Continuity

A function $f(x)$ is said to be continuous at a point $c$ if

$$\lim_{x \rightarrow c}f(x)=f(c).$$

Loosely put this means that the value of the function at point $c$ is equal to what we would guess it should be when approaching $c$ from either side.

Figure 4: Examples of a removable discontinuity and a jump discontinuity.

A function is said to be continuous on a interval $[a,b]$ if it is continuous at every point on the interval $[a,b]$ and also $\lim_{x \rightarrow a+}=f(a)$ and $\lim_{x \rightarrow b-}=f(b)$. There are several different ways a function may fail to be continuous on an interval. Figures 4 and 5 show functions which are discontinuous on the interval $[0,1]$. In all the cases the functions are not continuous at $x=0.5$ but as we can see they are discontinuous in many different ways.

Figure 5: Two more types of discontinuities

The first function in figure 4 has a removable discontinuity. The function is

$$f(x)=\frac{2x^2-x}{2x-1}.$$

This is not defined at $x=0.5$ but everywhere else $f(x)=x$. The function just has a hole at $x=0.5$. The second function in figure 4 is

$$f(x)= \left\{ \begin{array}{lc} 2x^2 & x \leq 0.5 \\ 2-x & x > 0.5 \end{array} \right.$$

This function has a jump discontinuity. The limit does not exist because the limit from the left is

$$\lim_{x \rightarrow 0.5-}=0.5$$

but the limit from the right is

$$\lim_{x\rightarrow 0.5+}=1.5.$$

We see that $f(x)$ jumps at this point. The first function in figure 5 is

$$f(x)=\frac{1}{(x-0.5)^2}.$$

This function is discontinuous because as $x$ appraoches $0.5$ the value of the function goes to infinity. The final discontinuous function is

$$f(x)=\sin(\frac{1}{x-0.5}).$$

As $x \rightarrow 0.5$ this function oscillates more and more wildly between $\pm1$ and so the limit as $x$ approaches $0.5$ is not defined. Even if we were to zoom in on the function near $x=0.5$ we would never be able to say what the value of the function was at $x=0.5$.

Now that we have discussed what discontinuous functions are like lets concentrate on continuous functions. If a function is continuous on the interval $[a,b]$ we say it comes from the class of functions $C^0([a,b])$. If a function and its first derivative are both continuous on the interval $[a,b]$ we say that $f(x) \in C^1([a,b])$. Likewise if the function and its first $n$ derivatives are all continuous on the interval $[a,b]$ we say that $f(x) \in C^n([a,b])$.

Figure 6: A function that is $C([0,1])$ but not $C^1([0,1])$. We can see that although the function is continuous it is not 'smooth' because there is a sharp bend at $x=0.5$. When we look at the derivative of the function we can see that it is not continuous because there is a jump in the derivative at $x=0.5$.

Figure 6 gives an example of a function that is in $C^0([0,1])$ but not in $C^1([0,1])$. The function is

$$f(x) = \left\{ \begin{array}{lc} 2x^2 & x \leq 0.5 \\ \frac{x}{4}+0.375 \end{array}. \right.$$

From the picture of $f(x)$ we can see that the function itself is continuous on $[0,1]$ but there is a sharp bend at $x=0.5$. When we look at the derivative of $f(x)$ there is a jump discontinuity at $x=0.5$ so the first derivative is not continuous. Thus $f(x)$ is in $C^0([0,1])$ but not in $C^1([0,1])$.

Figure 7: A $C^1([0,1])$ function but not a $C^2([0,1])$ function, $f(x)=(x-0.5)^{1.3}$. Figure 8 shows the first two derivatives of $f(x)$.

Figure 7 shows an example of a function that is in both $C^0([0,1])$ and $C^1([0,1])$ but not in $C^2([0,1])$. The function is

$$f(x)=(x-0.5)^{1.3}.$$

We can see from Figure 7 that the function is continuous on $[0,1]$ which means that it belongs to the class $C^0([0,1])$. Figure 8 shows the first and second derivatives of $f(x)$. We can see that the first derivative is also continuous which means that $f(x) \in C^1([0,1])$. However the first derivative has a sharp point to it at $x=0.5$ and the the second derivative is discontinuous at $x=0.5$ so the function is not in $C^2([0,1])$.

Figure 8: The first and second derivatives of $f(x)=(x-0.5)^{1.3}$. The first derivative of $f(x)$ is continuous so $f(x) \in C^1([0,1])$ but we can see a sharp point at $x=0.5$. This becomes a discontinuity in the second derivative so $f(x)$ is not in $C^2([0,1])$.

If the function and all its derivatives are continuous on an interval $[a,b]$ we say that the function is in $C^\infty([a,b])$. An example of functions that are $C^\infty([a,b])$ are the polynomials. All polynomials are continuous on any bounded interval. The derivative of a polynomial is another polynomial and so also continuous on the same interval. Note that $f(x)=0$ is a great function in this sense. Not only is it continuous but its derivative is the same as the function. So $f(x)=0$ is very definitely in $C^\infty([a,b])$.

Another useful fact about functions that are in the class $C^\infty([a,b])$ is that they are also in $C^n([a,b])$ for all values $n$. In fact there is an ordering of the sets

$$C^\infty([a,b]) \subset C^n([a,b]) \subset C^{n-1}([a,b]) \... ...dots \subset C^2([a,b]) \subset C^1([a,b]) \subset C^0([a,b]).$$