# Ngô Quốc Anh

## February 23, 2017

### In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose $(X, \|\cdot\|)$ is a normed space and $\{x_1,...,x_n\}$ is a set of linearly independent elements in $X$. Then $\{x_1,...,x_n\}$ is stable under a small perturbation in the sense that there exists some small number $\varepsilon>0$ such that for any $\|y_i\| < \varepsilon$ with $1 \leqslant i \leqslant n$, the all elements of $\{x_1+y_1,...,x_n+y_n\}$ are also linearly independent.

We prove this result by way of contradiction. Indeed, for any $\varepsilon>0$, there exist $n$ elements $y_i \in X$ with $\|y_i\| < \varepsilon$ such that all elements of $\{x_1+y_1,...,x_n+y_n\}$ are linearly dependent, that is, there exist real numbers $\alpha_i$ with $1 \leqslant i \leqslant n$ such that

$\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0$

with

$\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.$

## May 22, 2010

### The third and fouth fundamental results in the calculus of variation

Filed under: Giải tích 8 (MA5206), PDEs — Ngô Quốc Anh @ 17:56

Followed by this entry where the following questions have been discussed

1. Boundedness implies weakly convergence: If $E$ is a reflexive Banach space and $\{x_n\}_n \subset E$ is a bounded sequence. Then up to a subsequence $x_n$ converges weakly to some $x$ in $X$.
2. Weakly convergence becomes strongly convergence via compact operator: A compact operator $C : E \to X$ between Banach spaces maps every weakly convergent sequence in $E$ into one that converges strongly in $X$.

Now I shall discuss more results which appear frequently in the calculus of variation.

Let us recall over a minifold $M$, the Sobolev norm $H^1(M)$ (or $W^{1,2}(M)$) is defined by

$\displaystyle \|u\|_{H^1}^2=\|u\|_{L^2}^2+\|\nabla u\|_{L^2}^2$.

It is immediately to see that if $u_n \to u$ strongly in $H^1$, i.e. $\|u_n-u\|_{H^1}\to 0$ then $u_n \to u$ strongly in $L^2$ and $\nabla u_n \to \nabla u$ strongly in $L^2$.

It turns out to discuss what happen to weakly convergence. Actually, we shall prove the following important result, called the third fundamental result.

Weakly convergence in Sobolev spaces implies weakly convergence in $L^p$ spaces. We assume $u_n \rightharpoonup u$ in $H^1$. We shall prove both $u_n$ and $\nabla u_n$ converge weakly to $u$ and $\nabla u$ in $L^2$, respectively.

By the principle of uniform boundedness, any weakly convergence sequence is bounded in the norm. Consequently, $\{u_n\}_n$ and $\{\nabla u_n\}_n$ are bounded in $L^2$. By the weak compactness of balls in $L^2$, there is a subsequence $n_k$ such that

$\displaystyle u_{n_k} \rightharpoonup v, \quad \nabla u_{n_k} \rightharpoonup w$

in $L^2$ (i.e., both $v, w$ are in $L^2$). Since the weak convergence in $L^2$ implies the convergence in $\mathcal D'$ the dual space of $\mathcal D$-the space of test functions. It follows that $w = \nabla v$ and, hence, $v \in H^1$. It follows that $u\equiv v$ and thus

$\displaystyle u_{n_k} \rightharpoonup u, \quad \nabla u_{n_k} \rightharpoonup \nabla u$

in $L^2$ as desired.

Now we consider the reverse case. We shall prove the following

Weakly convergence in $L^p$ spaces plus the boundedness implies weakly convergence in Sobolev spaces. We assume $u_n \rightharpoonup u \in L^2$ in $L^2$ and $\|u_n\|_{H^1}$ is bounded. We shall prove that $u \in H^1$ and $u_n \rightharpoonup u$ in $H^1$.

Since $\{u_n\}$ is bounded in $H^1$, by the first fundamental result, $u_n \rightharpoonup v$ in $H^1$ for some $v \in H^1$. By the third fundamental result above, $u_n \rightharpoonup v$ in $L^2$. It follows from the uniqueness of weak limit that $u \equiv v$ which implies $u \in H^1$.

In order to prove $u_n \rightharpoonup u$ in $H^1$, we shall use the following result whose proof is based on the simple contradiction argument.

