How is $f(x)=x^p$ continuous for any real $p$?
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I have ben facing this hurdle for a long time: First it was how ordinary properties for $x^r$ ($r$ rational) extend to $x^p$ ($p$ real); but thanks to Tao's Analysis I, I have overcome most of that difficulty.
But this one remains: (And, from yesterday's my engagement in chatroom, I got to know that very many have same problem, and have to take it granted for lack of availability of proof)
How to show that $f:[0,+infty)tomathbb R$ defined as $f(x)=x^p$, is continuous for a fixed $pinmathbb R$?
Is this reasoning of mine correct:
Let us consider $ln (x^p)$ instead. Continuity of $ln x$ implies $pln x=ln(x^p)$ continuous. Now, baby Rudin's exer 26, chapter 4 says that: "Suppose $X,Y,Z$ are metric spaces, and $Y$ compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping $Y$ into $Z$, and put $h(x)=g(f(x))$ for $xin X$. Then $f$ continuous if $h$ continuous." So, if we restrict domain to $[a,b]$ for $a,bin[0,infty)$ for $a<b$, then we see $x^p$ continuous on $[a,b]$. And hence, $x^p$ continuous on $(0,+infty)$.
Is this my answer correct? How to extend this to $[0,+infty)$?
real-analysis continuity logarithms
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
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|
show 1 more comment
$begingroup$
I have ben facing this hurdle for a long time: First it was how ordinary properties for $x^r$ ($r$ rational) extend to $x^p$ ($p$ real); but thanks to Tao's Analysis I, I have overcome most of that difficulty.
But this one remains: (And, from yesterday's my engagement in chatroom, I got to know that very many have same problem, and have to take it granted for lack of availability of proof)
How to show that $f:[0,+infty)tomathbb R$ defined as $f(x)=x^p$, is continuous for a fixed $pinmathbb R$?
Is this reasoning of mine correct:
Let us consider $ln (x^p)$ instead. Continuity of $ln x$ implies $pln x=ln(x^p)$ continuous. Now, baby Rudin's exer 26, chapter 4 says that: "Suppose $X,Y,Z$ are metric spaces, and $Y$ compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping $Y$ into $Z$, and put $h(x)=g(f(x))$ for $xin X$. Then $f$ continuous if $h$ continuous." So, if we restrict domain to $[a,b]$ for $a,bin[0,infty)$ for $a<b$, then we see $x^p$ continuous on $[a,b]$. And hence, $x^p$ continuous on $(0,+infty)$.
Is this my answer correct? How to extend this to $[0,+infty)$?
real-analysis continuity logarithms
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
$endgroup$
2
$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
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Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
1
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
1
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17
|
show 1 more comment
$begingroup$
I have ben facing this hurdle for a long time: First it was how ordinary properties for $x^r$ ($r$ rational) extend to $x^p$ ($p$ real); but thanks to Tao's Analysis I, I have overcome most of that difficulty.
But this one remains: (And, from yesterday's my engagement in chatroom, I got to know that very many have same problem, and have to take it granted for lack of availability of proof)
How to show that $f:[0,+infty)tomathbb R$ defined as $f(x)=x^p$, is continuous for a fixed $pinmathbb R$?
Is this reasoning of mine correct:
Let us consider $ln (x^p)$ instead. Continuity of $ln x$ implies $pln x=ln(x^p)$ continuous. Now, baby Rudin's exer 26, chapter 4 says that: "Suppose $X,Y,Z$ are metric spaces, and $Y$ compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping $Y$ into $Z$, and put $h(x)=g(f(x))$ for $xin X$. Then $f$ continuous if $h$ continuous." So, if we restrict domain to $[a,b]$ for $a,bin[0,infty)$ for $a<b$, then we see $x^p$ continuous on $[a,b]$. And hence, $x^p$ continuous on $(0,+infty)$.
Is this my answer correct? How to extend this to $[0,+infty)$?
real-analysis continuity logarithms
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
$endgroup$
I have ben facing this hurdle for a long time: First it was how ordinary properties for $x^r$ ($r$ rational) extend to $x^p$ ($p$ real); but thanks to Tao's Analysis I, I have overcome most of that difficulty.
But this one remains: (And, from yesterday's my engagement in chatroom, I got to know that very many have same problem, and have to take it granted for lack of availability of proof)
How to show that $f:[0,+infty)tomathbb R$ defined as $f(x)=x^p$, is continuous for a fixed $pinmathbb R$?
Is this reasoning of mine correct:
Let us consider $ln (x^p)$ instead. Continuity of $ln x$ implies $pln x=ln(x^p)$ continuous. Now, baby Rudin's exer 26, chapter 4 says that: "Suppose $X,Y,Z$ are metric spaces, and $Y$ compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping $Y$ into $Z$, and put $h(x)=g(f(x))$ for $xin X$. Then $f$ continuous if $h$ continuous." So, if we restrict domain to $[a,b]$ for $a,bin[0,infty)$ for $a<b$, then we see $x^p$ continuous on $[a,b]$. And hence, $x^p$ continuous on $(0,+infty)$.
Is this my answer correct? How to extend this to $[0,+infty)$?
real-analysis continuity logarithms
real-analysis continuity logarithms
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
asked Dec 16 '18 at 3:08
SilentSilent
2,83632152
2,83632152
2
$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
$begingroup$
Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
1
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
1
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17
|
show 1 more comment
2
$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
$begingroup$
Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
1
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
1
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17
2
2
$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
$begingroup$
Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
$begingroup$
Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
1
1
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
1
1
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17
|
show 1 more comment
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$begingroup$
look at Baby Rudin chapter 1 problems 6,7. doesn't the fact that $x^p$ is surjective, together with monotonicity, establish continuity?
$endgroup$
– mathworker21
Dec 16 '18 at 3:14
$begingroup$
Wow! that was awesome observation.
$endgroup$
– Silent
Dec 16 '18 at 3:15
1
$begingroup$
Your question is wrong. $0^{-1}$ is ill-defined.
$endgroup$
– user21820
Dec 16 '18 at 12:43
$begingroup$
@user21820, yes i observed later, sorry. I thought not to edit. You may edit and correct it.
$endgroup$
– Silent
Dec 16 '18 at 14:08
1
$begingroup$
But once you restrict it to $(0,infty)$, then I don't see why you're making things so complicated. $x^p = exp(p·ln(x))$ which is clearly continuous with respect to $x$ since $ln$ is continuous on $(0,infty)$ and $exp$ is continuous on all reals.
$endgroup$
– user21820
Dec 16 '18 at 14:17