Prove that if $f$ is continuous in a then $|f|$ is also continuous. I have this exercise for homework of calculus I, and I was thinking that it could be treated by cases when $f>0$ and $f<0$, but when I make $f<0$ then it results $-f(a)$, following the continuous definition: $lim_{x\to a} f(x)=f(a)\dots$
Asked
Active
Viewed 9,238 times
0
-
1What is $a$, a real number? If so, the usual terminology is "$f$ is continuous at $a$." – Thomas Andrews Apr 24 '15 at 23:13
-
1Hint: $|,|a|-|b|,| \le |a-b|.$ – zhw. Apr 24 '15 at 23:17
-
Well, what is the relationship between $|f(x)-f(a)|$ and $\bigl||f(x)|-|f(a)|\bigr|$? Alternatively, what do you know about the absolute value function? – Cameron Buie Apr 24 '15 at 23:19
-
@zhw.: That hint is a bit confusing, I think, given the OP. – Cameron Buie Apr 24 '15 at 23:22
-
My notation was bad; I didn't even see the $a$ in the problem. I should have written $|,|s|-|t|,|\le |s-t|.$ – zhw. Apr 24 '15 at 23:49
1 Answers
2
There’s no reason to think that $f$ is either always positive or always negative.
HINT: Prove and use the following facts:
- If $f(a)\ne 0$, then there is an $\delta>0$ such that $f(x)$ and $f(a)$ have the same sign whenever $|x-a|<\delta$.
- If $f(a)=0$, $|f(x)-f(a)|=\big||f(x)|-f(a)\big|$ for all $x$.
Brian M. Scott
- 616,228
-
Thank you, I get the second point but I don´t understand very well what it follows the first one... could you explain me a little bit more, please? – Raquel nb Apr 24 '15 at 23:58
-
@Raquel: It implies that within the interval $(a-\delta,a+\delta)$ the function $|f|$ is either $f$ or $-f$. And continuity at $a$ depends only on what happens inside that interval. – Brian M. Scott Apr 25 '15 at 00:03