# The Banach Algebra of Borel Measures on Euclidean Space

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.

# The concept of generalised functions (distributions) and derivatives

In this post, we study the concept of generalised functions (a.k.a. distributions), and let's see how to evaluate the derivative no matter the function is differentiable or not.

# Elementary Properties of Cesàro Operator in L^2

We study the average of sum, in the sense of integral.

# Left Shift Semigroup and Its Infinitesimal Generator

## Left shift operator

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 $t$.

# Quasi-analytic Vectors and Hamburger Moment Problem (Operator Theory)

## Analytic and quasi-analytic vectors

Guided by researches in function theory, operator theorists gave the analogue to quasi-analytic classes. Let $A$ be an operator in a Banach space $X$. $A$ is not necessarily bounded hence the domain $D(A)$ is not necessarily to be the whole space. We say $x \in X$ is a $C^\infty$ vector if $x \in \bigcap_{n \geq 1}D(A^n)$. This is quite intuitive if we consider the differential operator. A vector is analytic if the series $\sum_{n=0}^{\infty}\lVert{A^n x}\rVert\frac{t^n}{n!}$ has a positive radius of convergence. Finally, we say $x$ is quasi-analytic for $A$ provided that $\sum_{n=0}^{\infty}\left(\frac{1}{\lVert A^n x \rVert}\right)^{1/n} = \infty$ or equivalently its nondecreasing majorant. Interestingly, if $A$ is symmetric, then $\lVert{A^nx}\rVert$ is log convex.

Based on the density of quasi-analytic vectors, we have an interesting result.

(Theorem) Let $A$ be a symmetric operator in a Hilbert space $\mathscr{H}$. If the set of quasi-analytic vectors spans a dense subset, then $A$ is essentially self-adjoint.

This theorem can be considered as a corollary to the fundamental theorem of quasi-analytic classes, by applying suitable Banach space techniques in lieu.

# Several ways to prove Hardy's inequality

Suppose $1 < p < \infty$ and $f \in L^p((0,\infty))$ (with respect to Lebesgue measure of course) is a nonnegative function, take $F(x) = \frac{1}{x}\int_0^x f(t)dt \quad 0 < x <\infty,$ we have Hardy's inequality $\def\lrVert[#1]{\lVert #1 \rVert}$ $\lrVert[F]_p \leq q\lrVert[f]_p$ where $\frac{1}{p}+\frac{1}{q}=1$ of course.

There are several ways to prove it. I think there are several good reasons to write them down thoroughly since that may be why you find this page. Maybe you are burnt out since it's left as exercise. You are assumed to have enough knowledge of Lebesgue measure and integration.

## Minkowski's integral inequality

Let $S_1,S_2 \subset \mathbb{R}$ be two measurable set, suppose $F:S_1 \times S_2 \to \mathbb{R}$ is measurable, then $\left[\int_{S_2} \left\vert\int_{S_1}F(x,y)dx \right\vert^pdy\right]^{\frac{1}{p}} \leq \int_{S_1} \left[\int_{S_2} |F(x,y)|^p dy\right]^{\frac{1}{p}}dx.$ A proof can be found at here by turning to Example A9. You may need to replace all measures with Lebesgue measure $m$.

Now let's get into it. For a measurable function in this place we should have $G(x,t)=\frac{f(t)}{x}$. If we put this function inside this inequality, we see \begin{aligned} \lrVert[F]_p &= \left[\int_0^\infty \left\vert \int_0^x \frac{f(t)}{x}dt \right\vert^p dx\right]^{\frac{1}{p}} \\ &= \left[\int_0^\infty \left\vert \int_0^1 f(ux)du \right\vert^p dx\right]^{\frac{1}{p}} \\ &\leq \int_0^1 \left[\int_0^\infty |f(ux)|^pdx\right]^{\frac{1}{p}}du \\ &= \int_0^1 \left[\int_0^\infty |f(ux)|^pudx\right]^{\frac{1}{p}}u^{-\frac{1}{p}}du \\ &= \lrVert[f]_p \int_0^1 u^{-\frac{1}{p}}du \\ &=q\lrVert[f]_p. \end{aligned} Note we have used change-of-variable twice and the inequality once.

## A constructive approach

I have no idea how people came up with this solution. Take $xF(x)=\int_0^x f(t)t^{u}t^{-u}dt$ where $0<u<1-\frac{1}{p}$. Hölder's inequality gives us \begin{aligned} xF(x) &= \int_0^x f(t)t^ut^{-u}dt \\ &\leq \left[\int_0^x t^{-uq}dt\right]^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \\ &=\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \end{aligned} Hence \begin{aligned} F(x)^p & \leq \frac{1}{x^p}\left\{\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}}\right\}^{p} \\ &= \left(\frac{1}{1-uq}\right)^{\frac{p}{q}}x^{\frac{p}{q}(1-uq)-p}\int_0^x f(t)^pt^{up}dt \\ &= \left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt \end{aligned}

