Mixing
When function has at most one zero?
$begingroup$
Let's define $n$-th function $f_{n}:mathbb{R}rightarrow mathbb{R}_{ge 0}$ , where as $n$ tends to infinity given functions tends to function $f(x)$.
Also for every $n$ $f_{n}(x)$ has at most one zero.
Here is the question:
When $f(x)$ has at most one zero?
I know that is not true in general for example: $f_{n}(x)=x^{2}+x+frac{1}{n}$.
But what kind of extra conditions given sequence of functions must satisfy to make it true?
I would like to make this discussion as a brainstorm.
Regards.
real-analysis sequences-and-series functions
asked Jan 26 at 13:35
mkultramkultra
758
$endgroup$
add a comment |
$begingroup$
Let's define $n$-th function $f_{n}:mathbb{R}rightarrow mathbb{R}_{ge 0}$ , where as $n$ tends to infinity given functions tends to function $f(x)$.
Also for every $n$ $f_{n}(x)$ has at most one zero.
Here is the question:
When $f(x)$ has at most one zero?
I know that is not true in general for example: $f_{n}(x)=x^{2}+x+frac{1}{n}$.
But what kind of extra conditions given sequence of functions must satisfy to make it true?
I would like to make this discussion as a brainstorm.
Regards.
real-analysis sequences-and-series functions
asked Jan 26 at 13:35
mkultramkultra
758
$endgroup$
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42
add a comment |
$begingroup$
Let's define $n$-th function $f_{n}:mathbb{R}rightarrow mathbb{R}_{ge 0}$ , where as $n$ tends to infinity given functions tends to function $f(x)$.
Also for every $n$ $f_{n}(x)$ has at most one zero.
Here is the question:
When $f(x)$ has at most one zero?
I know that is not true in general for example: $f_{n}(x)=x^{2}+x+frac{1}{n}$.
But what kind of extra conditions given sequence of functions must satisfy to make it true?
I would like to make this discussion as a brainstorm.
Regards.
real-analysis sequences-and-series functions
asked Jan 26 at 13:35
mkultramkultra
758
$endgroup$
Let's define $n$-th function $f_{n}:mathbb{R}rightarrow mathbb{R}_{ge 0}$ , where as $n$ tends to infinity given functions tends to function $f(x)$.
Also for every $n$ $f_{n}(x)$ has at most one zero.
Here is the question:
When $f(x)$ has at most one zero?
I know that is not true in general for example: $f_{n}(x)=x^{2}+x+frac{1}{n}$.
But what kind of extra conditions given sequence of functions must satisfy to make it true?
I would like to make this discussion as a brainstorm.
Regards.
real-analysis sequences-and-series functions
real-analysis sequences-and-series functions
asked Jan 26 at 13:35
mkultramkultra
758
asked Jan 26 at 13:35
mkultramkultra
758
asked Jan 26 at 13:35
mkultramkultra
758
asked Jan 26 at 13:35
mkultramkultra
758
asked Jan 26 at 13:35
mkultramkultra
758
758
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42
add a comment |
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think that it is easier to find a condition for at least one zero than for at most one zero. See the example of kccu in the comments.
If you talk about a sequence of functions, you should specify in which sence to you consider the convergence. I assume, you say, that $f_n$ converge pointwise to a function $f$, which means
$$
forall xinmathbb R~:~f(x)=lim_{ntoinfty} f_n(x).
$$
But through this definition, you can't controll the zeros of $f$ from the zeros of $f_n$. If $f$ has a zero $zinmathbb R$, than none of $f_n$ have to have a zero near $z$ or at all!
A far more trivial example is:
$$
f_n:mathbb Rtomathbb R_{geq 0},~f_n(x)=frac1n.
$$
Here, you get $fequiv 0$ and further $f_n$ converges uniformly to $f$. But none of $f_n$ has a zero, while $f$ has everywhere a zero.
Ok, if you like to have an example, where the functions $f_n$ are in some space like $L^p(mathbb R)$, you can get it:
Consider an arbitrary function $gin L^p(mathbb R)$ such that $g(x)>0$ for all $xinmathbb R$. Now define $f_n:=frac1ng$. Obviously, $f_n$ have no zeros for any $n$ and $f_n$ converge to $fequiv 0$ in $L^p(mathbb R)$.
