How to prove $x^{1/n} $ is uniformly continuos in $[0, a]$ where $a$ is a positive real.
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I proved that this funcion is not Lispchitz continuos making $y = 2x$ and making $x rightarrow 0$. But I'm stuck proving the uniformly continuity.
calculus continuity uniform-continuity
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
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add a comment |
$begingroup$
I proved that this funcion is not Lispchitz continuos making $y = 2x$ and making $x rightarrow 0$. But I'm stuck proving the uniformly continuity.
calculus continuity uniform-continuity
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
$endgroup$
add a comment |
$begingroup$
I proved that this funcion is not Lispchitz continuos making $y = 2x$ and making $x rightarrow 0$. But I'm stuck proving the uniformly continuity.
calculus continuity uniform-continuity
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
$endgroup$
I proved that this funcion is not Lispchitz continuos making $y = 2x$ and making $x rightarrow 0$. But I'm stuck proving the uniformly continuity.
calculus continuity uniform-continuity
calculus continuity uniform-continuity
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
asked Dec 14 '18 at 17:07
Tommy do NascimientoTommy do Nascimiento
14210
14210
add a comment |
add a comment |
1 Answer
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The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.
You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have
$$
|f(x)-f(y)| le C|x-y|^{1/n}.
$$
Can you prove that $C=1$ works?
Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
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$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.
You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have
$$
|f(x)-f(y)| le C|x-y|^{1/n}.
$$
Can you prove that $C=1$ works?
Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
$endgroup$
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
add a comment |
$begingroup$
The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.
You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have
$$
|f(x)-f(y)| le C|x-y|^{1/n}.
$$
Can you prove that $C=1$ works?
Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
$endgroup$
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
add a comment |
$begingroup$
The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.
You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have
$$
|f(x)-f(y)| le C|x-y|^{1/n}.
$$
Can you prove that $C=1$ works?
Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
$endgroup$
The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.
You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have
$$
|f(x)-f(y)| le C|x-y|^{1/n}.
$$
Can you prove that $C=1$ works?
Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
edited Dec 17 '18 at 9:09
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
answered Dec 14 '18 at 17:31
BigbearZzzBigbearZzz
8,69121652
8,69121652
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
add a comment |
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
$begingroup$
I did, thanks and since $f(x)$ is Lipschitz continuos in $lvert a, infty rvert$ then is uniformly continuos in all positive reals.
$endgroup$
– Tommy do Nascimiento
Dec 14 '18 at 18:35
add a comment |
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