How does one prove that a space is dense in another under some norm?
$begingroup$
This question arose while studying chapter 1 of Reed & Simon's book 'Methods of Modern Mathematical Physics'. It is about exercise 1.10 in particular. In this exercise, the following is asked (I reproduce all the relevant information given in the chapter):
Let $x_0,dots,x_n$ be a partition of the interval $[a,b]$, $x_0=a, x_n=b$. Let $chi_i(x)$ be the characteristic function of $[x_{i-1},x_i)$ except for $chi_n(x)$ which is the char. function of $[x_{n-1},x_n]$. A function on $[a,b]$ of the form $sum_{i=1}^n s_ichi_i(x)$ is called a step function. The set of all step functions for all possible finite partitions of $[a,b]$ is a normed linear space with the norm $$left|sum_{i=1}^n s_ichi_i(x)right|_infty=sup_{x in [a,b]}left|sum_{i=1}^n s_ichi_i(x)right|=max_{i=1,dots,n}{s_i}$$
We denote this space by $S[a,b]$.
Exercise 10: Prove that $S[a,b]$ is dense in the norm $|cdot|_infty$ in $PC[a,b]$, the space of all possible bounded piecewise continuous functions on $[a,b]$.
The definition of 'dense' was given earlier in the book:
Definition: A set $B$ in a metric space $M$ is called dense if every $xin M$ is a limit of elements of $B$.
I am quite confused about what one is supposed to do to solve the exercise: Is it asking me to show that the $|cdot|_infty$-norm of any function in $PC[a,b]$ is the limit of the $|cdot|_infty$-norm of some function in $S[a,b]$? That statement seems trivial since, with these definitions, the $|cdot|_infty$-norm of a bounded function is just its supremum, so any step function with a suitably chosen highest $s_i$ has the same $|cdot|_infty$-norm (or am I missing something here?).
Is it something else (plausibly something involving the metric naturally induced by this norm: $d(f,g)=|f-g|_infty$?) that I would have to show instead?
real-analysis functional-analysis
edited Dec 16 '18 at 7:12
davyjones
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
$endgroup$
add a comment |
$begingroup$
This question arose while studying chapter 1 of Reed & Simon's book 'Methods of Modern Mathematical Physics'. It is about exercise 1.10 in particular. In this exercise, the following is asked (I reproduce all the relevant information given in the chapter):
Let $x_0,dots,x_n$ be a partition of the interval $[a,b]$, $x_0=a, x_n=b$. Let $chi_i(x)$ be the characteristic function of $[x_{i-1},x_i)$ except for $chi_n(x)$ which is the char. function of $[x_{n-1},x_n]$. A function on $[a,b]$ of the form $sum_{i=1}^n s_ichi_i(x)$ is called a step function. The set of all step functions for all possible finite partitions of $[a,b]$ is a normed linear space with the norm $$left|sum_{i=1}^n s_ichi_i(x)right|_infty=sup_{x in [a,b]}left|sum_{i=1}^n s_ichi_i(x)right|=max_{i=1,dots,n}{s_i}$$
We denote this space by $S[a,b]$.
Exercise 10: Prove that $S[a,b]$ is dense in the norm $|cdot|_infty$ in $PC[a,b]$, the space of all possible bounded piecewise continuous functions on $[a,b]$.
The definition of 'dense' was given earlier in the book:
Definition: A set $B$ in a metric space $M$ is called dense if every $xin M$ is a limit of elements of $B$.
I am quite confused about what one is supposed to do to solve the exercise: Is it asking me to show that the $|cdot|_infty$-norm of any function in $PC[a,b]$ is the limit of the $|cdot|_infty$-norm of some function in $S[a,b]$? That statement seems trivial since, with these definitions, the $|cdot|_infty$-norm of a bounded function is just its supremum, so any step function with a suitably chosen highest $s_i$ has the same $|cdot|_infty$-norm (or am I missing something here?).
Is it something else (plausibly something involving the metric naturally induced by this norm: $d(f,g)=|f-g|_infty$?) that I would have to show instead?
real-analysis functional-analysis
edited Dec 16 '18 at 7:12
davyjones
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
$endgroup$
2
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53
add a comment |
$begingroup$
This question arose while studying chapter 1 of Reed & Simon's book 'Methods of Modern Mathematical Physics'. It is about exercise 1.10 in particular. In this exercise, the following is asked (I reproduce all the relevant information given in the chapter):
Let $x_0,dots,x_n$ be a partition of the interval $[a,b]$, $x_0=a, x_n=b$. Let $chi_i(x)$ be the characteristic function of $[x_{i-1},x_i)$ except for $chi_n(x)$ which is the char. function of $[x_{n-1},x_n]$. A function on $[a,b]$ of the form $sum_{i=1}^n s_ichi_i(x)$ is called a step function. The set of all step functions for all possible finite partitions of $[a,b]$ is a normed linear space with the norm $$left|sum_{i=1}^n s_ichi_i(x)right|_infty=sup_{x in [a,b]}left|sum_{i=1}^n s_ichi_i(x)right|=max_{i=1,dots,n}{s_i}$$
We denote this space by $S[a,b]$.
Exercise 10: Prove that $S[a,b]$ is dense in the norm $|cdot|_infty$ in $PC[a,b]$, the space of all possible bounded piecewise continuous functions on $[a,b]$.
