Mixing
If $f:{lambda a+(1-lambda)b:0leqlambda leq 1}to mathbb{R}$ is a linear function, then is it true that $f$ is...
$begingroup$
Question: Let $V$ be a real vector space and $a,bin V$ such that $aneq b.$
If $f:{lambda a+(1-lambda)b:0leqlambda leq 1}to mathbb{R}$ is a linear function, then is it true that $f$ is continuous?
I think it is true.
Since $f$ is linear, it suffices to show that $f$ is bounded.
Let $M = max{|f(a)|,|f(b)|}.$
Note that
$$|f(lambda a +(1-lambda)b)| = |lambda f(a) + (1-lambda) f(b)| leq lambda M + (1-lambda)M = M.$$
So $f$ is bounded.
Is my proof above correct?
real-analysis vector-spaces linear-transformations
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
$endgroup$
add a comment |
$begingroup$
Question: Let $V$ be a real vector space and $a,bin V$ such that $aneq b.$
If $f:{lambda a+(1-lambda)b:0leqlambda leq 1}to mathbb{R}$ is a linear function, then is it true that $f$ is continuous?
I think it is true.
Since $f$ is linear, it suffices to show that $f$ is bounded.
Let $M = max{|f(a)|,|f(b)|}.$
Note that
$$|f(lambda a +(1-lambda)b)| = |lambda f(a) + (1-lambda) f(b)| leq lambda M + (1-lambda)M = M.$$
So $f$ is bounded.
Is my proof above correct?
real-analysis vector-spaces linear-transformations
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
$endgroup$
$begingroup$
I would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
1
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
1
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07
add a comment |
$begingroup$
Question: Let $V$ be a real vector space and $a,bin V$ such that $aneq b.$
If $f:{lambda a+(1-lambda)b:0leqlambda leq 1}to mathbb{R}$ is a linear function, then is it true that $f$ is continuous?
I think it is true.
Since $f$ is linear, it suffices to show that $f$ is bounded.
Let $M = max{|f(a)|,|f(b)|}.$
Note that
$$|f(lambda a +(1-lambda)b)| = |lambda f(a) + (1-lambda) f(b)| leq lambda M + (1-lambda)M = M.$$
So $f$ is bounded.
Is my proof above correct?
real-analysis vector-spaces linear-transformations
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
$endgroup$
Question: Let $V$ be a real vector space and $a,bin V$ such that $aneq b.$
If $f:{lambda a+(1-lambda)b:0leqlambda leq 1}to mathbb{R}$ is a linear function, then is it true that $f$ is continuous?
I think it is true.
Since $f$ is linear, it suffices to show that $f$ is bounded.
Let $M = max{|f(a)|,|f(b)|}.$
Note that
$$|f(lambda a +(1-lambda)b)| = |lambda f(a) + (1-lambda) f(b)| leq lambda M + (1-lambda)M = M.$$
So $f$ is bounded.
Is my proof above correct?
real-analysis vector-spaces linear-transformations
real-analysis vector-spaces linear-transformations
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
asked Feb 3 at 3:23
IdonknowIdonknow
2,610950119
2,610950119
$begingroup$
I would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
1
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
1
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07
add a comment |
$begingroup$
I would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
1
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
1
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07
$begingroup$
I would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
$begingroup$
I would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
1
1
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
1
1
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $x=lambda_{x}a+(1-lambda_{x})b$ and $y=lambda_{y}a+(1-lambda_{y})b$ be points in $operatorname{dom}f$.
Note that
$$
x-y=left(lambda_{x}-lambda_{y}right)left(a-bright).
$$
Suppose $V$ is a normed vector space and that $aneq b$ (the case of $a=b$ is uninteresting since $operatorname{dom}f$ becomes a singleton).
By the above,
$$
left|lambda_{x}-lambda_{y}right|=frac{leftVert x-yrightVert }{leftVert a-brightVert }.
$$
Moreover,
$$
left|f(x)-f(y)right|=left|f(x-y)right|=left|lambda_{x}-lambda_{y}right|left|f(a-b)right|,
$$
and hence $f$ is continuous since $|lambda_{x}-lambda_{y}|rightarrow0$ as $leftVert x-yrightVert rightarrow0$.
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
$endgroup$
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
add a comment |
Your Answer
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%2f3098124%2fif-f-lambda-a1-lambdab0-leq-lambda-leq-1-to-mathbbr-is-a-linear%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
$begingroup$
Let $x=lambda_{x}a+(1-lambda_{x})b$ and $y=lambda_{y}a+(1-lambda_{y})b$ be points in $operatorname{dom}f$.
Note that
$$
x-y=left(lambda_{x}-lambda_{y}right)left(a-bright).
$$
Suppose $V$ is a normed vector space and that $aneq b$ (the case of $a=b$ is uninteresting since $operatorname{dom}f$ becomes a singleton).