And you can to it far worse! Consider a positive function $gin L^p(mathbb R)$ which is nonincreasing in $|x|$ and define $f_n(x):=g(x+n)$. Then you get $f_nto 0$ pointwise while $|f_n|_{L^p(mathbb R)}=|g|_{L^p(mathbb R)}notto 0$.
All in all, I don't think, there is a simple condition to controll the number of zeros of $f$ from above.
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
add a comment |
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%2f3088251%2fwhen-function-has-at-most-one-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think that it is easier to find a condition for at least one zero than for at most one zero. See the example of kccu in the comments.
If you talk about a sequence of functions, you should specify in which sence to you consider the convergence. I assume, you say, that $f_n$ converge pointwise to a function $f$, which means
$$
forall xinmathbb R~:~f(x)=lim_{ntoinfty} f_n(x).
$$
But through this definition, you can't controll the zeros of $f$ from the zeros of $f_n$. If $f$ has a zero $zinmathbb R$, than none of $f_n$ have to have a zero near $z$ or at all!
A far more trivial example is:
$$
f_n:mathbb Rtomathbb R_{geq 0},~f_n(x)=frac1n.
$$
Here, you get $fequiv 0$ and further $f_n$ converges uniformly to $f$. But none of $f_n$ has a zero, while $f$ has everywhere a zero.
Ok, if you like to have an example, where the functions $f_n$ are in some space like $L^p(mathbb R)$, you can get it:
Consider an arbitrary function $gin L^p(mathbb R)$ such that $g(x)>0$ for all $xinmathbb R$. Now define $f_n:=frac1ng$. Obviously, $f_n$ have no zeros for any $n$ and $f_n$ converge to $fequiv 0$ in $L^p(mathbb R)$.
And you can to it far worse! Consider a positive function $gin L^p(mathbb R)$ which is nonincreasing in $|x|$ and define $f_n(x):=g(x+n)$. Then you get $f_nto 0$ pointwise while $|f_n|_{L^p(mathbb R)}=|g|_{L^p(mathbb R)}notto 0$.
All in all, I don't think, there is a simple condition to controll the number of zeros of $f$ from above.
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
add a comment |
$begingroup$
I think that it is easier to find a condition for at least one zero than for at most one zero. See the example of kccu in the comments.
If you talk about a sequence of functions, you should specify in which sence to you consider the convergence. I assume, you say, that $f_n$ converge pointwise to a function $f$, which means
$$
forall xinmathbb R~:~f(x)=lim_{ntoinfty} f_n(x).
$$
But through this definition, you can't controll the zeros of $f$ from the zeros of $f_n$. If $f$ has a zero $zinmathbb R$, than none of $f_n$ have to have a zero near $z$ or at all!
A far more trivial example is:
$$
f_n:mathbb Rtomathbb R_{geq 0},~f_n(x)=frac1n.
$$
Here, you get $fequiv 0$ and further $f_n$ converges uniformly to $f$. But none of $f_n$ has a zero, while $f$ has everywhere a zero.
Ok, if you like to have an example, where the functions $f_n$ are in some space like $L^p(mathbb R)$, you can get it:
Consider an arbitrary function $gin L^p(mathbb R)$ such that $g(x)>0$ for all $xinmathbb R$. Now define $f_n:=frac1ng$. Obviously, $f_n$ have no zeros for any $n$ and $f_n$ converge to $fequiv 0$ in $L^p(mathbb R)$.
And you can to it far worse! Consider a positive function $gin L^p(mathbb R)$ which is nonincreasing in $|x|$ and define $f_n(x):=g(x+n)$. Then you get $f_nto 0$ pointwise while $|f_n|_{L^p(mathbb R)}=|g|_{L^p(mathbb R)}notto 0$.
All in all, I don't think, there is a simple condition to controll the number of zeros of $f$ from above.
