This blog post is intended to deliver a quick explanation of the
algebra of Borel measures on \(\mathbb{R}^n\). It will be broken into
pieces. All complex-valued complex Borel measures \(M(\mathbb{R}^n)\) clearly form a vector
space over \(\mathbb{C}\). The main
goal of this post is to show that this is a Banach space and also a
Banach algebra.

In fact, the \(\mathbb{R}^n\) case
can be generalised into any locally compact abelian group (see any
abstract harmonic analysis books), this is because what really matters
here is being locally compact and abelian. But at this moment we stick
to Euclidean spaces. Note since \(\mathbb{R}^n\) is \(\sigma\)-compact, all Borel measures are
regular.

To read this post you need to be familiar with some basic properties
of Banach algebra, complex Borel measures, and the most important,
Fubini's theorem.

Throughout we consider the Hilbert space \(L^2=L^2(\mathbb{R})\), the space of all
complex-valued functions with real variable such that \(f \in L^2\) if and only if \[
\lVert f \rVert_2^2=\int_{-\infty}^{\infty}|f(t)|^2dm(t)<\infty
\] where \(m\) denotes the
ordinary Lebesgue measure (in fact it's legitimate to consider Riemann
integral in this context).

For each \(t \geq 0\), we assign an
bounded linear operator \(Q(t)\) such
that \[
(Q(t)f)(s)=f(s+t).
\] This is indeed bounded since we have \(\lVert Q(t)f \rVert_2 = \lVert f \rVert_2\)
as the Lebesgue measure is translate-invariant. This is a left
translation operator with a single step \(t\).

(Gleason-Kahane-Żelazko) If \(\phi\) is a complex linear functional on a
unitary Banach algebra \(A\), such that
\(\phi(e)=1\) and \(\phi(x) \neq 0\) for every invertible \(x \in A\), then \[
\phi(xy)=\phi(x)\phi(y)
\] Namely, \(\phi\) is a complex
homomorphism.

Notations and remarks

Suppose \(A\) is a complex unitary
Banach algebra and \(\phi: A \to
\mathbb{C}\) is a linear functional which is not identically
\(0\) (for convenience), and if \[
\phi(xy)=\phi(x)\phi(y)
\] for all \(x \in A\) and \(y \in A\), then \(\phi\) is called a complex
homomorphism on \(A\). Note that a
unitary Banach algebra (with \(e\) as
multiplicative unit) is also a ring, so is \(\mathbb{C}\), we may say in this case \(\phi\) is a ring-homomorphism. For such
\(\phi\), we have an instant
proposition:

Proposition 0\(\phi(e)=1\) and \(\phi(x) \neq 0\) for every invertible \(x \in A\).

Proof. Since \(\phi(e)=\phi(ee)=\phi(e)\phi(e)\), we have
\(\phi(e)=0\) or \(\phi(e)=1\). If \(\phi(e)=0\) however, for any \(y \in A\), we have \(\phi(y)=\phi(ye)=\phi(y)\phi(e)=0\), which
is an excluded case. Hence \(\phi(e)=1\).

For invertible \(x \in A\), note
that \(\phi(xx^{-1})=\phi(x)\phi(x^{-1})=\phi(e)=1\).
This can't happen if \(\phi(x)=0\).
\(\square\)

The theorem reveals that Proposition \(0\) actually characterizes the complex
homomorphisms (ring-homomorphisms) among the linear functionals
(group-homomorphisms).

This theorem was proved by Andrew M. Gleason in 1967 and later
independently by J.-P. Kahane and W. Żelazko in 1968. Both of them
worked mainly on commutative Banach algebras, and the non-commutative
version, which focused on complex homomorphism, was by W. Żelazko. In
this post we will follow the third one.

Unfortunately, one cannot find an educational proof on the Internet
with ease, which may be the reason why I write this post and why you
read this.

Equivalences

Following definitions of Banach algebra and some logic manipulation,
we have several equivalences worth noting.

Subspace and ideal version

(Stated by Gleason) Let \(M\) be a linear subspace of codimension one
in a commutative Banach algebra \(A\)
having an identity. Suppose no element of \(M\) is invertible, then \(M\) is an ideal.

(Stated by Kahane and Żelazko) A subspace \(X \subset A\) of codimension \(1\) is a maximal ideal if and only if it
consists of non-invertible elements.

Spectrum version

(Stated by Kahane and Żelazko) Let \(A\) be a commutative complex Banach algebra
with unit element. Then a functional \(f \in
A^\ast\) is a multiplicative linear functional if and only if
\(f(x)=\sigma(x)\) holds for all \(x \in A\).

