is $f(x)= |{\sin x \over x}|$ uniformly continuous in $(-1,0) \cup (0,1) $?

Question

theorem : let $ f:(a,b) \to \Bbb{R}$ is continuous function and limits : $ lim _ {x \to a^+} f(x) $ and $ lim _ {x \to b^-} f(x) $ are exist , then f is uniformly continuous on (a,b)

if we use above theorem $f(x)= |{\sin x \over x}|$ is uniformly continuous in (-1,0) and (0,1) .then f is uniformly continuous in $(-1,0) \cup (0,1) $

is this way true ?

That function is NOT continuous on 0. – Theodoros Mpalis Mar 29 '16 at 15:19 — Theodoros Mpalis, Mar 29 '16 at 15:19
@TheodorosMpalis: Huh???... Are you sure about that? – Alex M. Jul 13 '16 at 18:08 — Alex M., Jul 13 '16 at 18:08

score 0 · Answer 1 · answered Mar 29 '16 at 15:14

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Define $g:[-1,1]\to\mathbb R$ by $$g(x) = \begin{cases}1, &\textrm{if }|x|=0\\ \tfrac{\sin x}{x},&\textrm{if }0<|x|<1\\ \sin 1, &\textrm{if }|x|=1\\ \end{cases} $$ Note that $g$ is a continuous function on a compact set, so is uniformly continuous.

Since $f$ is a restriction of $g$, it is uniformly continuous as well.

answered Mar 29 '16 at 15:14

MPW

43,638

f is not continuous on 0 though – Theodoros Mpalis Mar 29 '16 at 15:20

Christian Blatter · Answer 2 · 2016-03-29T18:48:05.373

0

Your function $f$ can be extended to $$f(x):=\int_0^1\cos(tx)\>dt\qquad(-\infty<x<\infty)\ ,$$ and is Lipschitz-continuous, hence uniformly continuous, on all of ${\mathbb R}$: $$|f(x)-f(y)|\leq\int_0^1|\cos(tx)-\cos(ty)|\>dt\leq\int_0^1 t|x-y|\>dt={1\over2}|x-y|\ .$$

edited Mar 29 '16 at 18:48

answered Mar 29 '16 at 15:36

Christian Blatter

226,825

is $f(x)= |{\sin x \over x}|$ uniformly continuous in $(-1,0) \cup (0,1) $?

2 Answers2