If $f:mathbb{R}to mathbb{R}$ is a continuous function with $|f|$ uniformly continuous, is $f$ necessarily...
$begingroup$
I don't think so. But, I can't find the counterexample.
The singularity shall be at infinity because $f$ is uniformly continuous over any bounded and closed intervals $[a,b]$. Also, $f$ must oscillate at infinity. Otherwise, the uniform continuity of $f$ is identical to $|f|$.
real-analysis
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
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add a comment |
$begingroup$
I don't think so. But, I can't find the counterexample.
The singularity shall be at infinity because $f$ is uniformly continuous over any bounded and closed intervals $[a,b]$. Also, $f$ must oscillate at infinity. Otherwise, the uniform continuity of $f$ is identical to $|f|$.
real-analysis
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
$endgroup$
1
$begingroup$
I can't imagine a counter-example. +1 for the question
$endgroup$
– Masacroso
Dec 1 '18 at 19:49
add a comment |
$begingroup$
I don't think so. But, I can't find the counterexample.
The singularity shall be at infinity because $f$ is uniformly continuous over any bounded and closed intervals $[a,b]$. Also, $f$ must oscillate at infinity. Otherwise, the uniform continuity of $f$ is identical to $|f|$.
real-analysis
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
$endgroup$
I don't think so. But, I can't find the counterexample.
The singularity shall be at infinity because $f$ is uniformly continuous over any bounded and closed intervals $[a,b]$. Also, $f$ must oscillate at infinity. Otherwise, the uniform continuity of $f$ is identical to $|f|$.
real-analysis
real-analysis
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
edited Dec 1 '18 at 19:44
Yuhang
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
asked Dec 1 '18 at 19:38
YuhangYuhang
835118
835118
1
$begingroup$
I can't imagine a counter-example. +1 for the question
$endgroup$
– Masacroso
Dec 1 '18 at 19:49
add a comment |
1
$begingroup$
I can't imagine a counter-example. +1 for the question
$endgroup$
– Masacroso
Dec 1 '18 at 19:49
1
1
$begingroup$
I can't imagine a counter-example. +1 for the question
$endgroup$
– Masacroso
Dec 1 '18 at 19:49
$begingroup$
I can't imagine a counter-example. +1 for the question
$endgroup$
– Masacroso
Dec 1 '18 at 19:49
add a comment |
1 Answer
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Yes. Let $epsilon>0$ be arbitrary. For some $delta>0$ uniform continuity tells us that $$|x-y|<deltaimplies left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}$$ Take $x<y$ arbitrary with $|x-y|<delta$. If $f(x)$ and $f(y)$ have the same sign, $$|f(x)-f(y)|=left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}<epsilon$$ and we are done. Otherwise, by continuity of $f$, for some $zin (x,y)$, $f(z)=0$. Then since $|x-z|<delta$ and $|z-y|<delta$, $|f(x)|$ and $|f(y)|$ are bounded by $frac{epsilon}{2}$, so that $$|f(x)-f(y)| le |f(x)|+|f(y)|<epsilon$$
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
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$begingroup$
Yes. Let $epsilon>0$ be arbitrary. For some $delta>0$ uniform continuity tells us that $$|x-y|<deltaimplies left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}$$ Take $x<y$ arbitrary with $|x-y|<delta$. If $f(x)$ and $f(y)$ have the same sign, $$|f(x)-f(y)|=left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}<epsilon$$ and we are done. Otherwise, by continuity of $f$, for some $zin (x,y)$, $f(z)=0$. Then since $|x-z|<delta$ and $|z-y|<delta$, $|f(x)|$ and $|f(y)|$ are bounded by $frac{epsilon}{2}$, so that $$|f(x)-f(y)| le |f(x)|+|f(y)|<epsilon$$
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
$endgroup$
add a comment |
$begingroup$
Yes. Let $epsilon>0$ be arbitrary. For some $delta>0$ uniform continuity tells us that $$|x-y|<deltaimplies left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}$$ Take $x<y$ arbitrary with $|x-y|<delta$. If $f(x)$ and $f(y)$ have the same sign, $$|f(x)-f(y)|=left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}<epsilon$$ and we are done. Otherwise, by continuity of $f$, for some $zin (x,y)$, $f(z)=0$. Then since $|x-z|<delta$ and $|z-y|<delta$, $|f(x)|$ and $|f(y)|$ are bounded by $frac{epsilon}{2}$, so that $$|f(x)-f(y)| le |f(x)|+|f(y)|<epsilon$$
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
$endgroup$
add a comment |
$begingroup$
Yes. Let $epsilon>0$ be arbitrary. For some $delta>0$ uniform continuity tells us that $$|x-y|<deltaimplies left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}$$ Take $x<y$ arbitrary with $|x-y|<delta$. If $f(x)$ and $f(y)$ have the same sign, $$|f(x)-f(y)|=left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}<epsilon$$ and we are done. Otherwise, by continuity of $f$, for some $zin (x,y)$, $f(z)=0$. Then since $|x-z|<delta$ and $|z-y|<delta$, $|f(x)|$ and $|f(y)|$ are bounded by $frac{epsilon}{2}$, so that $$|f(x)-f(y)| le |f(x)|+|f(y)|<epsilon$$
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
$endgroup$
Yes. Let $epsilon>0$ be arbitrary. For some $delta>0$ uniform continuity tells us that $$|x-y|<deltaimplies left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}$$ Take $x<y$ arbitrary with $|x-y|<delta$. If $f(x)$ and $f(y)$ have the same sign, $$|f(x)-f(y)|=left|left(|f(x)|-|f(y)|right)right|<frac{epsilon}{2}<epsilon$$ and we are done. Otherwise, by continuity of $f$, for some $zin (x,y)$, $f(z)=0$. Then since $|x-z|<delta$ and $|z-y|<delta$, $|f(x)|$ and $|f(y)|$ are bounded by $frac{epsilon}{2}$, so that $$|f(x)-f(y)| le |f(x)|+|f(y)|<epsilon$$
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
answered Dec 1 '18 at 20:23
doogliusdooglius
49624
49624
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I can't imagine a counter-example. +1 for the question
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– Masacroso
Dec 1 '18 at 19:49