Here \(\sigma(x)\) denotes the
spectrum of \(x\).

The connection

Clearly any maximal ideal contains no invertible element (if so, then
it contains \(e\), then it's the ring
itself). So it suffices to show that it has codimension 1, and if it
consists of non-invertible elements. Also note that every maximal ideal
is the kernel of some complex homomorphism. For such a subspace \(X \subset A\), since \(e \notin X\), we may define \(\phi\) so that \(\phi(e)=1\), and \(\phi(x) \in \sigma(x)\) for all \(x \in A\). Note that \(\phi(e)=1\) holds if and only if \(\phi(x) \in \sigma(x)\). As we will show,
\(\phi\) has to be a complex
homomorphism.

Tools to prove the theorem

Lemma 0 Suppose \(A\) is a unitary Banach algebra, \(x \in A\), \(\lVert x \rVert<1\), then \(e-x\) is invertible.

This lemma can be found in any functional analysis book introducing
Banach algebra.

Lemma 1 Suppose \(f\) is an entire function of one complex
variable, \(f(0)=1\), \(f'(0)=0\), and \[
0<|f(\lambda)| \leq e^{|\lambda|}
\] for all complex \(\lambda\),
then \(f(\lambda)=1\) for all \(\lambda \in \mathbb{C}\).

Note that there is an entire function \(g\) such that \(f=\exp(g)\). It can be shown that \(g=0\). Indeed, if we put \[
h_r(\lambda) = \frac{r^2g(\lambda)}{\lambda^2[2r-g(\lambda)]}
\] then we see \(h_r\) is
holomorphic in the open disk centred at \(0\) with radius \(2r\). Besides, \(|h_r(\lambda)| \leq 1\) if \(|\lambda|=r\). By the maximum modulus
theorem, we have \[
|h_r(\lambda)| \leq 1
\] whenever \(|\lambda| \leq
r\). Fix \(\lambda\) and let
\(r \to \infty\), by definition of
\(h_r(\lambda)\), we must have \(g(\lambda)=0\).

Jordan homomorphism

A mapping \(\phi\) from one ring
\(R\) to another ring \(R'\) is said to be a Jordan
homomorphism from \(R\) to
\(R'\) if \[
\phi(a+b)=\phi(a)+\phi(b)
\] and \[
\phi(ab+ba)=\phi(a)\phi(b)+\phi(b)\phi(a).
\] It's of course clear that every homomorphism is Jordan. Note
if \(R'\) is not of characteristic
\(2\), the second identity is
equivalent to \[
\phi(a^2)=\phi(a)^2.
\]To show the equivalence, one let \(b=a\) in the first case and puts \(a+b\) in place of \(a\) in the second case.

Since in this case \(R=A\) and \(R'=\mathbb{C}\), the latter of which is
commutative, we also write \[
\phi(ab+ba)=2\phi(a)\phi(b).
\] As we will show, the \(\phi\)
in the theorem is a Jordan homomorphism.

The proof

We will follow an unusual approach. By keep 'downgrading' the goal,
one will see this algebraic problem be transformed into a pure analysis
problem neatly.

To begin with, let \(N\) be the
kernel of \(\phi\).

Step
1 - It suffices to prove that \(\phi\)
is a Jordan homomorphism

If \(\phi\) is a complex
homomorphism, it is immediate that \(\phi\) is a Jordan homomorphism.
Conversely, if \(\phi\) is Jordan, we
have \[
\phi(xy+yx) =2\phi(x)\phi(y).
\] If \(x\in N\), the right hand
becomes \(0\), and therefore \[
xy+yx \in N \quad \text{if } x \in N, y \in A.
\] Consider the identity \[
(xy-yx)^2+(xy+yx)^2=2[x(yxy)+(yxy)x]
\]

Therefore \[
\begin{aligned}
\phi((xy-yx)^2+(xy+yx)^2)&=\phi((xy-yx)^2)+\phi((xy+yx)^2) \\
&=\phi(xy-yx)^2+\phi(xy+yx)^2 \\
&= \phi(xy-yx)^2 \\
&=2\phi[x(yxy)+(yxy)x] \\
&=0
\end{aligned}
\] Since \(x \in N\) and \(yxy \in A\), we see \(x(yxy)+(yxy)x \in N\). Therefore \(\phi(xy-yx)=0\) and \[
xy-yx \in N
\] if \(x \in N\) and \(y \in A\). Further we see \[
xy-yx+xy+yx=2xy \in N \quad \text {and}\quad xy+yx-xy+yx = 2yx \in N,
\] which implies that \(N\) is
an ideal. This may remind you of this classic diagram (we will not use
it since it is additive though):

For \(x,y \in A\), we have \(x \in \phi(x)e+N\) and \(y \in \phi(y)e+N\). As a result, \(xy \in \phi(x)\phi(y)e+N\), and therefore
\[
\phi(xy)=\phi(x)\phi(y)+0.
\]

Step 2 - It
suffices to prove that \(\phi(a^2)=0\)
if \(\phi(a)=0\).

