n}-x|<\delta$.

Conversely, suppose $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$ when $(x_{n})_{n=1}^{\infty}$ converges to x. We have to show that f is continuous at x. Suppose f is not continuous at x. That is to say, there is an $\varepsilon >0$ such that however small $\delta$ we may choose, there will be a y satisfying $|x-y|<\delta$ yet $|f(x)-f(y)|\geq \varepsilon$. So for every n, let $x_{n}$ be such a number for which $|x-x_{n}|<(1/n)$ and $|f(x)-f(x_{n})| \geq \varepsilon$. Now, we see that the sequence $(x_{n})_{n=1}^{\infty}$ converges to x. But $(f(x_{n}))_{n=1}^{\infty}$ does not converge to f(x) violating our hypothesis. So f must be continuous at x. QED.

More later,

Nalin Pithwa

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

Limits and Continuity — reblogging

We have seen many examples of functions earlier. Let us also consider the example of the range of r of a gun pointed at an angle $\theta$ to the horizon. Gun, shell and other conditions remaining the same, we know that for every angle $\theta$ we have a definite range $r(\theta)$, giving rise to the function r. Suppose we know that the target is at a distance d from the gun and that $r(\theta_{0})=d$ for some $\theta_{0}$. Then, we point our gun at an angle $\theta_{0}$ to the horizon to hit the target. A little deviation in $\theta$ in likely to cause an error in our hit. But, we also know that if our hit is within a certain distance from the target, then our objective is achieved (depending on the shell). Now, if want the hit to be within a distance $\varepsilon > 0$ from the target, can we adjust the deviation $\delta$ in $\theta$ around $\theta_{0}$ accordingly? Another way of putting the same question is to ask if for a small change in angle $\theta, can one get a small change in range$latex r(\theta)$? If yes, then we are allowed a little play in aiming our gun. If not, a small play might result in a wide miss. What we are asking is: for every $\varepsilon >0$, can we find a$\delta >0$such that $|r(\theta)-r(\theta_{0})|<\varepsilon$ whenever $|\theta -\theta_{0}|<0$? We shall come across similar questions in other situations too. For example, perhaps, you might be knowing how to obtain the values of $\pi$ through the series called Madhava-Gregory’s series. That method takes a long time to evaluate the value of $\pi$ correctly to, say, 4 places of decimal. There is also the formula: $\frac{\pi^{4}}{90}=\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots+\frac{1}{n^{4}}+\ldots$. To find the sum of the above series correct to, say, 4 places of decimal, it is enough to sum the first 25 terms. But the sum gives an approximate value of $\pi^{4}/90$. To get the value of$\pi$, we need to multiply the sum by $90$ and then extract its fourth root. Now the question is how would the error committed in evaluating $\pi^{4}/90$ be propagated in the subsequent calculations? Put in a different way, let us write $x=\frac{\pi^{4}}{90}$ and $f(x)=(90x)^{1/4}$, and let $x_{n}$ b/ e the approximate value of $\pi$ calculated by summing the first n terms of the series. So, the approximate value of$\pi\$ would be $f(x_{n})$. This leads naturally to the question: how  is the error $|f(x)-f(x_{n})|$ in the value of $f(x)$ related to the error $|x-x_{n}|$ in the value of x? Can we calculate $f(x)$ correct to the desired accuracy by calculating x sufficiently accurately? If not, then this method of calculation is not very useful. If a small perturbation in x causes an abrupt change in $f(x)$, then we should perhaps think of some other method of calculation. It may be noted that, philosophically, continuity forms the basis of large parts of the experimental sciences where it is tacitly assumed that small errors in measurement will not lead to drastic changes in conclusions.

These ideas lead to  the definition of continuity of functions.

Definition. let $f: \Re \rightarrow \Re$ be a function. W say that the function is continuous at $x_{0} \in \Re$ if for every $\varepsilon >0$, we can find a $\delta>0$ such that $|f(x)-f(x_{0}| , \varepsilon$ whenever $|x-x_{0}|<\delta$. This is to say that for continuous functions, f, the value of $f(x)$ can be restricted within the interval $(f(x_{0})-\varepsilon, f(x_{0})+\varepsilon)$ by restricting the value of x within $(x_{0}-\delta, x_{0}+\delta)$. Try to do draw a figure based on this!!

Note that for continuity at $x_{0}$. we need an interval containing $x_{0}$ to be contained in its domain, and hence, it is enough that the function f has its domain an interval in $[a,b]$. So we can define continuity of a function $f:[a,b] \rightarrow \Re$ in the same way as above. In the case of the end points a and b, we can only talk of  the right and left continuity, respectively.

More later,

Nalin Pithwa