An instance of quotient Space $X/M$

Question

Hi Guys i can't find a simple example (with analytic description i mean) that helps me to understand the meaning of quotient space.

I've understood the definition ($X$ normed linear space, $M$ closed subspace of $X$, for each $x,y \in X$ $xRy$ iff $x - y \in M$ we define $X/M$ as the set of all equivalence class is there any example of the form

$X/M = \left\{ x+M : x\in X \right\}$

i'm asking such thing because even if i've understood the definition i don't know how to apply to build an example of "quotient space".

Answer 1

I like the next one. Let $X=C[0,1]$ with the usual $\| \cdot \|_\infty$ norm and $M=\{ f \in C[0,1] : f(1/2)=0\}$. It is easy to see that $M$ is a closed subspace of $X$.

Can you now prove that $X/M \simeq \mathbb{C}$? Where $X/M$ is equipped with the quotient norm $(\|\cdot\|_{X/M})$ and $\mathbb C$ with the complex modulus norm $(|\cdot |)$.

HINT Prove that $\phi : X/M \to \mathbb C$ given by $$ \phi([f]):= f(1/2) $$ is an isometric isomorphism, that is that $\phi$ is a linear bijection and that $|\phi([f])|=\|[f]\|_{X/M}$

Spoiler solution

Note that $X/M=\{[f]: f \in X\}$ where $[f]=\{g \in X : f-g \in M\}=\{ g \in X : g(1/2)=f(1/2)\}$. Since $$\phi([\lambda f+g])=\lambda f(1/2)+g(1/2)=\lambda\phi([f])+\phi([g]),$$ $\phi$ is a linear map. Lets now check that $\phi$ is onto, clearly $\phi(X/M)\subset \mathbb C$, now if $z\in \mathbb C$ and define $f_z:[0,1] \to \mathbb{C}$ as $f_z(t):=z $ for all $t \in [0,1]$, clearly $f_z \in X$ and $\phi([f_z])=f_z(1/2)=z$, hence $\phi(X/M)\supset \mathbb C$ and therefore $\phi$ is onto. Finally, fix any $f \in X$, by quotient norm definition, $$ \| [f] \|_{X/M}=\inf\{ \|g\|_{\infty}: g \in [f]\}=\inf\{ \|g\|_{\infty}: g(1/2)=f(1/2)\} \geq |f(1/2)|=|\phi([f])|.$$ On the other hand, if $h(t):=f(1/2)$ then $\|h\|_{\infty} = |f(1/2)|=|\phi([f])|$ and thus $h \in [f]$, hence $$ \|[f]\|_{X/M}=\inf\{ \|g\|_{\infty}: g \in [f]\} \leq \|h\|_{\infty}=|\phi([f])|.$$ So indeed $|\phi([f])|=\|[f]\|_{X/M}$. This tells you that as normed spaces $X/M$ and $\mathbb{C}$ are exactly the same one, just with elements having different names, for instance the element $[f]\in X/M$ is fact the element $f(1/2)\in \mathbb{C}$, since they have the same norm they are essentially the same one, and that is why we say that $$(X/M, \|\cdot\|_{X/M}) \simeq (\mathbb C, |\cdot|)$$

Answer 2

Here is a perhaps easier to understand example. Let $(X,\mathcal M,\mu)$ be a measure space. We say that a function $f:X\to\mathbb R$ is an $L^1$ function, or $f\in L^1$, if $$\int_X |f|\mathsf d\mu <\infty. $$ Now, strictly speaking, this is an abuse of notation. To be more concrete, suppose $X=[0,1]$ and $\mu$ is Lebesgue measure restricted to $[0,1]$. Then the function $f:[0,1]\to\mathbb R$ defined by $f(x) = x$ is $L^1$, since $$\int_{[0,1]} |f|\mathsf d\mu = \int_0^1 x\mathsf dx = \frac12. $$ But changing the value of a function at countably many points does not affect its integral, since the Lebesgue measure of a countable set is zero. Hence the function $g:[0,1]\to\mathbb R$ with $$g(x) = \begin{cases} x,& x\ne\frac12\\ 1000,& x=\frac12\end{cases} $$ is $L^1$, since again $$\int_{[0,1]} |g|\mathsf d\mu = \frac12. $$ From a measure-theoretic standpoint, $f$ and $g$ are essentially the same function; we have $f=g$ a.e. (almost everywhere) since $$\mu(\{x\in[0,1] : f(x)\ne g(x)\}) = 0. $$ So the actual definition of $L^1(\mu)$ here is $$\left\{f:[0,1]\to\mathbb R \mid \int_X |f|\mathsf d\mu <\infty \right\} / \sim $$ where $\sim$ is the equivalence relation $$f\sim g\iff f=g \text{ a.e.} $$ In other words, the elements of $L^1(\mu)$ are not functions, but instead equivalence relations of functions.

Answer 3

Quotient Space

Given a Banach space $E$.

Construct the quotient:* $$E/M:=\{x+M:x\in E\}$$

As well as its norm: $$\|x+M\|:=\inf_{m\in M}\|x+m\|$$

By closedness indeed:* $$\|x+M\|=0\iff x+M=0+M$$

*Here closed subspace required!

Example

Given a measure space $\Omega$.

Consider the Banach algebra: $$\mathcal{B}_b(\Omega):=\{f\in\mathcal{B}(\Omega):\|f\|_\infty<\infty\}=:\mathcal{B}$$

Regard the closed ideal: $$\mathcal{I}_\mu(\Omega):=\{i\in\mathcal{B}_b(\Omega):i=0\bmod\mu\}=:\mathcal{I}$$

Construct the quotient: $$\mathcal{B}_b(\Omega)/\mathcal{I}_\mu(\Omega):=\{f+\mathcal{I}:f\in\mathcal{B}\}=:\mathcal{B}/\mathcal{I}$$

As well as its norm:* $$\|f+\mathcal{I}\|:=\inf_{i\in\mathcal{I}}\|f+i\|_\infty=:\mathrm{ess}-\sup_{\omega\in\Omega}\|f(\omega)\|$$

*This is essential supremum!

Answer 4

Problem

What makes the whole subject so difficult is that most of the time it is spoken in abstract terms like saying these are indeed equivalence classes. This may and most of the time does create more confusion. In fact, equivalence classes are much easier to handle than one might think of as soon as one demascates them as merely... ...sets.

Now, first ask yourself how would you define their distance to zero. Then explicitely write it out on a piece of paper. And finally think about what the essential supremum must look like in special.