Note we have used the fact that $\frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq$ and $\frac{p}{q}=p-1$. Fubini's theorem gives us the final answer: \begin{aligned} \int_0^\infty F(x)^pdx &\leq \int_0^\infty\left[\left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt\right]dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dx\int_0^x f(t)^pt^{up}x^{-up-1}dt \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dt\int_t^\infty f(t)^pt^{up}x^{-up-1}dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}\int_0^\infty f(t)^pdt. \end{aligned} It remains to find the minimum of $\varphi(u) = \left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}$. This is an elementary calculus problem. By taking its derivative, we see when $u=\frac{1}{pq}<1-\frac{1}{p}$ it attains its minimum $\left(\frac{p}{p-1}\right)^p=q^p$. Hence we get $\int_0^\infty F(x)^pdx \leq q^p\int_0^\infty f(t)^pdt,$ which is exactly what we want. Note the constant $q$ cannot be replaced with a smaller one. We simply proved the case when $f \geq 0$. For the general case, one simply needs to take absolute value.

## Integration by parts

This approach makes use of properties of $L^p$ space. Still we assume that $f \geq 0$ but we also assume $f \in C_c((0,\infty))$, that is, $f$ is continuous and has compact support. Hence $F$ is differentiable in this situation. Integration by parts gives $\int_0^\infty F^p(x)dx=xF(x)^p\vert_0^\infty- p\int_0^\infty xdF^p = -p\int_0^\infty xF^{p-1}(x)F'(x)dx.$ Note since $f$ has compact support, there are some $[a,b]$ such that $f >0$ only if $0 < a \leq x \leq b < \infty$ and hence $xF(x)^p\vert_0^\infty=0$. Next it is natural to take a look at $F'(x)$. Note we have $F'(x) = \frac{f(x)}{x}-\frac{\int_0^x f(t)dt}{x^2},$ hence $xF'(x)=f(x)-F(x)$. A substitution gives us $\int_0^\infty F^p(x)dx = -p\int_0^\infty F^{p-1}(x)[f(x)-F(x)]dx,$ which is equivalent to say $\int_0^\infty F^p(x)dx = \frac{p}{p-1}\int_0^\infty F^{p-1}(x)f(x)dx.$ Hölder's inequality gives us \begin{aligned} \int_0^\infty F^{p-1}(x)f(x)dx &\leq \left[\int_0^\infty F^{(p-1)q}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}} \\ &=\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}. \end{aligned} Together with the identity above we get $\int_0^\infty F^p(x)dx = q\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}$ which is exactly what we want since $1-\frac{1}{q}=\frac{1}{p}$ and all we need to do is divide $\left[\int_0^\infty F^pdx\right]^{1/q}$ on both sides. So what's next? Note $C_c((0,\infty))$ is dense in $L^p((0,\infty))$. For any $f \in L^p((0,\infty))$, we can take a sequence of functions $f_n \in C_c((0,\infty))$ such that $f_n \to f$ with respect to $L^p$-norm. Taking $F=\frac{1}{x}\int_0^x f(t)dt$ and $F_n = \frac{1}{x}\int_0^x f_n(t)dt$, we need to show that $F_n \to F$ pointwise, so that we can use Fatou's lemma. For $\varepsilon>0$, there exists some $m$ such that $\lrVert[f_n-f]_p < \frac{1}{n}$. Thus \begin{aligned} |F_n(x)-F(x)| &= \frac{1}{x}\left\vert \int_0^x f_n(t)dt - \int_0^x f(t)dt \right\vert \\ &\leq \frac{1}{x} \int_0^x |f_n(t)-f(t)|dt \\ &\leq \frac{1}{x} \left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}}\left[\int_0^x 1^qdt\right]^{\frac{1}{q}} \\ &=\frac{1}{x^{1/p}}\left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}} \\ &\leq \frac{1}{x^{1/p}}\lrVert[f_n-f]_p <\frac{\varepsilon}{x^{1/p}}. \end{aligned} Hence $F_n \to F$ pointwise, which also implies that $|F_n|^p \to |F|^p$ pointwise. For $|F_n|$ we have \begin{aligned} \int_0^\infty |F_n(x)|^pdx &= \int_0^\infty \left\vert\frac{1}{x}\int_0^x f_n(t)dt\right\vert^p dx \\ &\leq \int_0^\infty \left[\frac{1}{x}\int_0^x |f_n(t)|dt\right]^{p}dx \\ &\leq q\int_0^\infty |f_n(t)|^pdt \end{aligned} note the third inequality follows since we have already proved it for $f \geq 0$. By Fatou's lemma, we have \begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned}

# A Continuous Function Sending L^p Functions to L^1

Throughout, let $(X,\mathfrak{M},\mu)$ be a measure space where $\mu$ is positive.

## The question

If $f$ is of $L^p(\mu)$, which means $\lVert f \rVert_p=\left(\int_X |f|^p d\mu\right)^{1/p}<\infty$, or equivalently $\int_X |f|^p d\mu<\infty$, then we may say $|f|^p$ is of $L^1(\mu)$. In other words, we have a function \begin{aligned} \lambda: L^p(\mu) &\to L^1(\mu) \\ f &\mapsto |f|^p. \end{aligned} This function does not have to be one to one due to absolute value. But we hope this function to be fine enough, at the very least, we hope it is continuous.

