Properties of continuous functions

A function

$f(x)\;:K \to M$

is a rule that assigns each element $x \in K$ the only element $y \in M$ . The set where the function is defined is called the domain of the function. On the other hand, if a set S is from the domain of f(x) then

f(S)

denotes the image of S. In other words,

$f(S) = \{y\in M\;;\;\;\mbox{ such that } \exists\;x\in K\;\;\mbox{ and } f(x)=y \}$

In this section we are interested in real functions that are defined on the real line $\mathbb{R}$ and take real values. Our interest is focused on those functions that are called continuous.

Definition. 1 A real function f(x) is called continuous at a point

$x_0 \in \mathbb{R}$

  if, and only if,

$\forall \varepsilon > 0\;\;\;\exists\;\;\delta > 0\;\;\mbox{ such that } \mid x - x_0 \mid < \delta \;\;\Longrightarrow \;\;\; \mid f(x)- f(x_0) \mid < \varepsilon$

Real continuous functions have a wide range of important for application properties. The next theorem contains some of the basic algebraic properties of continuous functions.

  
Theorem 1. If the real functions f(x) and g(x) are continuous at a point x₀ 
 then so are the functions:

$f(x) \pm g(x),\;\;f(x)\cdot g(x) \;\;\mbox{ and }\;\; \frac{f(x)}{g(x)}\;\;\mbox{ for } x_0\;\;\mbox{ such that } g(x_0) \not= 0$

Proof. Let us show that $f(x) \pm g(x)$ is continuous at x₀. Since f(x) and g(x) are continuous at x₀, for any given

$\varepsilon > 0$

we can find

$\delta_f\;\;\mbox{ and } \delta_g$

such that

$\mid x - x_0 \mid < \delta_f \;\;\Longrightarrow \;\;\; \mid f(x)- f(x_0) \mid < \frac{\varepsilon }{2}$

and

$\mid x - x_0 \mid < \delta_g \;\;\Longrightarrow \;\;\; \mid g(x)- g(x_0) \mid < \frac{\varepsilon }{2}$

Hence, for any

$\varepsilon > 0$

we have

$\mid(f(x) \pm g(x)) - (f(x_0) \pm g(x_0)) \mid \le \mid f(x) - f(x_0) \mid + \mid g(x) - g(x_0) \mid < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon$

as long as

$\mid x - x_0 \mid < \min\{\delta_f, \delta_g \}$

Thus, for any

$\varepsilon > 0$

there exists

such that

$\mid(f(x) \pm g(x)) - (f(x_0) \pm g(x_0)) \mid < \varepsilon$

as long as

$\mid x - x_0 \mid < \delta$

We proved that

$f(x) \pm g(x)$

is continuous function at x₀.

Now we show that

$f(x) \cdot g(x)$

is continuous at x₀. First, we notice that the function g(x) locally bounded at x₀. Indeed, for $\varepsilon =1$ there exists such that

$\mid x - x_0 \mid < \delta_1 \;\;\Longrightarrow \;\;\mid g(x) - g (x_0)\mid < 1$

Hence,

$\mid x - x_0 \mid < \delta_1 \;\;\Longrightarrow \;\;\mid g(x) \mid < 1 + \mid g(x_0) \mid$

On the other hand, since f(x) and g(x) are continuous at x₀ we can choose

$\delta_f > 0 \;\;\mbox{ and }\;\;\delta_g > 0$

so that

$\mid x - x_0 \mid < \delta_f \;\;\Longrightarrow \;\;\mid f(x) - f(x_0) \mid < \frac{ \varepsilon }{2 \cdot G }$

and

$\mid x - x_0 \mid < \delta_g \;\;\Longrightarrow \;\;\mid f(x_0)\mid \cdot\mid g(x) - g(x_0) \mid < \frac{\varepsilon }{2 }$

Notice that

$\mid f(x) \cdot g(x) - f(x_0)\cdot g(x_0)\mid = \mid f(x) \cdot g(x)-f(x_0)\cdot g(x) + f(x_0)\cdot g(x) - f(x_0)\cdot g(x_0)\mid \le$

$\mid g(x) \mid \cdot \mid f(x)-f(x_0)\mid + \mid f(x_0) \mid \cdot \mid g(x) - g(x_0)\mid$

Now choose

$\delta = \min\{\delta_1,\; \delta_f, \; \delta_g \}$

Then it follows from

$\mid x - x_0 \mid < \delta$

that

$\mid f(x) \cdot g(x) - f(x_0)\cdot g(x_0)\mid < G\cdot \mid f(x)-f(x_0)\mid + \mid f(x_0) \mid \cdot \mid g(x) - g(x_0)\mid < G \cdot \frac{\varepsilon }{2\cdot G } + \frac{\varepsilon }{2 } = \varepsilon$

Hence, we established that

$\forall \varepsilon > 0\;\;\;\exists\;\;\delta>0$

such that

$\mid x - x_0 \mid < \delta\;\;\;\Longrightarrow \;\;\mid f(x)\cdot g(x) - f(x_0)\cdot g(x_0) \mid < \varepsilon$

That means $f(x)\cdot g(x)$ is continuous at x₀.