Again, if \(\phi\) is Jordan, we
have \(\phi(x^2)=\phi(x)^2\) for all
\(x \in A\). Conversely, if \(\phi(a^2)=0\) for all \(a \in N\), we may write \(x\) by \[
x=\phi(x)e+a
\] where \(a \in N\) for all
\(x \in A\). Therefore \[
\begin{aligned}
\phi(x^2)&=\phi((\phi(x)e+a)^2)=\phi(x)^2+2\phi(x)\phi(a)+\phi(a)^2=\phi(x)^2,
\end{aligned}
\] which also shows that \(\phi\) is Jordan.

Step
3 - It suffices to show that the following function is constant

Fix \(a \in N\), assume \(\lVert a \rVert = 1\) without loss of
generality, and define \[
f(\lambda)=\sum_{n=0}^{\infty}\frac{\phi(a^n)}{n!}\lambda^n
\] for all complex \(\lambda\).
If this function is constant (lemma 1), we immediately have \(f''(0)=\phi(a^2)=0\). This is
purely a complex analysis problem however.

Step
4 - It suffices to describe the behaviour of an entire function

Note in the definition of \(f\), we
have \[
\lvert \phi(a^n) \rvert \leq \lVert \phi \rVert \lVert a^n \rVert \leq
\lVert \phi \rVert \lVert a \rVert^n=\lVert \phi \rVert.
\] So we expect the norm of \(\phi\) to be finite, which ensures that
\(f\) is entire. By reductio ad
absurdum, if \(\lVert e-a \rVert <
1\) for \(a \in N\), by lemma 0,
we have \(e-e+a=a\) to be invertible,
which is impossible. Hence \(\lVert e-a \rVert
\geq 1\) for all \(a \in N\). On
the other hand, for \(\lambda \in
\mathbb{C}\), we have the following inequality: \[
\begin{aligned}
\lVert \lambda e-a \rVert = \lambda\lVert e-\lambda^{-1}a \rVert
&\geq|\lambda| \\
&= |\phi(\lambda e)-\phi(a)| \\
&= |\phi(\lambda e-a)|
\end{aligned}
\] Therefore \(\phi\) is
continuous with norm less than \(1\). The continuity of \(\phi\) is not assumed at the beginning but
proved here.

For \(f\) we have some immediate
facts. Since each coefficient in the series of \(f\) has finite norm, \(f\) is entire with \(f'(0)=\phi(a)=0\). Also, since \(\phi\) has norm \(1\), we also have \[
|f(\lambda)|=\left|\sum_{n=0}^{\infty}\frac{\phi(a^n)}{n!}\lambda^n\right|
\leq \sum_{n=0}^{\infty}\frac{|\lambda^n|}{n!}=e^{|\lambda|}.
\] All we need in the end is to show that \(f(\lambda) \neq 0\) for all \(\lambda \in \mathbb{C}\).

The series \[
E(\lambda)=\exp(a\lambda)=\sum_{n=0}^{\infty}\frac{(\lambda a)^n}{n!}
\] converges since \(\lVert a
\rVert=1\). The continuity of \(\phi\) shows now \[
f(\lambda)=\phi(E(\lambda)).
\] Note \[
E(-\lambda)E(\lambda)=\left(\sum_{n=0}^{\infty}\frac{(-\lambda
a)^n}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{(\lambda
a)^n}{n!}\right)=e.
\] Hence \(E(\lambda)\)is invertible for all \(\lambda \in
C\), hence \(f(\lambda)=\phi(E(\lambda)) \neq 0\). By
lemma 1, \(f(\lambda)=1\) is constant.
The proof is completed by reversing the steps. \(\square\)

References / Further reading

Walter Rudin, Real and Complex Analysis

Walter Rudin, Functional Analysis

Andrew M. Gleason, A Characterization of Maximal
Ideals

J.-P. Kahane and W. Żelazko, A Characterization of Maximal
Ideals in Commutative Banach Algebras

W. Żelazko A Characterization of Multiplicative linear
functionals in Complex Banach Algebras

We are going to show the completeness of \(X/N\) where \(X\) is a TVS and \(N\) a closed subspace. Alongside, a bunch
of useful analysis tricks will be demonstrated (and that's why you may
find this blog post a little tedious.). But what's more important, the
theorem proved here will be used in the future.