Let $X$ be a topological space. A sequence $\{x_n\} \subset X$  converges to $x \in X$ (in the topological of $X$) if and only if any subsequence of $\{x_n\}$ contains a sub-subsequence that converges to $x$.

Let us pick a particular subsequence of $\{u_n\}$ and rename it back to $\{u_n\}$ for simplicity. It suffices to prove that $\{u_n\}$ contains a subsequence that converges to $u$ weakly in $H^1$. Followed the proof of the third fundamental result, there is a subsequence of $\{u_n\}$ and a function $v \in H^1$ such that

$\displaystyle u_{n_k} \rightharpoonup v, \quad \nabla u_{n_k} \rightharpoonup v$

both in $L^2$. It follows from the definition of weakly convergence in $H^1$ that in fact we get

$\displaystyle u_{n_k} \rightharpoonup v$

in $H^1$. The reason is the following:

$\displaystyle (u_{n_k},\varphi)_{L^2}+(\nabla u_{n_k},\nabla\varphi)_{L^2} \to (v,\varphi)_{L^2} + (\nabla v, \varphi)_{L^2}$

for any $\varphi \in H^1$. Having this and the fact that weak limit is unique we deduce that $u \equiv v$. The latter now implies

$u_{n_k} \rightharpoonup u$

in $H^1$. The proof follows.

## May 15, 2010

### Two fundamental results in the calculus of variation

Filed under: Giải tích 8 (MA5206), PDEs — Ngô Quốc Anh @ 20:28

I suddenly think that I should post this entry ‘cos sometimes I don’t remember these stuffs. These results appear frequently in solving PDEs especially when using the direct method. For example, the simplest case is the following eigenvalue problem

$\displaystyle -{\rm div}(|\nabla u|^{p-2}\nabla u)=\lambda |u|^{p-2}u$

over a bounded domain $\Omega \subset \mathbb R^n$ with Dirichlet boundary condition. We assume $p. Our aim is to show the existence of the first eigenvalue $\lambda_1>0$. Obviously, our problem is to solve the following optimization

$\displaystyle\mathop {\inf }\limits_{u \in W^{1,2}_0(\Omega)} \left\{ {\int_\Omega {|\nabla u{|^p}dx} :\int_\Omega {|u{|^p}dx} = 1} \right\}$.

The direct method says that we firstly select a minimizing sequence, say $\{u_n\}_n$, then we need to prove $\{u_n\}$ is convergent. There are two steps in the above argument which lead to this entry. Our first claim is the following.

Boundedness implies weakly convergence. The first result says that

If $E$ is a reflexive Banach space and $\{x_n\}_n \subset E$ is a bounded sequence. Then up to a subsequence $x_n$ converges weakly to some $x$ in $X$.

The proof of this claim can be found in a book due to Brezis (Theorem III.27). Interestingly, its converse also holds by the Eberlein-Šmulian theorem.

Theorem (Eberlein-Šmulian). Suppose $E$ is a Banach space such that every bounded sequence $\{x_n\}_n$ contains a weakly convergent subsequence. Then $E$ is reflexive.

There was an elementary proof of this theorem. We refer the reader to a paper due to Whitley [here]. Let us get back to our optimization problem. Once we have a minimizing sequence $\{u_n\}_n \subset W^{1,p}_0(\Omega)$ it is clear to see that $\{u_n\}_n$ is bounded in $W^{1,p}_0(\Omega)$ since

$\displaystyle\int_\Omega {|{u_n}{|^p}dx} = 1$

and

$\displaystyle\int_\Omega {|\nabla {u_n}{|^p}dx} \leqslant C, \quad \forall n$.

By using the first claim, $u_n$ converges weakly to some $u \in W^{1,p}_0(\Omega)$. It is worth noticing that by saying $u_n \rightharpoonup u$ in $W^{1,p}_0(\Omega)$ we mean $u_n \rightharpoonup u$ in $L^p(\Omega)$ and $Du_n \rightharpoonup Du$ in $L^p(\Omega, \mathbb R^n)$. Now we need further argument

Weakly convergence becomes strongly convergence via compact operator. This second result says that

A compact operator $C : E \to X$ between Banach spaces maps every weakly convergent sequence in $E$ into one that converges strongly in $X$.