The statement that the ratio

$\frac{f(x)}{g(x)}$

is continuous at x_0 ( $g(x_0)\not= 0$ ) follows from the fact that product is continuous as long as each of the terms is continuous. The details are left for the reader as an exercise.

Now we turn to topological properties of continuous functions. For simplicity we assume that the real function f(x) in question is defined for all real numbers,

$f(x):\;\;\mathbb{R} \;\;\to \;\; \mathbb{R}$

The most basic topological properties of continuous functions are listed in the following theorem.

  
Theorem 2.  If f(x) is a continuous real function then the following statements hold.
1. If  then  is an open set.
2. If  is a compact set then so is its image, f(K).
3. If a < b and  then there exists  such that f(c) = 0.

Proof.

Statement 1. Let us take an arbitrary point

$x_0 \in f^{-1}((a,\;b))$

$(a,\;b)$ is an open interval. Hence, there exists $\varepsilon > 0$ such that

$(f(x_0)-\varepsilon ,\; f(x_0) +\varepsilon ) \subset (a,\;b)$

Since the function f(x) is continuous one can find Î´ > 0 for which

$\mid x - x_0 \mid < \delta\;\;\;\Longrightarrow \;\;\mid f(x) - f(x_0) \mid <\varepsilon$

That implies

$(x_0-\delta ,\; x_0 +\delta ) \subset f^{-1}((a,\;b))$

We proved that any point

$x_0 \in f^{-1}((a,\;b))$

possesses an open neighborhood

$(x_0-\delta ,\; x_0 +\delta ) \subset f^{-1}((a,\;b))$

It implies that

$f^{-1}((a,\;b))$

is an open set.

Statement 2. Consider an open covering

$f(K) \subset \cup_i F_i$

According to the first statement each

f ^-1(F_i)

is open and

$K \subset \cup_i f^{-1}(F_i)$

The set K is compact. That means any open cover has a finite subcover:

$K \subset \cup_{j=1}^m f^{-1}(F_{i_j})$

Hence,

$f(K) \subset \cup_ {j=1}^m F_{i_j}$

We proved that any open cover for f(K) admits a finite subcover. Thus, f(K) is compact.

Statement 3. Consider

$c = sup\{\;x\;;\;\mbox{ such that }f(x)\le 0 \}$

Since f(x) is continuous we have

f(c) = 0

Notice that

sup{S}

denotes the largest limit point of S.

We conclude this lecture by formulating two more properties of continuous functions:

  Superposition of continuous functions is continuous.
  Inverse of a continuous one-to-one function is continuous.

Theorem 3. If f(x) and g(x) are continuous functions on $\mathbb{R}$ then so is their superposition f(g(x)).

Proof. We need to show that

$\forall \; \varepsilon > 0 \; \exists \;\delta >0\;\;\mbox{ such that } \mid x - x_0 \mid <\delta \; \Longrightarrow \;\mid f(g(x)) -f(g(x_0))\mid < \varepsilon$

Since f(x) is a continuous function at g(x₀) we have

$\forall \; \varepsilon > 0 \; \exists \;\;\delta_f >0\;\;\mbox{ such that } \mid y - g(x_0) \mid < \delta_f \; \Longrightarrow \;\mid f(y) -f(g(x_0))\mid < \varepsilon$

On the other hand, the continuity of g(x) implies the existence of such $\delta > 0$ that

$\mid x - x_0 \mid < \delta_f \; \Longrightarrow \;\mid g(x) -g(x_0)\mid < \delta_f$

The proof is completed.

A function

$f\;:\;\;M\;\to \; N$

is called one-to-one if, and only if,

$x\not=y \;\; \Longrightarrow \; \;f(x) \not= f(y)$

and

$\forall \;y \in N \;\; \exists \; \; x\in M \;\;\mbox{ such that } f(x)=y$

Any one-to-one function f(x) has inverse function f ^-1(y) such that

f ^-1(f(x)) = x

Moreover, if f(x) is continuous then so is f ^-1(y). We leave the proof of this fact as an exercise to the reader.