If |f| is continuous at c, does it follow that f is continuous at c? Justify your answer.?

Let f: D--%26gt;R and define |f|:D--%26gt;R by |f|(x)=|f(x)|. Suppose that f is continuous at c belonging to D. Prove that |f| is continuous at c. If |f| is continuous at c, does it follow that f is continuous at c? Justify your answer.

If |f| is continuous at c, does it follow that f is continuous at c? Justify your answer.?
First, suppose f is cotinuous at c. According to the definition of continuity, for every eps %26gt;0 there exists a neighborhood U of C such that





|f(x) - f(c)| %26lt; eps (1)





for every x in U inter D. According to the triangle inequality, for such elements x we also have





|f(x) - f(c)| %26gt;= | |f(x) - |f(c) | (2) Combining (1) and (2), if follows that





| |f(x) - |f(c)| | %26lt; eps, valid for every x in U inted D, which is exactly th condition of continuity of |f| at c.





But the converse is not true. Take D = R, c = 0 and let f be given by





f(x) = -1 if x %26lt;0


f(x) = 1 if x %26gt;=0. Then, f is discontinuous at x = 0, but since |f(x)| =1 for every real x, if follows |f| is continuous at x = 0.





But if |f| is continuous at c and f(c) = 0, then f is continuous at c, as we can readily see.
Reply:No. Take the simple case


f(x) = 6 if x%26gt;=0 and f(x)=-6 if x%26lt;0


f(x) is not continuos at 0 while If(x)I =6 is continuos
Reply:Consider f(x)=-1 if x%26lt;0 and f(x)=1 if x%26gt;=0.


|f|(x)=1 is continuous yet f is not continuous at x=0.