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Suppose that $f(x)$ is continuous at $a$. Show that the function $|f(x)|$ is continuous at $a$.

Proof:

Since $f(x)$ is continuous at $a$ then, $$\lim_{x\to a}f(x)=f(a)$$ Show that $|f(x)|$ is continouos at $a$ $$\lim_{x\to a}|f(x)|=...$$ From here I can not figure a way to finish the proof. In my head $|f(x)|$ might not be continuous a $a$, such as if $f(a)$ is negative. Then $|f(a)|$ would be positive. Any help would be appreciated! Preferably relating to the Basic Limit Theorems of continuity if possible.

3 Answers 3

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I claim it converges to $|f(a)|$. Taking any sequence $x_n \to a$, we get by the inverse triangle inequality that $$ 0 \leq| \; |f(x_n)|-|f(a)| \; | \leq |f(x_n)-f(a)| $$ Now the right-hand side goes to 0, therefore we can use the sandwich lemma to conclude that $|f(x_n)| \to |f(a)|$

Edit: How does this proof continuity? We now know that $\lim_{n \to \infty}|\; |f(x_n)|-|f(a)|\;|=0$ For limits in the real numbers, we know that $\lim_{n \to \infty}|x_n -x|=0$ $\iff$ $\lim_{n \to \infty}x_n=x$. If you have not proven this in your class, its a good and simple exercise to proof it yourself.
Edit: Also a $\delta - \epsilon $ proof. Let $\epsilon > 0$ be given. Then there exists a $\delta > 0$ s.t. $|x-a|< \delta$ => $|f(x)-f(a)| < \epsilon$. Can you finish the proof using the inverse triangle inqueality again?

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    how does that prove continuity? I only know how to prove based on the basic limit theorems. – 2017-02-20
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    Not saying it is not correct! lol I just havent learned about convergence yet? Can you elaborate a little more please. – 2017-02-20
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    Well, what do you want? $\epsilon$-$\delta$ proof? – 2017-02-20
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    I edited it for you. Hope this helps. Also, giving us your definition of convergence is always helpful. – 2017-02-20
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    $|f(x_n)|-|f(a)| \; | \leq |f(x_n)-f(a)| \leq \epsilon$ is enough right? No need to use *sandwich lemma*. – 2017-02-20
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    Oh! I think I am starting to see that a little bit more. In essence, you are saying $|f(x)-L|<\epsilon$? – 2017-02-20
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    You can also do it via $\delta - \epsilon$. – 2017-02-20
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    The proof presented uses the Sequential Criterion for Continuity, which easy to use – 2017-02-20
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    Ok I am trying to wrap my head around this still, anyone have a good link to the Sequential Criterion for continuity? @ΘΣΦGenSan – 2017-02-20
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    This method is slightly to new for me, (+1) anyway for introducing me to a new topic thank you. – 2017-02-20
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    I also added some $\delta \epsilon$ proof start. Try it if you want to! – 2017-02-20
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    I definitely will! Thanks again – 2017-02-20
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I propose another approach, which is more complicated but also more general: to prove the continuity of $|f(x)|$, note that $|f(x)|=g(f(x))$ where $g(x)=|x|$ and recall that if $f,g$ are two functions such that $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g(f(x))$ is continuous at $a$. The proof of this assertion is routine $\epsilon$-$\delta$.

Proof of the previous assertion: for every $\epsilon>0$, since $g$ is continuous at $f(a)$ there is a $\delta_1 >0$ such that if $|f(x)-f(a)| < \delta_1$, then $$ |g(f(x))-g(f(a))| \le \epsilon. $$

Since $f$ is continuous at $a$, there exists a $\delta_2 > 0$ such that if $|x-a| < \delta_2$, then $$ |f(x)-f(a)| \le \delta_1. $$

In the last part I choose $\epsilon= \delta_1$ in the definition of continuity of $f$ at $a$, this is the key passage that provides the connection between the two limit definitions we are given. Since $\epsilon$ is arbitrary, note that this is exactly the definition of continuity of $g(f(x))$ at $a$.

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    (+1) as well, I am just confused on the $\epsilon-\delta$? – 2017-02-20
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    I mean, it suffices to write down the two definitions of continuity we are given (one for $f$ and one for $g$, one at $a$ and the other one at $f(a)$) choosing the right $\epsilon$s. If needed, I can expand :) Hope this helps. – 2017-02-20
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    If you could please! This is the second suggestion of $\epsilon-\delta$ and I am having trouble visualizing it. – 2017-02-20
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    Editing my answer in seconds :) – 2017-02-20
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First, show that the function $g(x)=|x|$ is continuous on reals, then use the fact that the composition of two continuous functions is also continuous ($|f(x)|= g \circ f (x)$).