Here, $f \sim g$ means that $f-g$ equals to $0$ almost everywhere with respect to $\mu$. It can be easily verified that this is a equivalence relation.

## Continuity

We still use $\varepsilon-\delta$ argument but it's in a metric space. Suppose $(X,d_1)$ and $(Y,d_2)$ are two metric spaces and $f:X \to Y$ is a function. We say $f$ is continuous at $x_0 \in X$ if for any $\varepsilon>0$, there exists some $\delta>0$ such that $d_2(f(x_0),f(x))<\varepsilon$ whenever $d_1(x_0,x)<\delta$. Further, we say $f$ is continuous on $X$ if $f$ is continuous at every point $x \in X$.

## Metrics

For $1\leq p<\infty$, we already have a metric by $d(f,g)=\lVert f-g \rVert_p$ given that $d(f,g)=0$ if and only if $f \sim g$. This is complete and makes $L^p$ a Banach space. But for $0<p<1$ (yes we are going to cover that), things are much more different, and there is one reason: Minkowski inequality holds reversely! In fact we have $\lVert f+g \rVert_p \geq \lVert f \rVert_p + \lVert g \rVert_p$ for $0<p<1$. In fact, $L^p$ space has too many weird things when $0<p<1$. Precisely,

For $0<p<1$, $L^p(\mu)$ is locally convex if and only if $\mu$ assumes finitely many values. (Proof.)

On the other hand, for example, $X=[0,1]$ and $\mu=m$ be the Lebesgue measure, then $L^p(\mu)$ has no open convex subset other than $\varnothing$ and $L^p(\mu)$ itself. However,

A topological vector space $X$ is normable if and only if its origin has a convex bounded neighbourhood. (See Kolmogorov's normability criterion.)

Therefore $L^p(m)$ is not normable, hence not Banach.

We have gone too far. We need a metric that is fine enough.

### Metric of $L^p$ when $0<p<1$

In this subsection we always have $0<p<1$.

Define $\Delta(f)=\int_X |f|^p d\mu$ for $f \in L^p(\mu)$. We will show that we have a metric by $d(f,g)=\Delta(f-g).$ Fix $y\geq 0$, consider the function $f(x)=(x+y)^p-x^p.$ We have $f(0)=y^p$ and $f'(x)=p(x+y)^{p-1}-px^{p-1} \leq px^{p-1}-px^{p-1}=0$ when $x > 0$ and hence $f(x)$ is nonincreasing on $[0,\infty)$, which implies that $(x+y)^p \leq x^p+y^p.$ Hence for any $f$, $g \in L^p$, we have $\Delta(f+g)=\int_X |f+g|^p d\mu \leq \int_X |f|^p d\mu + \int_X |g|^p d\mu=\Delta(f)+\Delta(g).$ This inequality ensures that $d(f,g)=\Delta(f-g)$ is a metric. It's immediate that $d(f,g)=d(g,f) \geq 0$ for all $f$, $g \in L^p(\mu)$. For the triangle inequality, note that $d(f,h)+d(g,h)=\Delta(f-h)+\Delta(h-g) \geq \Delta((f-h)+(h-g))=\Delta(f-g)=d(f,g).$ This is translate-invariant as well since $d(f+h,g+h)=\Delta(f+h-g-h)=\Delta(f-g)=d(f,g)$ The completeness can be verified in the same way as the case when $p>1$. In fact, this metric makes $L^p$ a locally bounded F-space.

## The continuity of $\lambda$

The metric of $L^1$ is defined by $d_1(f,g)=\lVert f-g \rVert_1=\int_X |f-g|d\mu.$ We need to find a relation between $d_p(f,g)$ and $d_1(\lambda(f),\lambda(g))$, where $d_p$ is the metric of the corresponding $L^p$ space.

### $0<p<1$

As we have proved, $(x+y)^p \leq x^p+y^p.$ Without loss of generality we assume $x \geq y$ and therefore $x^p=(x-y+y)^p \leq (x-y)^p+y^p.$ Hence $x^p-y^p \leq (x-y)^p.$ By interchanging $x$ and $y$, we get $|x^p-y^p| \leq |x-y|^p.$ Replacing $x$ and $y$ with $|f|$ and $|g|$ where $f$, $g \in L^p$, we get $\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \leq \int_X |f-g|^p d\mu.$ But $d_1(\lambda(f),\lambda(g))=\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \\ d_p(f,g)=\Delta(f-g)= d\mu \leq \int_X |f-g|^p d\mu$ and we therefore have $d_1(\lambda(f),\lambda(g)) \leq d_p(f,g).$ Hence $\lambda$ is continuous (and in fact, Lipschitz continuous and uniformly continuous) when $0<p<1$.