The main process

To make it clear, we should give a formal definition of \(F\)-space.

A topological space \(X\) is an
\(F\)-space if its topology \(\tau\) is induced by a complete invariant
metric \(d\).

A metric \(d\) on a vector space
\(X\) will be called invariant if for
all \(x,y,z \in X\), we have \[
d(x+z,y+z)=d(x,y).
\] By complete we mean every Cauchy sequence of \((X,d)\) converges.

Defining the quotient metric
\(\rho\)

The metric can be inherited to the quotient space naturally (we will
use this fact latter), that is

If \(X\) is a \(F\)-space, \(N\) is a closed subspace of a topological
vector space \(X\), then \(X/N\) is still a \(F\)-space.

Suppose \(d\) is a complete
invariant metric compatible with \(\tau_X\). The metric on \(X/N\) is defined by \[
\boxed{\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)}
\] ### \(\rho\) is a metric

Proof. First, if \(\pi(x)=\pi(y)\), that is, \(x-y \in N\), we see \[
\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)=d(x-y,x-y)=0.
\] If \(\pi(x) \neq \pi(y)\)
however, we shall show that \(\rho(\pi(x),\pi(y))>0\). In this case,
we have \(x-y \notin N\). Since \(N\) is closed, \(N^c\) is open, and \(x-y\) is an interior point of \(X-N\). Therefore there exists an open ball
\(B_r(x-y)\) centered at \(x-y\) with radius \(r>0\) such that \(B_r(x-y) \cap N = \varnothing\). Notice we
have \(d(x-y,z)>r\) since otherwise
\(z \in B_r(x-y)\). By putting \[
r_0=\sup\{r:B_r(x-y) \cap N = \varnothing\},
\] we see \(d(x-y,z) \geq r_0\)
for all \(z \in N\) and indeed \(r_0=\inf_{z \in N}d(x-y,z)>0\) (the
verification can be done by contradiction). In general, \(\inf_z d(x-y,z)=0\) if and only if \(x-y \in \overline{N}\).

Next, we shall show that \(\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))\),
and it suffices to assume that \(\pi(x) \neq
\pi(y)\). Sgince \(d\) is
translate invariant, we get \[
\begin{aligned}
d(x-y,z)&=d(x-y-z,0) \\
&=d(0,y-x+z) \\
&=d(-z,y-x) \\
&=d(y-x,-z).
\end{aligned}
\] Therefore the \(\inf\) of the
left hand is equal to the one of the right hand. The identity is
proved.

Finally, we need to verify the triangle inequality. Let \(r,s,t \in X\). For any \(\varepsilon>0\), there exist some \(z_\varepsilon\) and \(z_\varepsilon'\) such that \[
d(r-s,z_\varepsilon)<\rho(\pi(r),\pi(s))+\frac{\varepsilon}{2},\quad
d(s-t,z'_\varepsilon)<\rho(\pi(s),\pi(t))+\frac{\varepsilon}{2}.
\] Since \(d\) is invariant, we
see \[
\begin{aligned}
d(r-t,z_\varepsilon+z'_\varepsilon)&=d((r-s)+(s-t)-(z_\varepsilon+z'_\varepsilon),0)
\\
&=d([(r-s)-z_\varepsilon]+[(s-t)-z'_\varepsilon],0)
\\
&=d(r-s-z_\varepsilon,t-s+z'_\varepsilon)
\\
&\leq
d(r-s-z_\varepsilon,0)+d(t-s+z'_\varepsilon,0) \\
&=d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon)
\end{aligned}
\](I owe @LeechLattice for
the inequality above.)

Therefore \[
\begin{aligned}
d(r-t,z_\varepsilon+z'_\varepsilon)&\leq
d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \\
&<\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))+\varepsilon.
\end{aligned}
\](Warning: This does not imply that \(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))=\inf_z
d(r-t,z)\) since we don't know whether it is the lower bound or
not.)

If \(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))<\rho(\pi(r),\pi(t))\)
however, let \[
0<\varepsilon<\rho(\pi(r),\pi(t))-(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t)))
\] then there exists some \(z''_\varepsilon=z_\varepsilon+z'_\varepsilon\)
such that \[
d(r-t,z''_\varepsilon)<\rho(\pi(r),\pi(t))
\] which is a contradiction since \(\rho(\pi(r),\pi(t)) \leq d(r-t,z)\) for all
\(z \in N\).

(We are using the \(\varepsilon\) definition of \(\inf\). See here.)

