Improve the $(\epsilon, \delta)$ proof of absolute value function continuity

absolute valuecontinuityepsilon-deltainequalitysolution-verification

In Charles C. Pugh's Real Mathematical Analysis [2nd Ed] on page 76, continuity of $f : x \mapsto \lvert x \rvert$ on $\mathbb{R}$ is a corollary of another theorem. However, I wanted to see if I could use the $(\epsilon, \delta)$-condition to prove its continuity. My proof below seems overly complicated; if you can find any ways to improof it, please let me know!

Conclusion. We want to show that for any $\epsilon > 0$ and any $x \in \mathbb{R}$, there exists a $\delta >0$ such that if $y \in \mathbb{R}$ then the following holds:
$$ \lvert y – x \vert < \delta \implies \bigl\lvert |y| – |x| \bigr\rvert < \epsilon , $$
i.e., $\lvert * \rvert$ is continuous.

Proof. Choose $\delta := \epsilon$. We want to show that $\bigl\lvert |y| – |x| \bigr\rvert \leq \lvert y – x \vert < \delta = \epsilon$. There are 8 cases to consider:

  1. $|y| > |x|$
    a) $y > 0, x > 0$
    b) $y > 0, x < 0$
    c) $y < 0, x > 0$
    d) $y < 0, x < 0$
  2. The same four cases but with $|x| < |y|$.

Without loss of generality, we prove case 1. For each sub-case, $|y| > |x|$ implies $|y| – |x| > 0$, so $\bigl\lvert |y| – |x| \bigr\rvert = |y| – |x|$.

1.a) Note that $y – x > 0$. Hence

$$ \bigl\lvert |y| – |x| \bigr\rvert = y – x = | y – x | . $$

1.b) Note that since $x < 0$, we have $y + x < y – x$. Note also that $y – x > 0$. Hence

$$ \bigl\lvert |y| – |x| \bigr\rvert = y – (-x) = y + x < y – x = | y – x | . $$

1.c) Note that since $x > 0$, we have $-y + x > -y – x$. Note also that $x – y > 0$. Hence

$$ \bigl\lvert |y| – |x| \bigr\rvert = -y – x < x – y = | x – y | = | y – x |. $$

1.d) Note that since $-y < -x$ we have $x – y < 0$. Hence

$$ \bigl\lvert |y| – |x| \bigr\rvert = -y – (-x) = x – y < |x – y| = | y – x | . $$

Therefore all cases satisfy $\bigl\lvert |y| – |x| \bigr\rvert \leq \lvert y – x \vert < \epsilon$, as required.

I think it is possible to structure the proof so that, using symmetry, only two cases need to be proved.

Best Answer

Let's show that

$$ ||x|-|y|| \leq |x-y| $$

for any $x,y \in \mathbb{R}$. Indeed, by the triangle inequality, we have \begin{align} |x| &= |x - y + y| \\ &\leq |x-y| + |y|, \end{align} which implies $|x| - |y| \leq |x-y|$. Exchanging $x$ and $y$ leads to $|y| - |x| \leq |x-y|$. Hence, the inequality is established. Can you simplify your proof from here?

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