## $1 \leq p < \infty$

It's natural to think about Minkowski's inequality and Hölder's inequality in this case since they are critical inequality enablers. You need to think about some examples of how to create the condition to use them and get a fine result. In this section we need to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).$ This inequality is surprisingly easy to prove however. We will use nothing but the mean value theorem. Without loss of generality we assume that $x > y \geq 0$ and define $f(t)=t^p$. Then $\frac{f(x)-f(y)}{x-y}=f'(\zeta)=p\zeta^{p-1}$ where $y < \zeta < x$. But since $p-1 \geq 0$, we see $\zeta^{p-1} < x^{p-1} <x^{p-1}+y^{p-1}$. Therefore $f(x)-f(y)=x^p-y^p=p(x-y)\zeta^{p-1}<p(x-y)(x^{p-1}-y^{p-1}).$ For $x=y$ the equality holds.

Therefore \begin{aligned} d_1(\lambda(f),\lambda(g)) &= \int_X \left||f|^p-|g|^p\right|d\mu \\ &\leq \int_Xp\left||f|-|g|\right|(|f|^{p-1}+|g|^{p-1})d\mu \end{aligned} By Hölder's inequality, we have \begin{aligned} \int_X ||f|-|g||(|f|^{p-1}+|g|^{p-1})d\mu & \leq \left[\int_X \left||f|-|g|\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &\leq \left[\int_X \left|f-g\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &=\lVert f-g \rVert_p \left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q}. \end{aligned} By Minkowski's inequality, we have $\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \leq \left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q}$ Now things are clear. Since $1/p+1/q=1$, or equivalently $1/q=(p-1)/p$, suppose $\lVert f \rVert_p$, $\lVert g \rVert_p \leq R$, then $(p-1)q=p$ and therefore $\left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q} = \lVert f \rVert_p^{p-1}+\lVert g \rVert_p^{p-1} \leq 2R^{p-1}.$ Summing the inequalities above, we get \begin{aligned} d_1(\lambda(f),\lambda(g)) \leq 2pR^{p-1}\lVert f-g \rVert_p =2pR^{p-1}d_p(f,g) \end{aligned} hence $\lambda$ is continuous.

## Conclusion and further

We have proved that $\lambda$ is continuous, and when $0<p<1$, we have seen that $\lambda$ is Lipschitz continuous. It's natural to think about its differentiability afterwards, but the absolute value function is not even differentiable so we may have no chance. But this is still a fine enough result. For example we have no restriction to $(X,\mathfrak{M},\mu)$ other than the positivity of $\mu$. Therefore we may take $\mathbb{R}^n$ as the Lebesgue measure space here, or we can take something else.

It's also interesting how we use elementary Calculus to solve some much more abstract problems.

# The Big Three Pt. 6 - Closed Graph Theorem with Applications

(Before everything: elementary background in topology and vector spaces, in particular Banach spaces, is assumed.)

## A surprising result of Banach spaces

We can define several relations between two norms. Suppose we have a topological vector space $X$ and two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$. One says $\lVert \cdot \rVert_1$ is weaker than $\lVert \cdot \rVert_2$ if there is $K>0$ such that $\lVert x \rVert_1 \leq K \lVert x \rVert_2$ for all $x \in X$. Two norms are equivalent if each is weaker than the other (trivially this is a equivalence relation). The idea of stronger and weaker norms is related to the idea of the "finer" and "coarser" topologies in the setting of topological spaces.

So what about their limit? Unsurprisingly this can be verified with elementary $\epsilon-N$ arguments. Suppose now $\lVert x_n - x \rVert_1 \to 0$ as $n \to 0$, we immediately have $\lVert x_n - x \rVert_2 \leq K \lVert x_n-x \rVert_1 < K\varepsilon$

for some large enough $n$. Hence $\lVert x_n - x \rVert_2 \to 0$ as well. But what about the converse? We give a new definition of equivalence relation between norms.

(Definition) Two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ of a topological vector space are compatible if given that $\lVert x_n - x \rVert_1 \to 0$ and $\lVert x_n - y \rVert_2 \to 0$ as $n \to \infty$, we have $x=y$.

By the uniqueness of limit, we see if two norms are equivalent, then they are compatible. And surprisingly, with the help of the closed graph theorem we will discuss in this post, we have

(Theorem 1) If $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are compatible, and both $(X,\lVert\cdot\rVert_1)$ and $(X,\lVert\cdot\rVert_2)$ are Banach, then $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ are equivalent.

This result looks natural but not seemingly easy to prove, since one find no way to build a bridge between the limit and a general inequality. But before that, we need to elaborate some terminologies.

## Preliminaries

(Definition) For $f:X \to Y$, the graph of $f$ is defined by $G(f)=\{(x,f(x)) \in X \times Y:x \in X\}.$

If both $X$ and $Y$ are topological spaces, and the topology of $X \times Y$ is the usual one, that is, the smallest topology that contains all sets $U \times V$ where $U$ and $V$ are open in $X$ and $Y$ respectively, and if $f: X \to Y$ is continuous, it is natural to expect $G(f)$ to be closed. For example, by taking $f(x)=x$ and $X=Y=\mathbb{R}$, one would expect the diagonal line of the plane to be closed.

(Definition) The topological space $(X,\tau)$ is an $F$-space if $\tau$ is induced by a complete invariant metric $d$. Here invariant means that $d(x+z,y+z)=d(x,y)$ for all $x,y,z \in X$.

A Banach space is easily to be verified to be a $F$-space by defining $d(x,y)=\lVert x-y \rVert$.

