Math for Non-Geeks/ Composition of continuous functions

{{#invoke:Math for Non-Geeks/Seite|oben}}

Many functions are defined as the concatenation - the linking together of things, like in a chain - of other functions. Checking for continuity of such concatenated functions by using the classical epsilon-delta criterion for continuity is often tedious. However, one can prove that the concatenation of continuous functions is once again a continuous function. This is an important tool that simplifies the proof for continuity of compositions of functions.

Concatenation Theorems

The concatenation theorems for continuous functions are the following:

Motivation

Imagine, we are given a function containing sums, products and quotients like $f:ℝ\to {ℝ}_0^{+}$, $x\mapsto \left|\frac{1+x^3}{1+x^2}\right|$ . We would like to know whether this function is continuous at some argument $a\in ℝ$ . So we consider any sequence of arguments $(x_n{)}_{n\in \mathbb{N}}$ converging to $a$ and check whether there is always $\underset{n\to \mathrm{\infty }}{\lim }f(x_n)=f(a)$. For handling sums, products and quotients, the limit theorems for convergent sequences turn out to be very useful:

Those formulas could be applied because all subsequences converge (we showed this in the end of our calculations). As $a\in ℝ$ was chosen to be arbitrary, we directly obtain the continuity of the entire function $f$ . This proof of continuity is basically an application of the sequence criterion plus theorems for sequence limits. Since, thanks to the limit theorems, the limit can be pulled into the function, we can use it to establish continuity. And the above concatenation theorems shorten proofs of this kind even further. we consider the following functions:

• $a:ℝ\to ℝ:x\mapsto x$
• $b:ℝ\to ℝ:x\mapsto 1$
• $c:ℝ\to {ℝ}^{+}:x\mapsto |x|$

Now, $f$ may be written as a concatenation of those three functions:

As all three functions $a$, $b$ and $c$ are continuous, the concatenation theorems will directly imply continuity for $f$ . This argumentation is even shorter than the proof using the sequence criterion. So the continuity proof can be concluded in one sentence: $f$ is continuous because it is a concatenation of continuous functions. And indeed, any concatenation constructed out of continuous functions (i.e. by combining polynomials) is continuous.

Problem Example

The following problem illustrates just how easy it is to establish continuity of a function using the concatenation theorems:

<section begin="Problem:Continuity of a concatenated square root function" />Math for Non-Geeks: Template:Aufgabe<section end="Problem:Continuity of a concatenated square root function" />

General sketch of the proof

<section begin="Sketch of the proof" />Following the concatenation theorems, every composition of continuous functions is again continuous function. So if $f:D\to ℝ$ can be written as a concatenation of continuous functions, we can directly infer continuity of $f$ . A corresponding proof could be of the following form:

In a lecture, this proof scheme may of course only be applied if the concatenation theorems have already been treated before. In any way, it is a very efficient method to characterize continuous functions.<section end="Sketch of the proof" />

Corollary: Polynomial functions are continuous

Every polynomial function can be written as a concatenation of following the two functions:

• $f:ℝ\to ℝ:x\mapsto x$
• $g_c:ℝ\to ℝ:x\mapsto c$

$f$ is the identity $g_c$ the constant function with value $c$. These functions are continuous and hence, every polynomial function is continuous. For instance, we may construct the polynomial function $h:ℝ\to ℝ:x\mapsto 2x^3-4x+23$ out of $f$ and $g_c$ as follows:

Explicitly, this decomposition reads:

Further examples

Exactly as subsequences have to converge for a concatenation sequence to converge, we need that our building brick functions are continuous in order to obtain a continuous concatenation function. When using non-continuous functions for the concatenation, we do not know anything about the continuity of the outcome. For instance:

The function $f$is continuous at $x=0$, whereas $g$ is not. The product of both functions is again $g$, since $f(x)\cdot g(x)=1\cdot g(x)=g(x)$. Therefore the product (i.e. a concatenation) is discontinuous at $x=0$. By contrast to what one may intuitively expect, concatenating discontinuous functions may also yield us a continuous function. To illustrate this, let us consider:

This function maps rational numbers to $1$ an all other to $0$ . Concatenating $h$ with itself, we get the following function $h\circ h$:

$h\circ h$ is just a constant function. Hence it is continuous - although $h$ was actually nowhere continuous. So concatenating discontinuous functions may indeed yield us a continuous function.

Comparison with the epsilon-delta criterion

In the beginning of this article, we used the concatenation theorems in order to show that the function $\sqrt{5+x^2}$ is continuous. We will no perform a second proof "by hand", using the epsilon-delta criterion. The proof will cost us more work, but we will get an explicit information on the maximal initial error $\delta$ we have to choose, in order to stay below a given threshold $ϵ>0$ for the error of the outcome.

<section begin="Problem:Epsilon-delta proof of continuity for a square root function" />Math for Non-Geeks: Template:Aufgabe<section end="Problem:Epsilon-delta proof of continuity for a square root function" />

{{#invoke:Math for Non-Geeks/Seite|unten}}