# Continuous Functions

### Continuous Curves

In the last two posts we discussed about the system of real numbers and understood how the real numbers provide a model for the geometrical notion of a line which is continuous without any gaps. In this post and a few forthcoming ones we will try to create a model for continuous curves. Readers are already familiar with various continuous curves for example, circle, parabola and conic sections in general. The intuitive notion about the continuity of a curve is that we can draw in on a paper without lifting pen. This fact is so intuitive that many readers will feel that it is too much of an overkill to model it using some mathematical construct.

However as the reader will notice the mathematical model once formed is quite useful and it applies to many weird kinds of curves which are in fact not possible to draw easily on paper. To arrive at a precise notion of a continuous curve, we must be able to define in some precise way a curve and then later add feature of continuity to it. A basic notion is that a curve is made of points in a plane. But just any set of points can not be said to form a curve, let alone a continuous curve. It turns out that when we picture a curve in our mind we don't think of a set of isolated points, but rather intuitively assume that however small a portion of curve we look at, it must contain an infinity of points. Another intuitive property we associate with a curve is that if we take two points on a curve then the portion of the curve between these two points is a series of points starting from one end point and leading to the other end point.

Taking note of the above two properties we see that an interval possesses the same kind of properties and it represents a portion of the number line. To define a curve we need to have a set of points $C$ such that we can map each point of an interval $I$ to a unique point of $C$ such that no point of $C$ is left unmapped. Now if we use the language of coordinate geometry we observe that a point is represented by a pair of two real numbers which are called its coordinates. It is now easy to define the concept of curve as a set of points whose x-coordinates lie in some interval $I$ and whose y-coordinates are a function of the corresponding x-coordinate. To formalize we have the following definition of a curve:

Let $I$ be an interval of the real numbers and let $f$ be a function defined from $I$ to the set of real numbers $\mathbb{R}$. A curve is the set of points $C = \{ (x, y) \mid x \in I, y = f(x)\}$. We also say that the curve $C$ is the graph of the function $y = f(x)$ on interval $I$.

As a simple example $y = f(x) = x^{2}$ defined on $\mathbb{R}$ represents a parabola and $y = f(x) = ax + b$ represents a straight line.

Now that curves are being modeled using functions defined on intervals, it is only natural that continuous curves should be modeled by continuous functions. To define a continuous function we again go back to our intuition about continuous curves. Continuous curves can be drawn on paper without lifting pen and this means that we don't have to jump from one point to another point and in a very crude sense the next point on the curve is very close the previous point. Translating this in terms of function $y = f(x)$ defined on a certain interval $I$, this means that whenever we vary the value of $x$ by a little amount the value of $y = f(x)$ should also vary by a little amount. The above reasoning needs to be made precise. It turns out that in order to define continuous functions in an interval, it is first necessary to define continuity at a single point for a given function. Also if we are considering to define continuity at a point we must ensure that the function is defined not only at that point but also in a certain interval containing that point (if the function were just defined at only a single point, its graph would consist of an isolated point).

### Continuous Functions

Let $y = f(x)$ be a function defined on an interval $I$ and let $a$ be an interior point of $I$. The function $f(x)$ is said to be continuous at the point $a$ if for any arbitrarily given real number $\epsilon > 0$ there exists a real number $\delta > 0$ such that $|f(x) - f(a)| < \epsilon$ whenever $|x - a| < \delta$. If $f(x)$ is not continuous at point $a$ then we say that $f(x)$ is discontinuous at point $a$ or $a$ is a point of discontinuity of $f(x)$.

Clearly we can observe that $|f(x) - f(a)|$ represents the gap between value of function $f$ at point $a$ and its value at a nearby point $x$ and $|x - a|$ represents the gap between $x$ and $a$. So if the function $f$ is to be continuous at $a$, then we must be able to ensure that the value of $f$ changes by less than a given margin by choosing a suitable margin of values of $x$ near the point $a$. The definition might look complicated at the first sight to a beginner in these topics, but it is simplest definition which models our intuitive notion of a continuous curve. It is now a simple matter to define continuity of a function $f(x)$ in an interval $I$.

A function $f(x)$ defined on a certain interval $I$ is said to be continuous on (in) the interval $I$ if it is continuous at all the points of interval $I$. Continuity of $f(x)$ at any interior point of $I$ is defined above. Continuity of $f(x)$ at the left hand end point $a \in I$ requires that corresponding to any given $\epsilon > 0$, we should be able to find a real number $\delta > 0$ such that $|f(x) - f(a)| < \epsilon$ whenever $0 \leq x - a < \delta$. Similarly $f(x)$ is said to be continuous at the right hand end point $b \in I$ if corresponding to any given $\epsilon > 0$ there exists a number $\delta > 0$ such that $|f(x) - f(b)| < \epsilon$ whenever $0 \leq b - x < \delta$.

Common functions (e.g. $f(x) = x^{2}$ or $f(x) = 1/x$) which we normally see in practice are continuous at the points wherever they are defined. Constructing a discontinuous function in simplest cases requires that we have two distinct definitions of a function (i.e. one definition applying to certain points of the interval and another definition applying to the remaining points of the interval) and this looks quite artificial. Hence we will present an example of a function which is truly discontinuous.

