Proving that $d(x,ker f) = frac{|f(x)|}{|f|}$ if $f in X^*$
Exercise :
Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
$$d(x,V)=inf{|x-v|: v in V}$$
Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
$$d(x,ker f)=frac{|f(x)|}{|f|}$$
Attempt - thoughts :
So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :
$$|f(x)| leq M|x|$$
Also, the operator norm is given by :
$$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$
Now, the kernel of the function $f$ is defined as :
$$ker f = {x in X : f(x) = 0}$$
Essentialy, we need to calculate :
$$d(x,ker f) = inf{|x-v|:v in ker f}$$
Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
$$text{co}dim{ker f} = 1$$
I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.
Any hints, tips or thorough elaborations will be greatly appreciated !
real-analysis functional-analysis operator-theory
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
add a comment |
Exercise :
Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
$$d(x,V)=inf{|x-v|: v in V}$$
Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
$$d(x,ker f)=frac{|f(x)|}{|f|}$$
Attempt - thoughts :
So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :
$$|f(x)| leq M|x|$$
Also, the operator norm is given by :
$$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$
Now, the kernel of the function $f$ is defined as :
$$ker f = {x in X : f(x) = 0}$$
Essentialy, we need to calculate :
$$d(x,ker f) = inf{|x-v|:v in ker f}$$
Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
$$text{co}dim{ker f} = 1$$
I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.
Any hints, tips or thorough elaborations will be greatly appreciated !
real-analysis functional-analysis operator-theory
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
add a comment |
Exercise :
Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
$$d(x,V)=inf{|x-v|: v in V}$$
Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
$$d(x,ker f)=frac{|f(x)|}{|f|}$$
Attempt - thoughts :
So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :
$$|f(x)| leq M|x|$$
Also, the operator norm is given by :
$$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$
Now, the kernel of the function $f$ is defined as :
$$ker f = {x in X : f(x) = 0}$$
Essentialy, we need to calculate :
$$d(x,ker f) = inf{|x-v|:v in ker f}$$
Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
$$text{co}dim{ker f} = 1$$
I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.
Any hints, tips or thorough elaborations will be greatly appreciated !
real-analysis functional-analysis operator-theory
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
Exercise :
Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
$$d(x,V)=inf{|x-v|: v in V}$$
Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
$$d(x,ker f)=frac{|f(x)|}{|f|}$$
Attempt - thoughts :
So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :
$$|f(x)| leq M|x|$$
Also, the operator norm is given by :
$$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$
Now, the kernel of the function $f$ is defined as :
$$ker f = {x in X : f(x) = 0}$$
Essentialy, we need to calculate :
$$d(x,ker f) = inf{|x-v|:v in ker f}$$
Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
$$text{co}dim{ker f} = 1$$
I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.
Any hints, tips or thorough elaborations will be greatly appreciated !
real-analysis functional-analysis operator-theory
real-analysis functional-analysis operator-theory
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
edited Nov 18 at 16:30
Xander Henderson
14.1k103554
14.1k103554
asked Nov 18 at 14:08
Rebellos
14.4k31245
asked Nov 18 at 14:08
Rebellos
14.4k31245
asked Nov 18 at 14:08
Rebellos
14.4k31245
14.4k31245
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.
answered Nov 18 at 15:22
Matematleta
9,9522918
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
add a comment |
First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.
Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.
Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
$$|f(x)|= |f(x-v)|le|x-v|$$
which shows $|f(x)|le d(x,ker f)$.
For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.
Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
$$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
which tends to $|f(x)|$ as $ntoinfty$.
answered Nov 18 at 15:29
Berci
59.6k23672
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
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%2f3003578%2fproving-that-dx-ker-f-fracfx-f-if-f-in-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.
answered Nov 18 at 15:22
Matematleta
9,9522918
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
add a comment |
hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.
answered Nov 18 at 15:22
Matematleta
9,9522918
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
add a comment |
hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.
answered Nov 18 at 15:22
Matematleta
9,9522918
hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.
answered Nov 18 at 15:22
Matematleta
9,9522918
answered Nov 18 at 15:22
Matematleta
9,9522918
answered Nov 18 at 15:22
Matematleta
9,9522918
answered Nov 18 at 15:22
Matematleta
9,9522918
9,9522918
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
add a comment |
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
– Rebellos
Nov 18 at 22:23
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
– Matematleta
Nov 18 at 22:51
add a comment |
First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.
Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.
Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
$$|f(x)|= |f(x-v)|le|x-v|$$
which shows $|f(x)|le d(x,ker f)$.
For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.
Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
$$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
which tends to $|f(x)|$ as $ntoinfty$.
answered Nov 18 at 15:29
Berci
59.6k23672
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
add a comment |
First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.
Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.
Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
$$|f(x)|= |f(x-v)|le|x-v|$$
which shows $|f(x)|le d(x,ker f)$.
For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.
Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
$$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
which tends to $|f(x)|$ as $ntoinfty$.
answered Nov 18 at 15:29
Berci
59.6k23672
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
add a comment |
First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.
Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.
Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
$$|f(x)|= |f(x-v)|le|x-v|$$
which shows $|f(x)|le d(x,ker f)$.
For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.
Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
$$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
which tends to $|f(x)|$ as $ntoinfty$.
answered Nov 18 at 15:29
Berci
59.6k23672
First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.
Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.
Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
$$|f(x)|= |f(x-v)|le|x-v|$$
which shows $|f(x)|le d(x,ker f)$.
For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.
Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
$$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
which tends to $|f(x)|$ as $ntoinfty$.
answered Nov 18 at 15:29
Berci
59.6k23672
answered Nov 18 at 15:29
Berci
59.6k23672
answered Nov 18 at 15:29
Berci
59.6k23672
answered Nov 18 at 15:29
Berci
59.6k23672
59.6k23672
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
add a comment |
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
– Rebellos
Nov 18 at 22:20
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3003578%2fproving-that-dx-ker-f-fracfx-f-if-f-in-x%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
