Mixing
If $f(x)$ is continuous, then $f(x+t)$ is continuous.
$begingroup$
This might be an obvious question, but I couldn't prove it rigorously.
Suppose $f(x)$ is continuous at every $a$. Then, for every a for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|f(x)-f(a)right|<e).$
Let $g(x)=g(x+t)$.
I want to show that for every $a$ for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|g(x)-g(a)right|<e).$
begin{align*}
|g(x)-g(a)| &=|f(x+t)-f(a+t)|\
&=|f(x+t)-f(x)+f(x)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a+t)|\
&=|f(x+t)-f(x)|+|f(x)-f(a)+f(a)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a)|+|f(a)-f(a+t)|.
end{align*}
But I dont know how to go further. I've spent so much time on this simple problem, but still cant figure it out.
Thanks!
real-analysis continuity
edited Jan 30 at 22:48
Adam Higgins
613113
asked Jan 30 at 22:33
MadisonMadison
112
$endgroup$
|
show 1 more comment
$begingroup$
This might be an obvious question, but I couldn't prove it rigorously.
Suppose $f(x)$ is continuous at every $a$. Then, for every a for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|f(x)-f(a)right|<e).$
Let $g(x)=g(x+t)$.
I want to show that for every $a$ for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|g(x)-g(a)right|<e).$
begin{align*}
|g(x)-g(a)| &=|f(x+t)-f(a+t)|\
&=|f(x+t)-f(x)+f(x)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a+t)|\
&=|f(x+t)-f(x)|+|f(x)-f(a)+f(a)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a)|+|f(a)-f(a+t)|.
end{align*}
But I dont know how to go further. I've spent so much time on this simple problem, but still cant figure it out.
Thanks!
real-analysis continuity
edited Jan 30 at 22:48
Adam Higgins
613113
asked Jan 30 at 22:33
MadisonMadison
112
$endgroup$
2
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
1
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15
|
show 1 more comment
$begingroup$
This might be an obvious question, but I couldn't prove it rigorously.
Suppose $f(x)$ is continuous at every $a$. Then, for every a for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|f(x)-f(a)right|<e).$
Let $g(x)=g(x+t)$.
I want to show that for every $a$ for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|g(x)-g(a)right|<e).$
begin{align*}
|g(x)-g(a)| &=|f(x+t)-f(a+t)|\
&=|f(x+t)-f(x)+f(x)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a+t)|\
&=|f(x+t)-f(x)|+|f(x)-f(a)+f(a)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a)|+|f(a)-f(a+t)|.
end{align*}
But I dont know how to go further. I've spent so much time on this simple problem, but still cant figure it out.
Thanks!
real-analysis continuity
edited Jan 30 at 22:48
Adam Higgins
613113
asked Jan 30 at 22:33
MadisonMadison
112
$endgroup$
This might be an obvious question, but I couldn't prove it rigorously.
Suppose $f(x)$ is continuous at every $a$. Then, for every a for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|f(x)-f(a)right|<e).$
Let $g(x)=g(x+t)$.
I want to show that for every $a$ for every $e>0$, there exists $d>0$ such that $(|x-a|<d Rightarrow left|g(x)-g(a)right|<e).$
begin{align*}
|g(x)-g(a)| &=|f(x+t)-f(a+t)|\
&=|f(x+t)-f(x)+f(x)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a+t)|\
&=|f(x+t)-f(x)|+|f(x)-f(a)+f(a)-f(a+t)| \
& leq |f(x+t)-f(x)|+|f(x)-f(a)|+|f(a)-f(a+t)|.
end{align*}
But I dont know how to go further. I've spent so much time on this simple problem, but still cant figure it out.
Thanks!
real-analysis continuity
real-analysis continuity
edited Jan 30 at 22:48
Adam Higgins
613113
asked Jan 30 at 22:33
MadisonMadison
112
edited Jan 30 at 22:48
Adam Higgins
613113
asked Jan 30 at 22:33
MadisonMadison
112
edited Jan 30 at 22:48
Adam Higgins
613113
edited Jan 30 at 22:48
Adam Higgins
613113
edited Jan 30 at 22:48
Adam Higgins
613113
613113
asked Jan 30 at 22:33
MadisonMadison
112
asked Jan 30 at 22:33
MadisonMadison
112
asked Jan 30 at 22:33
MadisonMadison
112
112
2
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
1
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15
|
show 1 more comment
2
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
1
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15
2
2
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
1
1
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15
|
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%2f3094211%2fif-fx-is-continuous-then-fxt-is-continuous%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%2f3094211%2fif-fx-is-continuous-then-fxt-is-continuous%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
$begingroup$
Just use the fact that the function $x to x + t$ is continuous.
$endgroup$
– anomaly
Jan 30 at 22:34
1
$begingroup$
what @anomaly said coupled with the fact that continuity is preserved under composition of continuous functions
$endgroup$
– rapidracim
Jan 30 at 22:41
$begingroup$
Oh I see, it was so easy. Thanks a lot!
$endgroup$
– Madison
Jan 30 at 22:47
$begingroup$
Hint $|x-a|=|(x+t)-(a+t)|$ so apply continuity in points $X=x+t$ and $A=a+t$ then $|f(X)-f(A)|<epsilon$.
$endgroup$
– zwim
Jan 30 at 23:12
$begingroup$
Wow, yes, thanks!
$endgroup$
– Madison
Jan 31 at 2:15