\(\rho\) is translate invariant

Since \(\pi\) is surjective, we see
if \(u \in X/N\), there exists some
\(a \in X\) such that \(\pi(a)=u\). Therefore \[
\begin{aligned}
\rho(\pi(x)+u,\pi(y)+u) &=\rho(\pi(x)+\pi(a),\pi(y)+\pi(a)) \\
&=\rho(\pi(x+a),\pi(y+a)) \\
&=\inf_{z \in N}d(x+a-y-a,z) \\
&=\rho(\pi(x),\pi(y)).
\end{aligned}
\]

\(\rho\)
is well-defined

If \(\pi(x)=\pi(x')\) and \(\pi(y)=\pi(y')\), we have to show that
\(\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))\).
In fact, \[
\begin{aligned}
\rho(\pi(x),\pi(y)) &\leq
\rho(\pi(x),\pi(x'))+\rho(\pi(x'),\pi(y'))+\rho(\pi(y'),\pi(y))
\\
&=\rho(\pi(x'),\pi(y'))
\end{aligned}
\] since \(\rho(\pi(x),\pi(x'))=0\) as \(\pi(x)=\pi(x')\). Meanwhile \[
\begin{aligned}
\rho(\pi(x'),\pi(y')) &\leq \rho(\pi(x'),\pi(x)) +
\rho(\pi(x),\pi(y)) + \rho(\pi(y),\pi(y')) \\
&= \rho(\pi(x),\pi(y)).
\end{aligned}
\] therefore \(\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))\).

\(\rho\) is compatible with \(\tau_N\)

By proving this, we need to show that a set \(E \subset X/N\) is open with respect to
\(\tau_N\) if and only if \(E\) is a union of open balls. But we need
to show a generalized version:

If \(\mathscr{B}\) is a local base
for \(\tau\), then the collection \(\mathscr{B}_N\), which contains all sets
\(\pi(V)\) where \(V \in \mathscr{B}\), forms a local base for
\(\tau_N\).

Proof. We already know that \(\pi\) is continuous, linear and open.
Therefore \(\pi(V)\) is open for all
\(V \in \mathscr{B}\). For any open set
around \(E \subset X/N\) containing
\(\pi(0)\), we see \(\pi^{-1}(E)\) is open, and we have \[
\pi^{-1}(E)=\bigcup_{V\in\mathscr{B}}V
\] and therefore \[
E=\bigcup_{V \in \mathscr{B}}\pi(V).
\]

Now consider the local base \(\mathscr{B}\) containing all open balls
around \(0 \in X\). Since \[
\pi(\{x:d(x,0)<r\})=\{u:\rho(u,\pi(0))<r\}
\] we see \(\rho\) determines
\(\mathscr{B}_N\). But we have already
proved that \(\rho\) is invariant;
hence \(\mathscr{B}_N\) determines
\(\tau_N\).

If \(d\) is complete, then \(\rho\) is complete.

Once this is proved, we are able to claim that, if \(X\) is a \(F\)-space, then \(X/N\) is still a \(F\)-space, since its topology is induced by
a complete invariant metric \(\rho\).

Proof. Suppose \((x_n)\) is
a Cauchy sequence in \(X/N\), relative
to \(\rho\). There is a subsequence
\((x_{n_k})\) with \(\rho(x_{n_k},x_{n_{k+1}})<2^{-k}\).
Since \(\pi\) is surjective, we are
able to pick some \(z_k \in X\) such
that \(\pi(z_k) = x_{n_k}\) and such
that \[
d(z_{k},z_{k+1})<2^{-k}.
\] (The existence can be verified by contradiction still.) By the
inequality above, we see \((z_k)\) is
Cauchy (can you see why?). Since \(X\)
is complete, \(z_k \to z\) for some
\(z \in X\). By the
continuity of \(\pi\),
we also see \(x_{n_k} \to \pi(z)\) as
\(k \to \infty\). Therefore \((x_{n_k})\) converges. Hence \((x_n)\) converges since it has a convergent
subsequence. \(\rho\) is complete.

Remarks

This fact will be used to prove some corollaries in the open mapping
theorem. For instance, for any continuous linear map \(\Lambda:X \to Y\), we see \(\ker(\Lambda)\) is closed, therefore if
\(X\) is a \(F\)-space, then \(X/\ker(\Lambda)\) is a \(F\)-space as well. We will show in the
future that \(X/\ker(\Lambda)\) and
\(\Lambda(X)\) are homeomorphic if
\(\Lambda(X)\) is of the second
category.

