Tuesday, 19 November 2013

continuity - Can unbounded discontinuous functions be locally bounded?




Consider the function f(x)=x31+x3

enter image description here



Obviously this function is discontinuous at x=1 therefore discontinuous on R. Moreover, it is unbounded at the same point. Now, I would not say that this function is locally bounded either, as not all sets f(A) are bounded for a neighbourhood A about any x0R. Is this reasoning correct? Can an unbounded discontinuous function be locally bounded?


Answer



If I understand your definition correctly, the function f(x)={x1x<00x=0x+1x>0



fits the bill.




If you want a rational function, then no can do, because if p,q are polynomials with no common factors, then f(x)=p(x)q(x) is continuous if q(x)0 and is unbounded around x if q(x)=0.


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