Post History
#5: Post edited
by
Snoopy
·
2024-07-18T16:17:18Z (4 months ago)
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
- Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $$L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim.$$ Here, we only care about the set structure, so the quotient is a "[quotient set](https://en.wikipedia.org/wiki/Equivalence_class)."
- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $\iota(f)= [f]$
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
- **Short answer**.
- That means there exists an _injection_ $\iota:C([0,1])\to L^\infty([0,1])$ and one can identify $C([0,1])$ with the image $\iota(C([0,1]))$, which is a subset of $L^\infty([0,1])$.
- ---
- **Now a long one.**
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
- Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $$L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim.$$ Here, we only care about the set structure, so the quotient is a "[quotient set](https://en.wikipedia.org/wiki/Equivalence_class)."
- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $\iota(f)= [f]$
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
#4: Post edited
by
Snoopy
·
2024-01-10T03:20:15Z (11 months ago)
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $\iota(f)= [f]$
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
- Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $$L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim.$$ Here, we only care about the set structure, so the quotient is a "[quotient set](https://en.wikipedia.org/wiki/Equivalence_class)."
- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $\iota(f)= [f]$
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
#3: Post edited
by
Snoopy
·
2024-01-10T03:19:04Z (11 months ago)
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1]) /\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
$\iota(f)= [f]$- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
- Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."
- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $\iota(f)= [f]$
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
#2: Post edited
by
Snoopy
·
2024-01-10T03:18:27Z (11 months ago)
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set. Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1]) /\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
$$- \iota(f)= [f]
$$- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that$C([0,1])$ is not dense in $L^\infty([0,1])$,which really means$\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.- Such identification is so common in analysis that people just use the short version.
- We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
- Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1]) /\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."
- Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
- $
- \iota(f)= [f]
- $
- where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
- By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
- With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
- Such identification is so common in analysis that people just use the short version.
#1: Initial revision
by
Snoopy
·
2024-01-10T03:16:03Z (11 months ago)
We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set. Then $L^\infty([0,1]) =\mathscr{L}^\infty([0,1]) /\sim$. Here, we only care about the set structure, so the quotient is a "quotient set."
Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with
$$
\iota(f)= [f]
$$
where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a _bijection_ between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that
$C([0,1])$ is not dense in $L^\infty([0,1])$,
which really means
$\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
Such identification is so common in analysis that people just use the short version.