There are more properties that can be inherited by \(X/N\) from \(X\). For example, normability,
metrizability, local convexity. In particular, if \(X\) is Banach, then \(X/N\) is Banach as well. To do this, it
suffices to define the quotient norm by \[
\lVert \pi(x) \rVert = \inf\{\lVert x-z \rVert:z \in N\}.
\]

I'm assuming the reader has some abstract algebra and functional
analysis background. You may have learned this already in your linear
algebra class, but we are making our way to functional analysis
problems.

Motivation

The trouble with \(L^p\) spaces

Fix \(p\) with \(1 \leq p \leq \infty\). It's easy to see
that \(L^p(\mu)\) is a topological
vector space. But it is not a metric space if we define \[
d(f,g)=\lVert f-g \rVert_p.
\] The reason is, if \(d(f,g)=0\), we can only get \(f=g\) a.e., but they are not strictly
equal. With that being said, this function \(d\) is actually a pseudo metric. This is
unnatural. However, the relation \(\sim\) by \(f
\sim g \mathbb{R}ightarrow d(f,g)=0\) is a equivalence relation.
This inspires us to take the quotient set into consideration.

Vector spaces are groups
anyway

For a vector space \(V\), every
subspace of \(V\) is a normal subgroup.
There is no reason to prevent ourselves from considering the quotient
group and looking for some interesting properties. Further, a vector
space is an abelian group, therefore any subspace is automatically
normal.

Definition

Let \(N\) be a subspace of a vector
space \(X\). For every \(x \in X\), let \(\pi(x)\) be the coset of \(N\) that contains \(x\), that is \[
\pi(x)=x+N.
\] Trivially, \(\pi(x)=\pi(y)\)
if and only if \(x-y \in N\) (say,
\(\pi\) is well-defined since \(N\) is a vector space). This is a linear
function since we also have the addition and multiplication by \[
\pi(x)+\pi(y)=\pi(x+y) \quad \alpha\pi(x)=\pi(\alpha{x}).
\] These cosets are the elements of a vector space \(X/N\), which reads, the quotient space of
\(X\) modulo \(N\). The map \(\pi\) is called the canonical map as we all
know.

Examples

First, we shall treat \(\mathbb{R}^2\) as a vector space, and the
subspace \(\mathbb{R}\), which is
graphically represented by \(x\)-axis,
as a subspace (we will write it as \(X\)). For a vector \(v=(2,3)\), which is represented by \(AB\), we see the coset \(v+X\) has something special. Pick any \(u \in X\), for example, \(AE\), \(AC\), or \(AG\). We see \(v+u\) has the same \(y\) value. The reason is simple since we
have \(v+u=(2+x,3)\), where the \(y\) value remains fixed however \(u\) may vary.

With that being said, the set \(v+X\), which is not a vector space, can be
represented by \(\overrightarrow{AD}\).
This proceed can be generalized to \(\mathbb{R}^n\) with \(\mathbb{R}^m\) as a subspace with ease.

We now consider a fancy example. Consider all rational Cauchy
sequences, that is \[
(a_n)=(a_1,a_2,\cdots)
\] where \(a_k\in\mathbb{Q}\)
for all \(k\). In analysis class, we
learned two facts.

Any Cauchy sequence is bounded.

If \((a_n)\) converges, then \((a_n)\) is Cauchy.

However, the reverse of 2 does not hold in \(\mathbb{Q}\). For example, if we put \(a_k=(1+\frac{1}{k})^k\), we should have the
limit to be \(e\), but \(e \notin \mathbb{Q}\).

If we define the addition and multiplication term by term, namely
\[
(a_n)+(b_n)=(a_1+b_1,a_2+b_2,\cdots)
\] and \[
(\alpha a_n)=(\alpha a_1,\alpha a_2,\cdots)
\] where \(\alpha \in
\mathbb{Q}\), we get a vector space (the verification is easy).
The zero vector is defined by \[
(0)=(0,0,\cdots).
\] This vector space is denoted by \(\overline{\mathbb{Q}}\). The subspace
containing all sequences converges to \(0\) will be denoted by \(\overline{\mathbb{O}}\). Again, \((a_n)+\overline{\mathbb{O}}=(b_n)+\overline{\mathbb{O}}\)
if and only if \((a_n-b_n) \in
\overline{\mathbb{O}}\). Using the language of equivalence
relation, we also say \((a_n)\) and
\((b_n)\) are equivalent if \((a_n-b_n) \in \overline{\mathbb{O}}\). For
example, the two following sequences are equivalent: \[
(1,1,1,\cdots,1,\cdots)\quad\quad (0.9,0.99,0.999,\cdots).
\] Actually, we will get \(\mathbb{R}
\simeq \overline{\mathbb{Q}}/\overline{\mathbb{O}}\) in the end.
But to make sure that this quotient space is exactly the one we meet in
our analysis class, there are a lot of verifications should be done.