The definition of 'dense' was given earlier in the book:
Definition: A set $B$ in a metric space $M$ is called dense if every $xin M$ is a limit of elements of $B$.
I am quite confused about what one is supposed to do to solve the exercise: Is it asking me to show that the $|cdot|_infty$-norm of any function in $PC[a,b]$ is the limit of the $|cdot|_infty$-norm of some function in $S[a,b]$? That statement seems trivial since, with these definitions, the $|cdot|_infty$-norm of a bounded function is just its supremum, so any step function with a suitably chosen highest $s_i$ has the same $|cdot|_infty$-norm (or am I missing something here?).
Is it something else (plausibly something involving the metric naturally induced by this norm: $d(f,g)=|f-g|_infty$?) that I would have to show instead?
real-analysis functional-analysis
edited Dec 16 '18 at 7:12
davyjones
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
$endgroup$
This question arose while studying chapter 1 of Reed & Simon's book 'Methods of Modern Mathematical Physics'. It is about exercise 1.10 in particular. In this exercise, the following is asked (I reproduce all the relevant information given in the chapter):
Let $x_0,dots,x_n$ be a partition of the interval $[a,b]$, $x_0=a, x_n=b$. Let $chi_i(x)$ be the characteristic function of $[x_{i-1},x_i)$ except for $chi_n(x)$ which is the char. function of $[x_{n-1},x_n]$. A function on $[a,b]$ of the form $sum_{i=1}^n s_ichi_i(x)$ is called a step function. The set of all step functions for all possible finite partitions of $[a,b]$ is a normed linear space with the norm $$left|sum_{i=1}^n s_ichi_i(x)right|_infty=sup_{x in [a,b]}left|sum_{i=1}^n s_ichi_i(x)right|=max_{i=1,dots,n}{s_i}$$
We denote this space by $S[a,b]$.
Exercise 10: Prove that $S[a,b]$ is dense in the norm $|cdot|_infty$ in $PC[a,b]$, the space of all possible bounded piecewise continuous functions on $[a,b]$.
The definition of 'dense' was given earlier in the book:
Definition: A set $B$ in a metric space $M$ is called dense if every $xin M$ is a limit of elements of $B$.
I am quite confused about what one is supposed to do to solve the exercise: Is it asking me to show that the $|cdot|_infty$-norm of any function in $PC[a,b]$ is the limit of the $|cdot|_infty$-norm of some function in $S[a,b]$? That statement seems trivial since, with these definitions, the $|cdot|_infty$-norm of a bounded function is just its supremum, so any step function with a suitably chosen highest $s_i$ has the same $|cdot|_infty$-norm (or am I missing something here?).
Is it something else (plausibly something involving the metric naturally induced by this norm: $d(f,g)=|f-g|_infty$?) that I would have to show instead?
real-analysis functional-analysis
real-analysis functional-analysis
edited Dec 16 '18 at 7:12
davyjones
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
edited Dec 16 '18 at 7:12
davyjones
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
edited Dec 16 '18 at 7:12
davyjones
393213
edited Dec 16 '18 at 7:12
davyjones
393213
edited Dec 16 '18 at 7:12
davyjones
393213
393213
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
asked Jul 7 '14 at 23:40
DanuDanu
1,30911120
1,30911120
2
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53
add a comment |
2
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53
2
2
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)
So one way of rephrasing the question is as follows:
Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n - f|_infty = 0$.
This is not the same as the statement:
[WRONG] Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n|_infty = |f|_infty$.
The second statement is much weaker than the first.
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
$endgroup$
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%2f859619%2fhow-does-one-prove-that-a-space-is-dense-in-another-under-some-norm%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
$begingroup$
Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)
So one way of rephrasing the question is as follows:
Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n - f|_infty = 0$.
This is not the same as the statement:
[WRONG] Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n|_infty = |f|_infty$.
The second statement is much weaker than the first.
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
$endgroup$
add a comment |
$begingroup$
Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)
So one way of rephrasing the question is as follows:
Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n - f|_infty = 0$.
This is not the same as the statement:
[WRONG] Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n|_infty = |f|_infty$.
The second statement is much weaker than the first.
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
$endgroup$
add a comment |
$begingroup$
Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)
So one way of rephrasing the question is as follows:
Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n - f|_infty = 0$.
This is not the same as the statement:
[WRONG] Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n|_infty = |f|_infty$.
The second statement is much weaker than the first.
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
$endgroup$
Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)
So one way of rephrasing the question is as follows:
Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n - f|_infty = 0$.
This is not the same as the statement:
[WRONG] Show that for each $f in PC[a,b]$ there exists a sequence ${f_n} subset S[a,b]$ such that $lim_{n to infty} |f_n|_infty = |f|_infty$.
The second statement is much weaker than the first.
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
edited Jul 8 '14 at 0:05
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
answered Jul 7 '14 at 23:55
Nate EldredgeNate Eldredge
63.3k682171
63.3k682171
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%2f859619%2fhow-does-one-prove-that-a-space-is-dense-in-another-under-some-norm%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
$begingroup$
Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm)
$endgroup$
– qaphla
Jul 7 '14 at 23:44
$begingroup$
@qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)?
$endgroup$
– Danu
Jul 7 '14 at 23:53