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
add a comment |
$begingroup$
I think that it is easier to find a condition for at least one zero than for at most one zero. See the example of kccu in the comments.
If you talk about a sequence of functions, you should specify in which sence to you consider the convergence. I assume, you say, that $f_n$ converge pointwise to a function $f$, which means
$$
forall xinmathbb R~:~f(x)=lim_{ntoinfty} f_n(x).
$$
But through this definition, you can't controll the zeros of $f$ from the zeros of $f_n$. If $f$ has a zero $zinmathbb R$, than none of $f_n$ have to have a zero near $z$ or at all!
A far more trivial example is:
$$
f_n:mathbb Rtomathbb R_{geq 0},~f_n(x)=frac1n.
$$
Here, you get $fequiv 0$ and further $f_n$ converges uniformly to $f$. But none of $f_n$ has a zero, while $f$ has everywhere a zero.
Ok, if you like to have an example, where the functions $f_n$ are in some space like $L^p(mathbb R)$, you can get it:
Consider an arbitrary function $gin L^p(mathbb R)$ such that $g(x)>0$ for all $xinmathbb R$. Now define $f_n:=frac1ng$. Obviously, $f_n$ have no zeros for any $n$ and $f_n$ converge to $fequiv 0$ in $L^p(mathbb R)$.
And you can to it far worse! Consider a positive function $gin L^p(mathbb R)$ which is nonincreasing in $|x|$ and define $f_n(x):=g(x+n)$. Then you get $f_nto 0$ pointwise while $|f_n|_{L^p(mathbb R)}=|g|_{L^p(mathbb R)}notto 0$.
All in all, I don't think, there is a simple condition to controll the number of zeros of $f$ from above.
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
I think that it is easier to find a condition for at least one zero than for at most one zero. See the example of kccu in the comments.
If you talk about a sequence of functions, you should specify in which sence to you consider the convergence. I assume, you say, that $f_n$ converge pointwise to a function $f$, which means
$$
forall xinmathbb R~:~f(x)=lim_{ntoinfty} f_n(x).
$$
But through this definition, you can't controll the zeros of $f$ from the zeros of $f_n$. If $f$ has a zero $zinmathbb R$, than none of $f_n$ have to have a zero near $z$ or at all!
A far more trivial example is:
$$
f_n:mathbb Rtomathbb R_{geq 0},~f_n(x)=frac1n.
$$
Here, you get $fequiv 0$ and further $f_n$ converges uniformly to $f$. But none of $f_n$ has a zero, while $f$ has everywhere a zero.
Ok, if you like to have an example, where the functions $f_n$ are in some space like $L^p(mathbb R)$, you can get it:
Consider an arbitrary function $gin L^p(mathbb R)$ such that $g(x)>0$ for all $xinmathbb R$. Now define $f_n:=frac1ng$. Obviously, $f_n$ have no zeros for any $n$ and $f_n$ converge to $fequiv 0$ in $L^p(mathbb R)$.
And you can to it far worse! Consider a positive function $gin L^p(mathbb R)$ which is nonincreasing in $|x|$ and define $f_n(x):=g(x+n)$. Then you get $f_nto 0$ pointwise while $|f_n|_{L^p(mathbb R)}=|g|_{L^p(mathbb R)}notto 0$.
All in all, I don't think, there is a simple condition to controll the number of zeros of $f$ from above.
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
edited Jan 26 at 14:03
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 26 at 13:53
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
add a comment |
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
I think you mean to say "none of $f_n$" in the places where you say "any of $f_n$".
$endgroup$
– kccu
Jan 26 at 13:56
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
$begingroup$
You are right. Thanks.
$endgroup$
– Mundron Schmidt
Jan 26 at 13:58
add a comment |
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%2f3088251%2fwhen-function-has-at-most-one-zero%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
$begingroup$
I don't think that your counterexample works. $f_5(x)=x^2+x+frac{1}{5}$ has two zeros. However, you can make the counterexample $f_n(x)=frac{1}{n}x$. For every $n$, $f_n(x)$ has exactly one zero, but the limit has infinitely many zeros.
$endgroup$
– kccu
Jan 26 at 13:42