Continuous Function: Definition, Examples, Properties

Continuous functions is an important topic in Real Analysis. On this page, we will study about continuous functions along with its several properties and examples.

Definition

Definition of a Continuous Function: Let f: D → ℝ be a function and let c ∈ D. We say f is continuous at x=c, if

lim_x→c f(x) = f(c).

That is, for every ε>0 there exists a δ>0 such that

|f(x)-f(c)| < ε for all x in D satisfying 0<|x-c|<δ.

If f is continuous at all points on D, we say f is a continuous function on D.

Example: f(x)=x² is a continuous function on ℝ.

Sequential Criterion for Continuity

Let f: D → ℝ be a function. If f is continuous at c ∈ D then for every sequence {x_n} → c, we have {f(x_n)} → f(c).

Proof:

As lim_n→∞ x_n = c, we have |x_n-c| < δ for all n ≥ N.

Combining this with the above definition of continuity, it follows that

|f(x_n)-f(c)| < ε for all n ≥ N.

Hence, lim_n→∞ f(x_n) = f(c).

Applications of Sequential Criterion

$\boxed{\color{blue}\textbf{Example 1}:}$ Consider the function $f(x) = \begin{cases} \cos(\frac{1}{x}), & x \neq 0 \\ 0, & x=0. \end{cases}$

Note f(x) is not continuous at x=0. Why?

Ans:

$\left\{\dfrac{1}{2\pi n} \right\} \to 0$ but $\left\{f \left(\dfrac{1}{2\pi n}\right)\right\} \not\to 0=f(0)$.

$\boxed{\color{blue}\textbf{Example 2}:}$ Consider the function $f(x) = \begin{cases} 1, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q}. \end{cases}$

Note f(x) is nowhere continuous (why?).

Ans:

Let c ∈ ℚ. Consider a sequence {x_n}$ in ℝ – ℚ converging to c. But, {f(x_n)} $\not\to$ 1 = f(c). So, discontinuous at c.

This function is known as Dirichlet’s function which is everywhere discontinuous.

Neighbourhood Property for Continuous Functions

Let f: I → ℝ be a continuous function. Let c ∈ I with f(c) ≠ 0. Then there exists a δ >0 such that in the neighbourhood of c in which f(x) keeps the same sign as that of f(c).

Proof:

Suppose f(c)>0.

Choose an ε>0 such that f(c)-ε > 0.

Since f is continuous at c, |f(x)-f(c)| < ε for all x ∈ (c-δ, c+δ) ∩ I = I_c (say).

⇒ f(c)-ε < f(x) < f(c)+ε on I_c.

As f(c)-ε>0 by assumption, we complete the proof when f(c)>0.

Similarly, we can prove the result when f(c) is negative.

Applications of Neighbourhood Property

$\boxed{\color{blue}\textbf{Example 1}:}$ Let f: ℝ → ℝ be a continuous function. Let S={x: f(x) >0}. As an application of Neighbourhood Property, we can show that the set S is open.

Proof:

Suppose f(x)>0 for all x ∈ ℝ. Then S = ℝ, so open.

Next, assume  f(x) ≤ 0 for all x ∈ ℝ. Then S = ∅, so open.

Finally, assume  S ⊂ ℝ. 

Take c ∈ S. By neighbourhood property, there is an open neighbourhood c ∈(c-δ,c+δ) ⊆ S. Thus, c is an interior point, making the set S open.

In a similar way, T = {x: f(x) <0} is an open set in ℝ.

Hence, the set {x: f(x) =0} = ℝ – (S ∪ T) is a closed set. 

Theorems

Theorem 1: If f: [a,b] → ℝ is a continuous function, then it is bounded.

Proof:

Let c ∈ [a,b] and ε>0. As f is continuous at c, we deduce that

|f(x)| ≤ |f(c)| + |f(x)-f(c)|

⇒ |f(x)| < |f(c)| + ε for all x ∈ (c-δ, c+δ):=I_c (say).

So bounded.

Consider {I_c: c ∈ [a,b]}. This forms an open cover of [a, b].

By Heine-Borel theorem, we have

[a, b] ⊆ $\bigcup_{i=1}^n I_{c_i}$.

On each $I_{c_i}$, |f(x)| ≤ M_i.

Set M:= max{M₁, M₂, …, M_n}.

Thus, |f(x)| ≤ M for all x ∈ [a,b]. So bounded.

Theorem 2: Let f: I=[a, b] → ℝ be a continuous function. Then, there are points c and d in [a, b] such that

f(c) = $\sup_{x \in [a,b]}$ f(x) $\quad$ and $\quad$ f(d) = $\inf_{x \in [a,b]}$ f(x).

Proof:

Since f is bounded, M = sup f(I) and m = inf f(I) exist.

Suppose f(x) < M on I.

Consider φ(x) = $\frac{1}{M-f(x)}$. It is continuous on I, so bounded.

Thus, on I, we have that

0 < $\dfrac{1}{M-f(x)}$ < B for some B>0.

It follows f(x) < M-$\frac{1}{B}$ for all x ∈ [a,b], a contradiction to the fact that M=sup f(I).

Therefore, there exists a point c in [a,b] such that f(c)=M.

Remarks:

1. If f is continuous on (a, b), then it may not be bounded. For example, f(x) = $\frac{1}{x}$ on (0, 1). 
2. Also, consider f(x) = $\sqrt{x}$ on [0, ∞).  
3. sup and inf may not be attained if f is continuous on open and bounded interval (a, b). For example, f(x)=x² on (2, 3).
4. Let f: [a,b] → ℝ be continuous. Then f([a,b]) =[m, M].
5. The continuous image of an open bounded interval may not be an open bounded interval. For example, f(x)=x² on (-1, 1). Note, Image(f) = [0,1).

Real Analysis Sample Questions

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Power Series and Radius of Convergence

Well Ordering Principle

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