How is $f(x)=x^p$ continuous for any real $p$?
$begingroup$
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
2,80932151
$endgroup$
|
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
2,80932151
$endgroup$
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
$begingroup$
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
$begingroup$
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
2,80932151
$endgroup$
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
2,80932151
asked Dec 16 '18 at 3:08
SilentSilent
2,80932151
asked Dec 16 '18 at 3:08
SilentSilent
2,80932151
asked Dec 16 '18 at 3:08
SilentSilent
2,80932151
asked Dec 16 '18 at 3:08
SilentSilent
2,80932151
2,80932151
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
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042195%2fhow-is-fx-xp-continuous-for-any-real-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042195%2fhow-is-fx-xp-continuous-for-any-real-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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