(Open mapping theorem) See this post

By definition of closed set, we have a practical criterion on whether $G(f)$ is closed.

(Proposition 1) $G(f)$ is closed if and only if, for any sequence $(x_n)$ such that the limits $x=\lim_{n \to \infty}x_n \quad \text{ and }\quad y=\lim_{n \to \infty}f(x_n)$ exist, we have $y=f(x)$.

In this case, we say $f$ is closed. For continuous functions, things are trivial.

(Proposition 2) If $X$ and $Y$ are two topological spaces and $Y$ is Hausdorff, and $f:X \to Y$ is continuous, then $G(f)$ is closed.

Proof. Let $G^c$ be the complement of $G(f)$ with respect to $X \times Y$. Fix $(x_0,y_0) \in G^c$, we see $y_0 \neq f(x_0)$. By the Hausdorff property of $Y$, there exists some open subsets $U \subset Y$ and $V \subset Y$ such that $y_0 \in U$ and $f(x_0) \in V$ and $U \cap V = \varnothing$. Since $f$ is continuous, we see $W=f^{-1}(V)$ is open in $X$. We obtained a open neighborhood $W \times U$ containing $(x_0,y_0)$ which has empty intersection with $G(f)$. This is to say, every point of $G^c$ has a open neighborhood contained in $G^c$, hence a interior point. Therefore $G^c$ is open, which is to say that $G(f)$ is closed. $\square$

REMARKS. For $X \times Y=\mathbb{R} \times \mathbb{R}$, we have a simple visualization. For $\varepsilon>0$, there exists some $\delta$ such that $|f(x)-f(x_0)|<\varepsilon$ whenever $|x-x_0|<\delta$. For $y_0 \neq f(x_0)$, pick $\varepsilon$ such that $0<\varepsilon<\frac{1}{2}|f(x_0)-y_0|$, we have two boxes ($CDEF$ and $GHJI$ on the picture), namely $B_1=\{(x,y):x_0-\delta<x<x_0+\delta,f(x_0)-\varepsilon<y<f(x_0)+\varepsilon\}$ and $B_2=\{(x,y):x_0-\delta<x<x_0+\delta,y_0-\varepsilon<y<y_0+\varepsilon\}.$ In this case, $B_2$ will not intersect the graph of $f$, hence $(x_0,y_0)$ is an interior point of $G^c$.

The Hausdorff property of $Y$ is not removable. To see this, since $X$ has no restriction, it suffices to take a look at $X \times X$. Let $f$ be the identity map (which is continuous), we see the graph $G(f)=\{(x,x):x \in X\}$ is the diagonal. Suppose $X$ is not Hausdorff, we reach a contradiction. By definition, there exists some distinct $x$ and $y$ such that all neighborhoods of $x$ contain $y$. Pick $(x,y) \in G^c$, then all neighborhoods of $(x,y) \in X \times X$ contain $(x,x)$ so $(x,y) \in G^c$ is not a interior point of $G^c$, hence $G^c$ is not open.

Also, as an immediate consequence, every affine algebraic variety in $\mathbb{C}^n$ and $\mathbb{R}^n$ is closed with respect to Euclidean topology. Further, we have the Zariski topology $\mathcal{Z}$ by claiming that, if $V$ is an affine algebraic variety, then $V^c \in \mathcal{Z}$. It's worth noting that $\mathcal{Z}$ is not Hausdorff (example?) and in fact much coarser than the Euclidean topology although an affine algebraic variety is both closed in the Zariski topology and the Euclidean topology.

## The closed graph theorem

After we have proved this theorem, we are able to prove the theorem about compatible norms. We shall assume that both $X$ and $Y$ are $F$-spaces, since the norm plays no critical role here. This offers a greater variety but shall not be considered as an abuse of abstraction.

(The Closed Graph Theorem) Suppose

1. $X$ and $Y$ are $F$-spaces,

2. $f:X \to Y$ is linear,

3. $G(f)$ is closed in $X \times Y$.

Then $f$ is continuous.

In short, the closed graph theorem gives a sufficient condition to claim the continuity of $f$ (keep in mind, linearity does not imply continuity). If $f:X \to Y$ is continuous, then $G(f)$ is closed; if $G(f)$ is closed and $f$ is linear, then $f$ is continuous.

Proof. First of all we should make $X \times Y$ an $F$-space by assigning addition, scalar multiplication and metric. Addition and scalar multiplication are defined componentwise in the nature of things: $\alpha(x_1,y_1)+\beta(x_2,y_2)=(\alpha x_1+\beta x_2,\alpha y_1 + \beta y_2).$ The metric can be defined without extra effort: $d((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2)+d_Y(y_1,y_2).$ Then it can be verified that $X \times Y$ is a topological space with translate invariant metric. (Potentially the verifications will be added in the future but it's recommended to do it yourself.)

Since $f$ is linear, the graph $G(f)$ is a subspace of $X \times Y$. Next we quote an elementary result in point-set topology, a subset of a complete metric space is closed if and only if it's complete, by the translate-invariance of $d$, we see $G(f)$ is an $F$-space as well. Let $p_1: X \times Y \to X$ and $p_2: X \times Y \to Y$ be the natural projections respectively (for example, $p_1(x,y)=x$). Our proof is done by verifying the properties of $p_1$ and $p_2$ on $G(f)$.