The proof of this relies on the contradiction argument and the fact that once a sequence converges strongly to some limit, this limit is unique. However, the converse is no long true. For example, by the Schur theorem, a sequence $\{x_n\}$ in $\ell^1$ converges weakly, it also converges strongly. We take $C$ to be the identity $I$ in $\ell^1$. Since $\ell^1$ has infinity dimensional, by using the Riesz theorem, $I$ cannot be compact.

Using this claim we deduce that $u_n$ converges strongly to $u$ in $L^q$ for any $1\leqslant q. By using the Minkowski and Holder inequalities we can show that $u$ satisfies the constraint. It now follows from the weakly lower semi-continuous of norm that $u$ indeed satisfies the equation. The proof follows.

## May 14, 2010

### Symmetrization: Schwarz symmetrization

Filed under: Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:34

Given a measurable subset $E \subset \mathbb R^N$, we denote its $N$-dimensional Lebesgue measure by $|E|$. We will denote by $E^\star$ the open ball centered at the origin and having the same measure as $E$, i.e. $|E^\star|=|E|$. The norm of vector $x \in \mathbb R^n$ will be denoted by $|x|$. Finally, we will denote by $\omega_N$ the volume of the unit ball in $\mathbb R^N$. It is worth recalling that

$\displaystyle \omega_N=\frac{\pi^\frac{N}{2}}{\Gamma \left(\frac{N}{2}+1\right)}$

where $\Gamma$ us the usual gamma function.

Definition (Schwarz symmetrization). Let $\Omega \subset \mathbb R^N$ be a bounded domain. Let $u : \Omega \to \mathbb R$ be a measurable function. Then, its Schwarz symmetrization (or the spherically symmetric and decreasing rearrangement) is the function $u^\star : \Omega^\star \to \mathbb R$ defined by

$u^\star(x)=u^\sharp (\omega_N|x|^N), \quad x \in \Omega^\star$.

Observe that if $R$ is the radius of $\Omega^\star$, then

$\displaystyle\begin{gathered} \int_{{\Omega ^ \star }} {{u^ \star }(x)dx} = \int_{{\Omega ^ \star }} {{u^\sharp }({\omega _N}{{\left| x \right|}^N})dx} \hfill \\ \qquad= \int_0^R {{u^\sharp }({\omega _N}{{\left| x \right|}^N})N{\omega _N}{\tau ^{N - 1}}d\tau } \hfill \\ \qquad= \int_0^{|{\Omega ^ \star }|} {{u^\sharp }(s)ds} \hfill \\ \qquad= \int_0^{|\Omega |} {{u^\sharp }(s)ds} . \hfill \\ \end{gathered}$

We obviously have the following properties of Schwarz symmetrization (more…)

## May 11, 2010

### Compact embedding of Hölder spaces

Filed under: Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 2:51

Hölder continuous. In mathematics, a real or complex-valued function $f$ on $d$-dimensional Euclidean space satisfies a Hölder condition, or is Hölder continuous, when there are nonnegative real constants $C, \alpha$, such that

$\displaystyle | f(x) - f(y) | \leqslant C |x - y|^{\alpha}$

for all $x$ and $y$ in the domain of $f$. More generally, the condition can be formulated for functions between any two metric spaces. The number $\alpha$ is called the exponent of the Hölder condition. If $\alpha = 1$, then the function satisfies a Lipschitz condition. If $\alpha = 0$, then the function simply is bounded.

Hölder spaces. Hölder spaces consisting of functions satisfying a Hölder condition are basic in areas of functional analysis relevant to solving partial differential equations, and in dynamical systems. The Hölder space $C^{k,\alpha}(\Omega)$, where $\Omega$ is an open subset of some Euclidean space and $k \geqslant 0$ an integer, consists of those functions on $\Omega$ having derivatives  up to order $k$ and such that the $k$-th partial derivatives are Hölder continuous with exponent $\alpha$, where $0 <\alpha \leqslant 1$. This is a locally convex topological vector space.

If the Hölder coefficient

$\displaystyle | f |_{C^{0,\alpha}} = \sup_{x,y \in \Omega} \frac{| f(x) - f(y) |}{|x-y|^\alpha}$,

is finite, then the function $f$ is said to be (uniformly) Hölder continuous with exponent $\alpha$ in $\Omega$. In this case, Hölder coefficient serves as a seminorm. If the Hölder coefficient is merely bounded on compact subsets of $\Omega$, then the function $f$ is said to be locally Hölder continuous with exponent $\alpha$ in $\Omega$.

