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Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.

Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|<\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?

I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution!

Martin Sleziak
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user989
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1 Answers1

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Not quite: In order to show that $|f|$ is continuous, one needs to estimate $\Big||f(x) - |f(c)|\Big|$, not just show that $|f(x)| - |f(c)|$ can be bounded above (how do we know that $|f(x)| - |f(c)|$ is not some really large negative number?). But happily, this is also a consequence of the reverse triangle inequality, which states in full that

$$\Big||a| - |b|\Big| \le |a - b|$$

Therefore, we can just say that

$$\Big||f(x)| - |f(c)|\Big| \le |f(x) - f(c)| \le \epsilon$$

for some appropriately constrained $x, c$.