Example 20: $|1-x+|x||$ is a continuous function

Basics you need

Solution

1. The domain, and the inner and outer functions

   $1 - x + |x|$ and its modulus can be computed for every real $x$⁠, so the domain of $f$ is $\R$⁠. Following the book, define

   $$g(x) = 1 - x + |x|, \qquad h(x) = |x|.$$

   Then

   $$\begin{aligned} (h \circ g)(x) &= h(g(x)) \\ &= h(1 - x + |x|) \\ &= |1 - x + |x|| = f(x). \end{aligned}$$
   The last operation in computing $f(x)$ is the outer modulus, so $h = |\cdot|$ is the outer function, and everything inside it, $1 - x + |x|$⁠, is the inner function $g$⁠. (The inner $g$ contains its own $|x|$⁠; that is part of $g$⁠, not of $h$⁠.)
   Basics: composite functions

2. The pieces of $g$ are continuous

   $g(x) = (1 - x) + |x|$ is a sum of two functions:

   • $1 - x$ is a polynomial, continuous at every real number (Example 14);
   • $|x|$ is continuous at every real number (Example 7).
   To use Theorem 1(1) we need each summand to be continuous at the point; both are continuous at every real number.
   Example 14 and Example 7.

3. So $g$ is continuous (Theorem 1)

   By Theorem 1(1), the sum $g = (1 - x) + |x|$ is continuous at every real number $c$⁠:

   $$\lim_{x\to c} g(x) = 1 - c + |c| = g(c).$$
   This is the book's sentence "$g$ being a sum of a polynomial function and the modulus function is continuous".
   Theorem 1(1), Section 5.2.1. Lesson: algebra of continuous functions

4. The outer function is continuous at $g(c)$

   $h(x) = |x|$ is continuous at every real number (Example 7), in particular at the number $g(c)$⁠.

   Theorem 2 needs the outer function continuous at the inner value $g(c)$⁠. Because $|x|$ is continuous everywhere, this holds whatever $g(c)$ is.

5. Apply Theorem 2

   $g$ is continuous at $c$ (Step 3) and $h$ is continuous at $g(c)$ (Step 4), so by Theorem 2 $f = h \circ g$ is continuous at $c$⁠. As $c$ was any real number, $f$ is a continuous function.

   As $x \to c$⁠, the inner value $g(x)$ approaches $g(c)$⁠, and then its modulus approaches $|g(c)| = f(c)$⁠.
   Theorem 2, Section 5.2.1. Lesson: algebra of continuous functions

6. A check: write $f$ without modulus signs

   For $x \ge 0$⁠, $|x| = x$⁠, so $g(x) = 1 - x + x = 1$⁠. For $x < 0$⁠, $|x| = -x$⁠, so $g(x) = 1 - 2x$⁠, which is greater than 1 because $-2x > 0$⁠. So $g(x) \ge 1 > 0$ for every $x$⁠, the outer modulus changes nothing, and

   $$f(x) = \begin{cases} 1, & \text{if } x \ge 0 \\ 1 - 2x, & \text{if } x < 0 \end{cases}$$

   Each piece is a polynomial. A point $c \ne 0$ has an interval around it, of radius $|c|$⁠, that stays on one side of 0, so near $c$ the function $f$ is a single polynomial and is continuous at $c$ (Example 14). At the junction $x = 0$⁠:

   $$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-}(1 - 2x) = 1,$$ $$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} 1 = 1,$$ $$f(0) = |1 - 0 + 0| = 1.$$

   All three equal 1, so $f$ is continuous at 0 as well, confirming Step 5.

   The case analysis is not needed for the proof, but it shows concretely what Theorems 1 and 2 guarantee: the two pieces meet at the same height. The corner at $(0, 1)$ is allowed; continuity only forbids gaps and jumps.
   Basics: piecewise functions

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