We shall give more definitions for calculation. The multiplication of
two Cauchy sequences is defined term by term à la the addition. For
\(\overline{\mathbb{Q}}/\overline{\mathbb{O}}\)
we have \[
((a_n)+\overline{\mathbb{O}})+((b_n)+\overline{\mathbb{O}})=(a_n+b_n) +
\overline{\mathbb{O}}
\] and \[
((a_n)+\overline{\mathbb{O}})((b_n)+\overline{\mathbb{O}})=(a_nb_n)+\overline{\mathbb{O}}.
\] As for inequality, a partial order has to be defined. We say
\((a_n) > (0)\) if there exists some
\(N>0\) such that \(a_n>0\) for all \(n \geq N\). By \((a_n) > (b_n)\) we mean \((a_n-b_n)>(0)\) of course. For cosets,
we say \((a_n)+\overline{\mathbb{O}}>\overline{\mathbb{O}}\)
if \((x_n) > (0)\) for some \((x_n) \in (a_n)+\overline{\mathbb{O}}\).
This is well defined. That is, if \((x_n)>(0)\), then \((y_n)>(0)\) for all \((y_n) \in
(a_n)+\overline{\mathbb{O}}\).

With these operations being defined, it can be verified that \(\overline{\mathbb{Q}}/\overline{\mathbb{O}}\)
has the desired properties, for example, the least-upper-bound property.
But this goes too far from the topic, we are not proving it here. If you
are interested, you may visit here
for more details.

Finally, we are trying to make \(L^p\) a Banach space. Fix \(p\) with \(1 \leq
p < \infty\). There is a seminorm defined for all Lebesgue
measurable functions on \([0,1]\) by
\[
p(f)=\lVert f \rVert_p=\left\{\int_{0}^{1}|f(t)|^pdt\right\}^{1/p}
\]\(L^p\) is a vector space
containing all functions \(f\) with
\(p(f)<\infty\). But it's not a
normed space by \(p\), since \(p(f)=0\) only implies \(f=0\) almost everywhere. However, the set
\(N\) which contains all functions that
equal \(0\) is also a vector space. Now
consider the quotient space by \[
\tilde{p}(\pi(f))=p(f),
\] where \(\pi\) is the
canonical map of \(L^p\) into \(L^p/N\). We shall prove that \(\tilde{p}\) is well-defined here. If \(\pi(f)=\pi(g)\), we have \(f-g \in N\), therefore \[
0=p(f-g)\geq |p(f)-p(g)|,
\] which forces \(p(f)=p(g)\).
Therefore in this case we also have \(\tilde{p}(\pi(f))=\tilde{p}(\pi(g))\). This
indeed ensures that \(\tilde{p}\) is a
norm, and \(L^p/N\) a Banach
space. There are some topological facts required to prove this, we are
going to cover a few of them.

Topology of quotient space

Definition

We know if \(X\) is a topological
vector space with a topology \(\tau\),
then the addition and scalar multiplication are continuous. Suppose now
\(N\) is a closed subspace of \(X\). Define \(\tau_N\) by \[
\tau_N=\{E \subset X/N:\pi^{-1}(E)\in \tau\}.
\] We are expecting \(\tau_N\)
to be properly-defined. And fortunately, it is. Some interesting
techniques will be used in the following section.

\(\tau_N\) is a vector topology

There will be two steps to get this done.

\(\tau_N\) is a topology.

It is trivial that \(\varnothing\)
and \(X/N\) are elements of \(\tau_N\). Other properties are immediate as
well since we have \[
\pi^{-1}(A \cap B) = \pi^{-1}(A) \cap \pi^{-1}(B)
\] and \[
\pi^{-1}(\cup A_\alpha)=\cup\pi^{-1}( A_{\alpha}).
\] That said, if we have \(A,B\in
\tau_N\), then \(A \cap B \in
\tau_N\) since \(\pi^{-1}(A \cap
B)=\pi^{-1}(A) \cap \pi^{-1}(B) \in \tau\).

Similarly, if \(A_\alpha \in
\tau_N\) for all \(\alpha\), we
have \(\cup A_\alpha \in \tau_N\).
Also, by definition of \(\tau_N\),
\(\pi\) is continuous.

\(\tau_N\) is a vector topology.

First, we show that a point in \(X/N\), which can be written as \(\pi(x)\), is closed. Notice that \(N\) is assumed to be closed, and \[
\pi^{-1}(\pi(x))=x+N
\] therefore has to be closed.