For simplicity one can simply define $p_1$ on $G(f)$ instead of the whole space $X \times Y$, but we make it a global projection on purpose to emphasize the difference between global properties and local properties. One can also write $p_1|_{G(f)}$ to dodge confusion.

Claim 1. $p_1$ (with restriction on $G(f)$) defines an isomorphism between $G(f)$ and $X$.

For $x \in X$, we see $p_1(x,f(x)) = x$ (surjectivity). If $p_1(x,f(x))=0$, we see $x=0$ and therefore $(x,f(x))=(0,0)$, hence the restriction of $p_1$ on $G$ has trivial kernel (injectivity). Further, it's trivial that $p_1$ is linear.

Claim 2. $p_1$ is continuous on $G(f)$.

For every sequence $(x_n)$ such that $\lim_{n \to \infty}x_n=x$, we have $\lim_{n \to \infty}f(x_n)=f(x)$ since $G(f)$ is closed, and therefore $\lim_{n \to \infty}p_1(x_n,f(x_n)) =x$. Meanwhile $p_1(x,f(x))=x$. The continuity of $p_1$ is proved.

Claim 3. $p_1$ is a homeomorphism with restriction on $G(f)$.

We already know that $G(f)$ is an $F$-space, so is $X$. For $p_1$ we have $p_1(G(f))=X$ is of the second category (since it's an $F$-space and $p_1$ is one-to-one), and $p_1$ is continuous and linear on $G(f)$. By the open mapping theorem, $p_1$ is an open mapping on $G(f)$, hence is a homeomorphism thereafter.

Claim 4. $p_2$ is continuous.

This follows the same way as the proof of claim 2 but much easier since there is no need to care about $f$.

Now things are immediate once one realises that $f=p_2 \circ p_1|_{G(f)}^{-1}$, which implies that $f$ is continuous. $\square$

## Applications

Before we go for theorem 1 at the beginning, we drop an application on Hilbert spaces.

Let $T$ be a bounded operator on the Hilbert space $L_2([0,1])$ so that if $\phi \in L_2([0,1])$ is a continuous function so is $T\phi$. Then the restriction of $T$ to $C([0,1])$ is a bounded operator of $C([0,1])$.

Now we go for the identification of norms. Define \begin{aligned} f:(X,\lVert\cdot\rVert_1) &\to (X,\lVert\cdot\rVert_2) \\ x &\mapsto x \end{aligned} i.e. the identity map between two Banach spaces (hence $F$-spaces). Then $f$ is linear. We need to prove that $G(f)$ is closed. For the convergent sequence $(x_n)$ $\lim_{n \to \infty}\lVert x_n -x \rVert_1=0,$ we have $\lim_{n \to \infty} \lVert f(x_n)-x \rVert_2=\lim_{n \to \infty}\lVert x_n -x\rVert_2=\lim_{n \to \infty}\lVert f(x_n)-f(x)\rVert_2=0.$ Hence $G(f)$ is closed. Therefore $f$ is continuous, hence bounded, we have some $K$ such that $\lVert x \rVert_2 =\lVert f(x) \rVert_1 \leq K \lVert x \rVert_1.$ By defining \begin{aligned} g:(X,\lVert\cdot\rVert_2) &\to (X,\lVert\cdot\rVert_1) \\ x &\mapsto x \end{aligned} we see $g$ is continuous as well, hence we have some $K'$ such that $\lVert x \rVert_1 =\lVert g(x) \rVert_2 \leq K'\lVert x \rVert_2$ Hence two norms are weaker than each other.

## The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it's time to make a list of the series. It's been around half a year.

## References

• Walter Rudin, Functional Analysis
• Peter Lax, Functional Analysis
• Jesús Gil de Lamadrid, Some Simple Applications of the Closed Graph Theorem

# A proof of the ordinary Gleason-Kahane-Żelazko theorem for complex functionals

## The Theorem

(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. $\square$

• 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
• I. N. Herstein, Jordan Homomorphisms

# The Big Three Pt. 5 - The Hahn-Banach Theorem (Dominated Extension)

The Hahn-Banach theorem has been a central tool for functional analysis and therefore enjoys a wide variety, many of which have a numerous uses in other fields of mathematics. Therefore it's not possible to cover all of them. In this post we are covering two 'abstract enough' results, which are sometimes called the dominated extension theorem. Both of them will be discussed in real vector space where topology is not endowed. This allows us to discuss any topological vector space.

Another interesting thing is, we will be using axiom of choice, or whatever equivalence you may like, for example Zorn's lemma or well-ordering principle. Before everything, we need to examine more properties of vector spaces.

## Vector space

It's obvious that every complex vector space is also a real vector space. Suppose $X$ is a complex vector space, and we shall give the definition of real-linear and complex-linear functionals.

An addictive functional $\Lambda$ on $X$ is called real-linear (complex-linear) if $\Lambda(\alpha x)=\alpha\Lambda(x)$ for every $x \in X$ and for every real (complex) scalar $\alpha$.