By the above,
$$
left|lambda_{x}-lambda_{y}right|=frac{leftVert x-yrightVert }{leftVert a-brightVert }.
$$
Moreover,
$$
left|f(x)-f(y)right|=left|f(x-y)right|=left|lambda_{x}-lambda_{y}right|left|f(a-b)right|,
$$
and hence $f$ is continuous since $|lambda_{x}-lambda_{y}|rightarrow0$ as $leftVert x-yrightVert rightarrow0$.
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
$endgroup$
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
add a comment |
$begingroup$
Let $x=lambda_{x}a+(1-lambda_{x})b$ and $y=lambda_{y}a+(1-lambda_{y})b$ be points in $operatorname{dom}f$.
Note that
$$
x-y=left(lambda_{x}-lambda_{y}right)left(a-bright).
$$
Suppose $V$ is a normed vector space and that $aneq b$ (the case of $a=b$ is uninteresting since $operatorname{dom}f$ becomes a singleton).
By the above,
$$
left|lambda_{x}-lambda_{y}right|=frac{leftVert x-yrightVert }{leftVert a-brightVert }.
$$
Moreover,
$$
left|f(x)-f(y)right|=left|f(x-y)right|=left|lambda_{x}-lambda_{y}right|left|f(a-b)right|,
$$
and hence $f$ is continuous since $|lambda_{x}-lambda_{y}|rightarrow0$ as $leftVert x-yrightVert rightarrow0$.
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
$endgroup$
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
add a comment |
$begingroup$
Let $x=lambda_{x}a+(1-lambda_{x})b$ and $y=lambda_{y}a+(1-lambda_{y})b$ be points in $operatorname{dom}f$.
Note that
$$
x-y=left(lambda_{x}-lambda_{y}right)left(a-bright).
$$
Suppose $V$ is a normed vector space and that $aneq b$ (the case of $a=b$ is uninteresting since $operatorname{dom}f$ becomes a singleton).
By the above,
$$
left|lambda_{x}-lambda_{y}right|=frac{leftVert x-yrightVert }{leftVert a-brightVert }.
$$
Moreover,
$$
left|f(x)-f(y)right|=left|f(x-y)right|=left|lambda_{x}-lambda_{y}right|left|f(a-b)right|,
$$
and hence $f$ is continuous since $|lambda_{x}-lambda_{y}|rightarrow0$ as $leftVert x-yrightVert rightarrow0$.
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
$endgroup$
Let $x=lambda_{x}a+(1-lambda_{x})b$ and $y=lambda_{y}a+(1-lambda_{y})b$ be points in $operatorname{dom}f$.
Note that
$$
x-y=left(lambda_{x}-lambda_{y}right)left(a-bright).
$$
Suppose $V$ is a normed vector space and that $aneq b$ (the case of $a=b$ is uninteresting since $operatorname{dom}f$ becomes a singleton).
By the above,
$$
left|lambda_{x}-lambda_{y}right|=frac{leftVert x-yrightVert }{leftVert a-brightVert }.
$$
Moreover,
$$
left|f(x)-f(y)right|=left|f(x-y)right|=left|lambda_{x}-lambda_{y}right|left|f(a-b)right|,
$$
and hence $f$ is continuous since $|lambda_{x}-lambda_{y}|rightarrow0$ as $leftVert x-yrightVert rightarrow0$.
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
answered Feb 3 at 4:28
parsiadparsiad
18.8k32554
18.8k32554
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
add a comment |
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
But this only works on the subspace generated by $a,b$. In general the assertion cannot be true since any, also discontinuous, linear mapping is continuous in finite dimensional subspaces.
$endgroup$
– Jens Schwaiger
Feb 3 at 7:12
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
$begingroup$
Since OP wrote $f : {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } rightarrow mathbb{R}$, I assumed $operatorname{dom}f = {lambda a + (1 - lambda b) b: 0 leq lambda leq 1 } subset operatorname{span}(a, b)$.
$endgroup$
– parsiad
Feb 3 at 21:59
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%2f3098124%2fif-f-lambda-a1-lambdab0-leq-lambda-leq-1-to-mathbbr-is-a-linear%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 would be careful, since from what I remember a linear transformation on a vector space is (locally) bounded if and only if it is continuous (I think this is the theorem you are quoting), however the domain of $f$ is not a vector space in this case so I'm not sure if the theorem holds here.
$endgroup$
– clmundergrad
Feb 3 at 3:31
1
$begingroup$
Is $V$ equipped with a norm? Or is it just a general topological vector space?
$endgroup$
– parsiad
Feb 3 at 3:56
1
$begingroup$
The question has no meaning unless you specify a topology on $V$.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 5:07