separately continuous functions $f: mathbb{R}^2 rightarrow mathbb{R}$ but nowhere continuous
Is there a functions $f: mathbb{R}^2 rightarrow mathbb{R}$ separately continuous but nowhere continuous?
There is a general question Here and this is unanswered.
real-analysis
asked Nov 29 '18 at 20:58
machlearmachlear
111
add a comment |
Is there a functions $f: mathbb{R}^2 rightarrow mathbb{R}$ separately continuous but nowhere continuous?
There is a general question Here and this is unanswered.
real-analysis
asked Nov 29 '18 at 20:58
machlearmachlear
111
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08
add a comment |
Is there a functions $f: mathbb{R}^2 rightarrow mathbb{R}$ separately continuous but nowhere continuous?
There is a general question Here and this is unanswered.
real-analysis
asked Nov 29 '18 at 20:58
machlearmachlear
111
Is there a functions $f: mathbb{R}^2 rightarrow mathbb{R}$ separately continuous but nowhere continuous?
There is a general question Here and this is unanswered.
real-analysis
real-analysis
asked Nov 29 '18 at 20:58
machlearmachlear
111
asked Nov 29 '18 at 20:58
machlearmachlear
111
asked Nov 29 '18 at 20:58
machlearmachlear
111
asked Nov 29 '18 at 20:58
machlearmachlear
111
asked Nov 29 '18 at 20:58
machlearmachlear
111
111
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08
add a comment |
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in
each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.
Let $f colon mathbb{R}^2 rightarrow mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.
Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = begin{cases} 1-|x| & text{ if } |x| leq 1 \ 0 & text{ otherwise} end{cases}$$
Moreover define following partion of unity:
$$g_n(x) := frac{g(x-n)}{sum_{m in mathbb{Z}} g(x+m)}$$
Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $sum_{n in mathbb{Z}}g(x+n) =1$. We define
$$f_n(x,y) := sum_{k=-infty}^infty f(x,k cdot 2^{-n}) g_k(2^nx).$$
Then $f_n$ is a continuous function and $f_n rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.
We can construct an examples, which are (for example) not continuous on $mathbb{Q} times mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is
$$f(x,y) := begin{cases} frac{xy}{x^2+y^2} & text{ if } (x,y) ne (0,0) \ 0 & text{ if } (x,y) = (0,0) end{cases}.$$
Note that because of $|xy| le x^2 +y^2$, we always have $|f(x,y)| le 1$. Let $(q_n)_{n in mathbb{N}}$ be an enumeration of $mathbb{Q}$. Define
$$g(x,y) = sum_{n,k=1}^infty frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$
This function is separately continuous, but not continuous in any point of $mathbb{Q} times mathbb{Q}$.
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
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%2f3019208%2fseparately-continuous-functions-f-mathbbr2-rightarrow-mathbbr-but-now%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
The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in
each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.
Let $f colon mathbb{R}^2 rightarrow mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.
Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = begin{cases} 1-|x| & text{ if } |x| leq 1 \ 0 & text{ otherwise} end{cases}$$
Moreover define following partion of unity:
$$g_n(x) := frac{g(x-n)}{sum_{m in mathbb{Z}} g(x+m)}$$
Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $sum_{n in mathbb{Z}}g(x+n) =1$. We define
$$f_n(x,y) := sum_{k=-infty}^infty f(x,k cdot 2^{-n}) g_k(2^nx).$$
Then $f_n$ is a continuous function and $f_n rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.
We can construct an examples, which are (for example) not continuous on $mathbb{Q} times mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is
$$f(x,y) := begin{cases} frac{xy}{x^2+y^2} & text{ if } (x,y) ne (0,0) \ 0 & text{ if } (x,y) = (0,0) end{cases}.$$
Note that because of $|xy| le x^2 +y^2$, we always have $|f(x,y)| le 1$. Let $(q_n)_{n in mathbb{N}}$ be an enumeration of $mathbb{Q}$. Define
$$g(x,y) = sum_{n,k=1}^infty frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$
This function is separately continuous, but not continuous in any point of $mathbb{Q} times mathbb{Q}$.
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
add a comment |
The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in
each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.
Let $f colon mathbb{R}^2 rightarrow mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.
Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = begin{cases} 1-|x| & text{ if } |x| leq 1 \ 0 & text{ otherwise} end{cases}$$
Moreover define following partion of unity:
$$g_n(x) := frac{g(x-n)}{sum_{m in mathbb{Z}} g(x+m)}$$
Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $sum_{n in mathbb{Z}}g(x+n) =1$. We define
$$f_n(x,y) := sum_{k=-infty}^infty f(x,k cdot 2^{-n}) g_k(2^nx).$$
Then $f_n$ is a continuous function and $f_n rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.
We can construct an examples, which are (for example) not continuous on $mathbb{Q} times mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is
$$f(x,y) := begin{cases} frac{xy}{x^2+y^2} & text{ if } (x,y) ne (0,0) \ 0 & text{ if } (x,y) = (0,0) end{cases}.$$
Note that because of $|xy| le x^2 +y^2$, we always have $|f(x,y)| le 1$. Let $(q_n)_{n in mathbb{N}}$ be an enumeration of $mathbb{Q}$. Define
$$g(x,y) = sum_{n,k=1}^infty frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$
This function is separately continuous, but not continuous in any point of $mathbb{Q} times mathbb{Q}$.
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
add a comment |
The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in
each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.
Let $f colon mathbb{R}^2 rightarrow mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.
Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = begin{cases} 1-|x| & text{ if } |x| leq 1 \ 0 & text{ otherwise} end{cases}$$
Moreover define following partion of unity:
$$g_n(x) := frac{g(x-n)}{sum_{m in mathbb{Z}} g(x+m)}$$
Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $sum_{n in mathbb{Z}}g(x+n) =1$. We define
$$f_n(x,y) := sum_{k=-infty}^infty f(x,k cdot 2^{-n}) g_k(2^nx).$$
Then $f_n$ is a continuous function and $f_n rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.
We can construct an examples, which are (for example) not continuous on $mathbb{Q} times mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is
$$f(x,y) := begin{cases} frac{xy}{x^2+y^2} & text{ if } (x,y) ne (0,0) \ 0 & text{ if } (x,y) = (0,0) end{cases}.$$
Note that because of $|xy| le x^2 +y^2$, we always have $|f(x,y)| le 1$. Let $(q_n)_{n in mathbb{N}}$ be an enumeration of $mathbb{Q}$. Define
$$g(x,y) = sum_{n,k=1}^infty frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$
This function is separately continuous, but not continuous in any point of $mathbb{Q} times mathbb{Q}$.
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in
each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.
Let $f colon mathbb{R}^2 rightarrow mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.
Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = begin{cases} 1-|x| & text{ if } |x| leq 1 \ 0 & text{ otherwise} end{cases}$$
Moreover define following partion of unity:
$$g_n(x) := frac{g(x-n)}{sum_{m in mathbb{Z}} g(x+m)}$$
Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $sum_{n in mathbb{Z}}g(x+n) =1$. We define
$$f_n(x,y) := sum_{k=-infty}^infty f(x,k cdot 2^{-n}) g_k(2^nx).$$
Then $f_n$ is a continuous function and $f_n rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.
We can construct an examples, which are (for example) not continuous on $mathbb{Q} times mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is
$$f(x,y) := begin{cases} frac{xy}{x^2+y^2} & text{ if } (x,y) ne (0,0) \ 0 & text{ if } (x,y) = (0,0) end{cases}.$$
Note that because of $|xy| le x^2 +y^2$, we always have $|f(x,y)| le 1$. Let $(q_n)_{n in mathbb{N}}$ be an enumeration of $mathbb{Q}$. Define
$$g(x,y) = sum_{n,k=1}^infty frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$
This function is separately continuous, but not continuous in any point of $mathbb{Q} times mathbb{Q}$.
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
edited Dec 1 '18 at 7:43
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
answered Nov 30 '18 at 21:26
p4schp4sch
4,770217
4,770217
add a comment |
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%2f3019208%2fseparately-continuous-functions-f-mathbbr2-rightarrow-mathbbr-but-now%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
By "separately continuous", do you mean that $xmapsto f(x,y_0)$ and $ymapsto f(x_0, y)$ are continuous for fixed $x_0,y_0$?
– Arthur
Nov 29 '18 at 21:02
The answer is on the page you linked: $mathbb R$ is Hausdorff, locally compact and metrizable, so the answer is no
– Federico
Nov 29 '18 at 21:03
Martin claims in the comments to that question that such a function is not possible when the spaces are completely metrizable.
– jgon
Nov 29 '18 at 21:03
I just want to find the solution in $f: mathbb{R}^2 rightarrow mathbb{R}$ case. The general problem, there are some concepts I do not know the definition.
– machlear
Nov 29 '18 at 21:08