For *-linear functionals, we have two important but easy theorems.

If $u$ is the real part of a complex-linear functional $f$ on $X$, then $u$ is real-linear and $f(x)=u(x)-iu(ix) \quad (x \in X).$

Proof. For complex $f(x)=u(x)+iv(x)$, it suffices to denote $v(x)$ correctly. But $if(x)=iu(x)-v(x),$ we see $\Im(f(x)=v(x)=-\Re(if(x))$. Therefore $f(x)=u(x)-i\Re(if(x))=u(x)-i\Re(f(ix))$ but $\Re(f(ix))=u(ix)$, we get $f(x)=u(x)-iu(ix).$ To show that $u(x)$ is real-linear, note that $f(x+y)=u(x+y)+iv(x+y)=f(x)+f(y)=u(x)+u(y)+i(v(x)+v(y)).$ Therefore $u(x)+u(y)=u(x+y)$. Similar process can be applied to real scalar $\alpha$. $\square$

Conversely, we are able to generate a complex-linear functional by a real one.

If $u$ is a real-linear functional, then $f(x)=u(x)-iu(ix)$ is a complex-linear functional

Proof. Direct computation. $\square$

Suppose now $X$ is a complex topological vector space, we see a complex-linear functional on $X$ is continuous if and only if its real part is continuous. Every continuous real-linear $u: X \to \mathbb{R}$ is the real part of a unique complex-linear continuous functional $f$.

### Sublinear, seminorm

Sublinear functional is 'almost' linear but also 'almost' a norm. Explicitly, we say $p: X \to \mathbb{R}$ a sublinear functional when it satisfies \begin{aligned} p(x)+p(y) &\leq p(x+y) \\ p(tx) &= tp(x) \\ \end{aligned} for all $t \geq 0$. As one can see, if $X$ is normable, then $p(x)=\lVert x \rVert$ is a sublinear functional. One should not be confused with semilinear functional, where inequality is not involved. Another thing worth noting is that $p$ is not restricted to be nonnegative.

A seminorm on a vector space $X$ is a real-valued function $p$ on $X$ such that \begin{aligned} p(x+y) &\leq p(x)+p(y) \\ p(\alpha x)&=|\alpha|p(x) \end{aligned} for all $x,y \in X$ and scalar $\alpha$.

Obviously a seminorm is also a sublinear functional. For the connection between norm and seminorm, one shall note that $p$ is a norm if and only if it satisfies $p(x) \neq 0$ if $x \neq 0$.

## Dominated extension theorems

Are the results will be covered in this post. Generally speaking, we are able to extend a functional defined on a subspace to the whole space as long as it's dominated by a sublinear functional. This is similar to the dominated convergence theorem, which states that if a convergent sequence of measurable functions are dominated by another function, then the convergence holds under the integral operator.

(Hahn-Banach) Suppose

1. $M$ is a subspace of a real vector space $X$,
2. $f: M \to \mathbb{R}$ is linear and $f(x) \leq p(x)$ on $M$ where $p$ is a sublinear functional on $X$

Then there exists a linear $\Lambda: X \to \mathbb{R}$ such that $\Lambda(x)=f(x)$ for all $x \in M$ and $-p(-x) \leq \Lambda(x) \leq p(x)$ for all $x \in X$.

### Step 1 - Extending the function by one dimension

With that being said, if $f(x)$ is dominated by a sublinear functional, then we are able to extend this functional to the whole space with a relatively proper range.

Proof. If $M=X$ we have nothing to do. So suppose now $M$ is a nontrivial proper subspace of $X$. Choose $x_1 \in X-M$ and define $M_1=\{x+tx_1:x \in M,t \in R\}.$ It's easy to verify that $M_1$ satisfies all axioms of vector space (warning again: no topology is endowed). Now we will be using the properties of sublinear functionals.

Since $f(x)+f(y)=f(x+y) \leq p(x+y) \leq p(x-x_1)+p(x_1+y)$ for all $x,y \in M$, we have $f(x)-p(x-x_1) \leq p(x_1+y) -f(y).$ Let $\alpha=\sup_{x}\{f(x)-p(x-x_1):x \in M\}.$ By definition, we naturally get $f(x)-\alpha \leq p(x-x_1)$ and $f(y)+\alpha \leq p(x_1+y).$ Define $f_1$ on $M_1$ by $f_1(x+tx_1)=f(x)+t\alpha.$ So when $x +tx_1 \in M$, we have $t=0$, and therefore $f_1=f$.

To show that $f_1 \leq p$ on $M_1$, note that for $t>0$, we have $f(x/t)-\alpha \leq p(x/t-x_1),$ which implies $f(x)-t\alpha=f_1(x-t\alpha)\leq p(x-tx_1).$ Similarly, $f(y/t)+\alpha \leq p(y/t+x_1),$ and therefore $f(y)+t\alpha=f_1(y+tx_1) \leq p(y+tx_1).$ Hence $f_1 \leq p$.