## May 2, 2010

### Symmetrization: The Decreasing Rearrangement

Given a measurable subset $E \subset \mathbb R^N$, we denote its $N$-dimensional Lebesgue measure by $|E|$.

Let $\Omega$ be a bounded measurable set. Let $u :\Omega \to \mathbb R$ be a measurable function. For $t \in \mathbb R$, the level set $\{u>t\}$ is defined as

$\displaystyle \{u>t\}=\{x\in \Omega: u(x)>t\}$.

The sets $\{u, $\{u \geqslant t\}$, $\{u=t\}$ and so on are defined by analogy. Then the distribution function of $u$ is given by

$\displaystyle \mu_u(t)=|\{u>t\}|$.

This function is a monotonically decreasing function of $t$ and for $t \geq {\rm esssup}(u)$ we have $\mu_u(t)=0$ while for $t\leqslant {\rm essinf}(u)$, we have $\mu_u(t)=|\Omega|$. Thus the range of $\mu_u$ is the interval $[0, |\Omega|]$.

Definition (Decreasing rearrangement). Let $\Omega \subset \mathbb R^N$ be bounded and let $u :\Omega \to \mathbb R$ be a measurable function. Then the (unidimensional) decreasing rearrangement of $u$, denoted by $u^\sharp$, is defined on $[0, |\Omega|]$ by

$\displaystyle {u^\sharp }(s) = \begin{cases} {\rm esssup} (u),& s = 0, \hfill \\ \mathop {\inf }\limits_t \left\{ {t:{\mu _u}(t) < s} \right\}, & s > 0. \hfill \\ \end{cases}$

Essentially, $u^\sharp$ is just the inverse function of the distribution function $\mu_u$ of $u$. The following properties of the decreasing rearrangement are immediate from its definition.

Proposition 1. Let $u : \Omega \to \mathbb R^N$ where $\Omega \subset \mathbb R^N$ is bounded. Then $u^\sharp$ is a nonincreasing and left-continuous function.

Proposition 2. The mapping $u \mapsto u^\sharp$ is non-decreasing, i.e. if $u\leqslant v$ in the sense that $u(x) \leqslant v(x)$ for all $x$, where $u$ and $v$ are real-valued functions on $\Omega$ then $u^\sharp \leqslant v^\sharp$.

We now see that $u^\sharp$ is indeed a rearrangement of $u$.

Proposition 3. The function $u : \Omega \to \mathbb R$ and $u^\sharp : [0,|\Omega|] \to \mathbb R$ are equimeasurable (i.e. they have the same distribution function), i.e. for all $t$

$\displaystyle |\{u >t\}|=|\{u^\sharp >t\}|$.

## April 17, 2010

### A Theorem of Banach and Saks

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:39

According to Banach and Saks, every bounded sequence in $L^p$ or $\ell^p$ ($1) has a subsequence whose Cesaro-means converge strongly. More generally, every uniformly convex Banach space possesses this so-called Banach-Saks property, as shown by Kakutani. In particular, every Hilbert space has this property. In nonlinear analysis, by utilizing a duality mapping some assertions which are valid in the case of Hilbert spaces are extended to the case of special classes of Banach spaces. Especially in the case of Banach spaces with a uniformly convex conjugate space, such extentions are often obtained since a duality mapping is uniformly strongly continuous on each bounded subset of such a Banach space.

The Banach-Saks theorem in $L^2$ states that

Theorem (Banach-Saks for $L^2$ spaces). Given in $L^2$ a sequence $\{f_n\}_n$ which converges weakly to an element $f$, we can select a subsequence $\{f_{n_k}\}_k$ such that the arithmetic means

$\displaystyle\frac{{{f_{{n_1}}} + {f_{{n_2}}} + \cdots + {f_{{n_k}}}}}{k}$

converge in strongly to $f$.

This theorem is due to the two Polish geometers S.  Banach and S. Saks, whose work and, in particular, the importance of whose research in the topics treated in this book are widely acknowledged.

## February 22, 2010

### The Poincaré inequality: W^{1,p} vs. W_0^{1,p}

Filed under: Giải Tích 6 (MA5205), Giải tích 8 (MA5206), Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 1:50

In mathematics, the Poincaré inequality is a result in the theory of Sobolev spaces, named after the French mathematician Henri Poincaré. The inequality allows one to obtain bounds on a function using bounds on its derivatives and the geometry of its domain of definition. Such bounds are of great importance in the modern, direct methods of the calculus of variations. A very closely related result is the Friedrichs’ inequality.