Let us consider the function $f(x) = \sin (1/x)$. This is clearly defined for all values of $x$ except $0$. Hence it is discontinuous at $x = 0$. But the strange part is that even if we define it to have some value let's say $k$ when $x = 0$, then we can't make it continuous at $x = 0$. To understand that we just need to look at the values of the function $f(x)$ at points $x = 2/((4n + 1)\pi)$ (the value is $1$ at these points) and its values at points $x = 2/((4n + 3)\pi)$ (the value is now $-1$). Clearly if we take any interval around point $x = 0$, it will have some points from both the series of values of $x$ quoted above and hence the function will assume both the values $1$ and $-1$ and therefore $|f(x) - f(0)| = |f(x) - k|$ will be sometimes $|1 - k|$ and sometimes $|1 + k|$ and both of these cannot be made less than some positive $\epsilon$ at the same time. The functions keeps on jumping between values $-1$ and $1$ in any interval around $0$.

To modify the same example into a continuous function let's take $f(x) = x\sin(1/x)$. This is clearly not defined for $x = 0$, but for rest of the values of $x$ it is well defined. Lets give it a value $0$ at $x = 0$ i.e. $f(0) = 0$. It can be now seen that $f(x)$ is continuous at $x = 0$. Quite simply $|f(x) - f(0)| = |f(x)| = |x\sin(1/x)| \leq |x|$ which can definitely be made less than any positive $\epsilon$ by taking $| x - 0|$ less than a corresponding positive $\delta$ (in fact $\delta = \epsilon$ will suffice here).

### Properties of Continuous Functions

It turns out that the continuous function have many nice properties and we will study these now. We need to introduce a simple but technical term here before we proceed any further. If $a$ is an interior point of some interval $I$ then we say that $I$ is a neighborhood of point $a$.

First thing we must understand is that a function $f(x)$ continuous at a point $a$ is bounded in some neighborhood of $a$. The term bounded here refers to the fact that the values of function $f(x)$ in that neighborhood of $a$ form a set which is both bounded above and below.

This is simply because we can find a $\delta > 0$ such that $|f(x) - f(a)| < 1$ whenever $|x - a| < \delta$ In other words $f(a) - 1 < f(x) < f(a) + 1$ whenever $a - \delta < x < a + \delta$ so that $f(a) - 1$ is a lower bound (and $f(a) + 1$ an upper bound) for the values of $f(x)$ in the neighborhood $(a - \delta, a + \delta)$ of point $a$.

Another very important property is the following:
Let $f(x)$ be continuous at point $x = a$ and assume that $f(a) \neq 0$. Then $f(x)$ maintains its sign in a neighborhood of point $x = a$.

Again this is simple enough to prove. Assume that $f(a) > 0$. Then we can find a $\delta > 0$ such that $|f(x) - f(a)| < f(a) / 2$ whenever $|x - a| < \delta$. Thus $f(a) - f(a) / 2 < f(x) < f(a) + f(a)/2$ whenever $x \in (a - \delta, a + \delta)$. In other words $0 < f(a) / 2 < f(x)$ for all $x \in (a - \delta, a + \delta)$. Similarly we can handle the case when $f(a) < 0$.

We are now ready to establish the following properties:
Let $f(x)$ and $g(x)$ be two functions which are defined in a neighborhood of a point $x = a$ and let us also assume further that both the functions are continuous at $x = a$. Then
• $f(x) \pm g(x)$ is also continuous at $x = a$
• $k\cdot f(x)$ is also continuous at $x = a$
• $f(x)\cdot g(x)$ is also continuous at $x = a$
• $f(x) / g(x)$ is also continuous at $x = a$ provided $g(a) \neq 0$
We will establish the case of $f(x) \cdot g(x)$ and in the process the reader will get familiar enough with the technique to prove the other cases. Since $f(x), g(x)$ are continuous at $x = a$ they are both bounded in some neighborhood of $x = a$. Hence there is a neighborhood $I$ which can be taken in the form $(a - h, a + h)$ where $h > 0$ and there exist two positive numbers $F, G$ such that $|f(x)| < F$ and $|g(x)| < G$ whenever $x \in (a - h, a + h)$.

Let there be an arbitrary positive number $\epsilon$. We now know that there is a $\delta_{1} > 0$ and a $\delta_{2} > 0$ such that $|f(x) - f(a)| < \epsilon / G$ when $|x - a| < \delta_{1}$ and $|g(x) - g(a)| < \epsilon / F$ whenever $|x - a| < \delta_{2}$. Clearly if we call the lesser of the $\delta_{1}$ and $\delta_{2}$ as $\delta$ then both the inequalities (one for $f(x)$ and another for $g(x)$) hold in the neighborhood $(a - \delta, a + \delta)$.

Then \begin{align}|f(x)g(x) - f(a)g(a)| &= |f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)|\notag\\ &\leq |f(x)g(x) - f(x)g(a)| + |f(x)g(a) - f(a)g(a)|\notag\\ &= |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)|\notag\\ &< F \cdot (\epsilon / F) + G \cdot (\epsilon / G) = 2\epsilon\notag \end{align} Since $\epsilon > 0$ was arbitrary it follows that the function $f(x)g(x)$ is continuous at $x = a$.

Now the function $f(x) = x$ is clearly continuous everywhere and hence by the above properties we can see that all the polynomials in $x$ are continuous everywhere. Similarly it is an easy exercise to prove that $\sin x$ and $\cos x$ are continuous everywhere.

The above properties which we have studied are all local properties of continuous functions (i.e. they talk about continuity at a single points and this depends upon values of function in a neighborhood of that point). In the next post we will extending some of these properties to continuous functions defined on an interval.