In fact, \(F \subset X/N\) is \(\tau_N\)-closed if and only if \(\pi^{-1}(F)\) is \(\tau\)-closed. To prove this, one needs to
notice that \(\pi^{-1}(F^c)=(\pi^{-1}(F))^{c}\).

Suppose \(V\) is open, then \[
\pi^{-1}(\pi(V))=N+V
\] is open. By definition of \(\tau_N\), we have \(\pi(V) \in \tau_N\). Therefore \(\pi\) is an open mapping.

If now \(W\) is a neighbourhood of
\(0\) in \(X/N\), there exists a neighbourhood \(V\) of \(0\) in \(X\) such that \[
V + V \subset \pi^{-1}(W).
\] Hence \(\pi(V)+\pi(V) \subset
W\). Since \(\pi\) is open,
\(\pi(V)\) is a neighbourhood of \(0\) in \(X/N\), this shows that the addition is
continuous.

The continuity of scalar multiplication will be shown in a direct way
(so can the addition, but the proof above is intended to offer some
special technique). We already know, the scalar multiplication on \(X\) by \[
\begin{aligned}
\varphi:\Phi \times X &\to X \\
(\alpha,x) &\mapsto \alpha{x}
\end{aligned}
\] is continuous, where \(\Phi\)
is the scalar field (usually \(\mathbb{R}\) or \(\mathbb{C}\). Now the scalar multiplication
on \(X/N\) is by \[
\begin{aligned}
\psi: \Phi \times X/N &\to X/N \\
(\alpha,x+N) &\mapsto \alpha{x}+N.
\end{aligned}
\] We see \(\psi(\alpha,x+N)=\pi(\varphi(\alpha,x))\).
But the composition of two continuous functions is continuous, therefore
\(\psi\) is continuous.

A commutative diagram
by quotient space

We are going to talk about a classic commutative diagram that you
already see in algebra class.

There are some assumptions.

\(X\) and \(Y\) are topological vector spaces.

\(\Lambda\) is linear.

\(\pi\) is the canonical map.

\(N\) is a closed subspace of \(X\) and \(N
\subset \ker\Lambda\).

Algebraically, there exists a unique map \(f: X/N \to Y\) by \(x+N \mapsto \Lambda(x)\). Namely, the
diagram above is commutative. But now we are interested in some analysis
facts.

\(f\) is linear.

This is obvious. Since \(\pi\) is
surjective, for \(u,v \in
X/N\), we are able to find some \(x,y
\in X\) such that \(\pi(x)=u\)
and \(\pi(y)=v\). Therefore we have
\[
\begin{aligned}
f(u+v)=f(\pi(x)+\pi(y))&=f(\pi(x+y)) \\
&=\Lambda(x+y) \\
&=\Lambda(x)+\Lambda(y) \\
&= f(\pi(x))+f(\pi(y)) \\
&=f(u)+f(v)
\end{aligned}
\] and \[
\begin{aligned}
f(\alpha{u})=f(\alpha\pi(x))&=f(\pi(\alpha{x})) \\
&= \Lambda(\alpha{x}) \\
&= \alpha\Lambda(x) \\
&= \alpha{f(\pi(x))} \\
&= \alpha{f(u)}.
\end{aligned}
\]

\(\Lambda\) is open if and only if
\(f\) is open.

If \(f\) is open, then for any open
set \(U \subset X\), we have \[
\Lambda(U)=f(\pi(U))
\] to be an open set since \(\pi\) is open, and \(\pi(U)\) is an open set.

If \(f\) is not open, then there
exists some \(V \subset X/N\) such that
\(f(V)\) is closed. However, since
\(\pi\) is continuous, we have \(\pi^{-1}(V)\) to be open. In this case, we
have \[
f(\pi(\pi^{-1}(V)))=f(V)=\Lambda(\pi^{-1}(V))
\] to be closed. \(\Lambda\) is
therefore not open. This shows that if \(\Lambda\) is open, then \(f\) is open.

\(\Lambda\) is continuous if and
only if \(f\) is continuous.

If \(f\) is continuous, for any open
set \(W \subset Y\), we have \(\pi^{-1}(f^{-1}(W))=\Lambda^{-1}(W)\) to be
open. Therefore \(\Lambda\) is
continuous.

Conversely, if \(\Lambda\) is
continuous, for any open set \(W \subset
Y\), we have \(\Lambda^{-1}(W)\)
to be open. Therefore \(f^{-1}(W)=\pi(\Lambda^{-1}(W))\) has to be
open since \(\pi\) is open.