This topic will cover two versions of the Poincaré inequality, one is for $W^{1,p}(\Omega)$ spaces and the other is for $W_o^{1,p}(\Omega)$ spaces.

The classical Poincaré inequality for $W^{1,p}(\Omega)$ spaces. Assume that $1\leq p \leq \infty$ and that $\Omega$ is a bounded open subset of the $n$dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ and $p$, such that for every function $u$ in the Sobolev space $W^{1,p}(\Omega)$,

$\displaystyle \| u - u_{\Omega} \|_{L^{p} (\Omega)} \leq C \| \nabla u \|_{L^{p} (\Omega)}$,

where

$\displaystyle u_{\Omega} = \frac{1}{|\Omega|} \int_{\Omega} u(y) \, \mathrm{d} y$

is the average value of $u$ over $\Omega$, with $|\Omega|$ standing for the Lebesgue measure of the domain $\Omega$.

Proof. We argue by contradiction. Were the stated estimate false, there would exist for each integer $k = 1,...$ a function $u_k \in W^{1,p}(\Omega)$ satisfying

$\displaystyle \| u_k - (u_k)_{\Omega} \|_{L^{p} (\Omega)} \geq k \| \nabla u_k \|_{L^{p} (\Omega)}$.

We renormalize by defining

$\displaystyle {v_k} = \frac{{{u_k} - {{({u_k})}_\Omega }}}{{{{\left\| {{u_k} - {{({u_k})}_\Omega }} \right\|}_{{L^p}(\Omega )}}}}, \quad k \geqslant 1$.

Then

$\displaystyle {({v_k})_\Omega } = 0, \quad {\left\| {{v_k}} \right\|_{{L^p}(\Omega )}} = 1$

and therefore

$\displaystyle\| \nabla v_k \|_{L^{p} (\Omega)} \leqslant \frac{1}{k}$.

In particular the functions $\{v_k\}_{k\geq 1}$ are bounded in $W^{1,p}(\Omega)$.

By mean of the Rellich-Kondrachov Theorem, there exists a subsequence ${\{ {v_{{k_j}}}\} _{j \geqslant 1}} \subset {\{ {v_k}\} _{k \geqslant 1}}$ and a function $v \in L^p(\Omega)$ such that

$\displaystyle v_{k_j} \to v$ in $L^p(\Omega)$.

Passing to a limit, one easily gets

$\displaystyle v_\Omega = 0, \quad {\left\| {{v}} \right\|_{{L^p}(\Omega )}} = 1$.

On the other hand, for each $i=\overline{1,n}$ and $\varphi \in C_0^\infty(\Omega)$,

$\displaystyle\int_\Omega {v{\varphi _{{x_i}}}dx} = \mathop {\lim }\limits_{{k_j} \to \infty } \int_\Omega {{v_{{k_j}}}{\varphi _{{x_i}}}dx} = - \mathop {\lim }\limits_{{k_j} \to \infty } \int_\Omega {{v_{{k_j},{x_i}}}\varphi dx} = 0$.

Consequently, $v\in W^{1,p}(\Omega)$ with $\nabla v=0$ a.e. Thus $v$ is constant since $\Omega$ is connected. Since $v_\Omega=0$ then $v \equiv 0$. This contradicts to $\|v\|_{L^p(\Omega)}=1$.

The Poincaré inequality for $W_0^{1,2}(\Omega)$ spaces. Assume that $\Omega$ is a bounded open subset of the $n$-dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ such that for every function $u$ in the Sobolev space $W_0^{1,2}(\Omega)$,

$\displaystyle \| u \|_{L^2(\Omega)} \leq C \| \nabla u \|_{L^2(\Omega)}$.

Proof. Assume $\Omega$ can be enclosed in a cube

$\displaystyle Q=\{ x \in \mathbb R^n: |x_i| \leqslant a, 1\leqslant i \leqslant n\}$.

Then for any $x \in Q$, we have

$\displaystyle\begin{gathered} {u^2}(x) = {\left( {\int_{ - a}^{{x_1}} {{u_{{x_1}}}(t,{x_2},...,{x_n})dt} } \right)^2} \hfill \\ \qquad\leqslant ({x_1} + a)\int_{ - a}^{{x_1}} {{{({u_{{x_1}}})}^2}dt} \hfill \\ \qquad\leqslant 2a\int_{ - a}^a {{{({u_{{x_1}}})}^2}dt} . \hfill \\ \end{gathered}$.