### Step 2 - An application of Zorn's lemma

#### Side note: Why Zorn's lemma

It seems that we can never stop using step 1 to extend $M$ to a larger space, but we have to extend. (If $X$ is a finite dimensional space, then this is merely a linear algebra problem.) This meets exactly what William Timothy Gowers said in his blog post:

If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.

-- How to use Zorn's lemma

And we will show that, as W. T. Gowers said,

If the resulting partial order satisfies the chain condition and if a maximal element must be a structure of the kind one is trying to build, then the proof is complete.

To apply Zorn's lemma, we need to construct a partially ordered set. Let $\mathscr{P}$ be the collection of all ordered pairs $(M',f')$ where $M'$ is a subspace of $X$ containing $M$ and $f'$ is a linear functional on $M'$ that extends $f$ and satisfies $f' \leq p$ on $M'$. For example we have $(M,f) , (M_1,f_1) \subset \mathscr{P}.$ The partial order $\leq$ is defined as follows. By $(M',f') \leq (M'',f'')$, we mean $M' \subset M''$ and $f' = f''$ on $M'$. Obviously this is a partial order (you should be able to check this).

Suppose now $\mathcal{F}$ is a chain (totally ordered subset of $\mathscr{P}$). We claim that $\mathcal{F}$ has an upper bound (which is required by Zorn's lemma). Let $M_0=\bigcup_{(M',f') \in \mathcal{F}}M'$ and $f_0(y)=f(y)$ whenever $(M',f') \in \mathcal{F}$ and $y \in M'$. It's easy to verify that $(M_0,f_0)$ is the upper bound we are looking for. But $\mathcal{F}$ is arbitrary, therefore by Zorn's lemma, there exists a maximal element $(M^\ast,f^\ast)$ in $\mathscr{P}$. If $M^* \neq X$, according to step 1, we are able to extend $M^\ast$, which contradicts the maximality of $M^\ast$. And $\Lambda$ is defined to be $f^\ast$. By the linearity of $\Lambda$, we see $-p(-x) \leq -\Lambda(-x)=\Lambda{x}.$ The theorem is proved. $\square$

### How this proof is constructed

This is a classic application of Zorn's lemma (well-ordering principle, or Hausdorff maximality theorem). First, we showed that we are able to extend $M$ and $f$. But since we do not know the dimension or other properties of $X$, it's not easy to control the extension which finally 'converges' to $(X,\Lambda)$. However, Zorn's lemma saved us from this random exploration: Whatever happens, the maximal element is there, and take it to finish the proof.

## Generalisation onto the complex field

Since inequality is appeared in the theorem above, we need more careful validation.

(Bohnenblust-Sobczyk-Soukhomlinoff) Suppose $M$ is a subspace of a vector space $X$, $p$ is a seminorm on $X$, and $f$ is a linear functional on $M$ such that $|f(x)| \leq p(x)$ for all $x \in M$. Then $f$ extends to a linear functional $\Lambda$ on $X$ satisfying $|\Lambda (x)| \leq p(x)$ for all $x \in X$.

Proof. If the scalar field is $\mathbb{R}$, then we are done, since $p(-x)=p(x)$ in this case (can you see why?). So we assume the scalar field is $\mathbb{C}$.

Put $u = \Re f$. By dominated extension theorem, there is some real-linear functional $U$ such that $U(x)=u$ on $M$ and $U \leq p$ on $X$. And here we have $\Lambda(x)=U(x)-iU(ix)$ where $\Lambda(x)=f(x)$ on $M$.

To show that $|\Lambda(x)| \leq p(x)$ for $x \neq 0$, by taking $\alpha=\frac{|\Lambda(x)|}{\Lambda(x)}$, we have $U(\alpha{x})=\Lambda(\alpha{x})=|\Lambda(x)|\leq p(\alpha x)=p(x)$ since $|\alpha|=1$ and $p(\alpha{x})=|\alpha|p(x)=p(x)$. $\square$

## Extending Hahn-Banach theorem under linear transform

To end this post, we state a beautiful and useful extension of the Hahn-Banach theorem, which is done by R. P. Agnew and A. P. Morse.

(Agnew-Morse) Let $X$ denote a real vector space and $\mathcal{A}$ be a collection of linear maps $A_\alpha: X \to X$ that commute, or namely $A_\alpha A_\beta=A_\beta A_\alpha$ for all $A_\alpha,A_\beta \in \mathcal{A}$. Let $p$ be a sublinear functional such that $p(A_\alpha{x})=p(x)$ for all $A_\alpha \in \mathcal{A}$. Let $Y$ be a subspace of $X$ on which a linear functional $f$ is defined such that

1. $f(y) \leq p(y)$ for all $y \in Y$.
2. For each mapping $A$ and $y \in Y$, we have $Ay \in Y$.
3. Under the hypothesis of 2, we have $f(Ay)=f(y)$.

Then $f$ can be extended to $X$ by $\Lambda$ so that $-p(-x) \leq \Lambda(x) \leq p(x)$ for all $x \in X$, and $\Lambda(A_\alpha{x})=\Lambda{x}.$

To prove this theorem, we need to construct a sublinear functional that dominates $f$. For the whole proof, see Functional Analysis by Peter Lax.

## The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it's time to make a list of the series. It's been around half a year.