Thus

$\displaystyle\int_{ - a}^a {{u^2}(x)dx} \leqslant 4{a^2}\int_{ - a}^a {{{({u_{{x_1}}})}^2}dt}$.

Integration over $x_2,...,x_n$ from $-a$ to $a$ gives the result.

The Poincaré inequality for $W_0^{1,p}(\Omega)$ spaces. Assume that $1\leq p and that $\Omega$ is a bounded open subset of the $n$-dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ and $p$, such that for every function $u$ in the Sobolev space $W_0^{1,p}(\Omega)$,

$\displaystyle \| u \|_{L^{p^\star} (\Omega)} \leq C \| \nabla u \|_{L^{p} (\Omega)}$,

where $p^\star$ is defined to be $\frac{np}{n-p}$.

Proof. The proof of this version is exactly the same to the proof of $W^{1,p}(\Omega)$ case.

Remark. The point $u =0$ on the boundary of $\Omega$ is important. Otherwise, the constant function will not satisfy the Poincaré inequality. In order to avoid this restriction, a weight has been added like the classical Poincaré inequality for $W^{1,p}(\Omega)$ case. Sometimes, the Poincaré inequality for $W_0^{1,p}(\Omega)$ spaces is called the Sobolev inequality.

## October 28, 2009

### The weak and weak* topologies: A few words

Filed under: Giải tích 8 (MA5206), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space $X$. The dual space of $X$, denoted by $X'$, is the collection of all continuous linear functionals, i.e., the set of all mapping $\ell : X \to \mathbb R$ satisfying

$\ell(ax)=a \ell (x)$, $\ell(x+y)=\ell(x)+\ell(y)$

and

$\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x)$ when $\displaystyle\lim_{n \to \infty} \|x_n - x\|=0$.

Definition 1. In $X$, the strong topology is the norm topology, i.e., we can talk about an open set of $X$, for example $U$ in the following sense: $U \subset X$ is said to be open if and only if for each $x_0 \in U$, there exists $\varepsilon>0$ such that $\{ x \in X: \|x-x_0\|<\varepsilon\} \subset U$.

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional $\ell$ is said to be bounded if there is a positive number $c$ such that $|\ell (x)| \leq c\|x\|$ for all $x \in X$.

Now we assume $\ell$ is continuous but not bounded; then for any choice of $c=n$, one has $\ell(x_n) > n \|x_n\|$. Clearly, $x_n$ can be replaced by any multiple of $x_n$; if we normalize $x_n$ so that

$\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}$

then $x_n \to 0$ but $\ell (x_n) \to \infty$. This shows the lack of boundedness implies the lack of continuity.

Now we assume $\ell$ is bounded. For arbitray $x_n$ and $x$, one gets

$|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|$;

this shows that boundedness implies continuity.

Definition 2. In $X$, the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form

$\{ x : a< \ell(x) < b\}$.

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls

$\{ x : \|x\|

opens in the strong topology, are not open is the weak topology.

Definition 3. In $X'$ the dual space of $X$, the weak* topology is the crudest topology in which all linear functionals

$x: X' \to \mathbb R, x(\ell) := \ell(x)$

are continuous.

If $X'$ is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in $X'$ is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in $X$ is compact in the weak topology if and only if $X$ is reflexive.

## March 4, 2009

### Show that the closure of a convex set also is convex

Filed under: Các Bài Tập Nhỏ, Giải tích 8 (MA5206), Linh Tinh — Ngô Quốc Anh @ 23:29

Today on the way, my Chinese friend and I have discussed the following question: Let  denote a convex subset of a locally convex topological linear space . Show that the closure  of  also is convex.

I have suggested the following solution.

We define $f: X \times X \times \mathbb{R} \to X$ as following $f: \left( {x,y,\lambda } \right) \mapsto \lambda x + \left( {1 - \lambda } \right)y$. Then  is continuous and $f\left( {K \times K \times \left[ {0,1} \right]} \right) \subset K$ since  is convex. We then obtain that $f\left( {\overline K \times \overline K \times \left[ {0,1} \right]} \right) \subset \overline K$ due to the continuity of , that is $\lambda \overline K + \left( {1 - \lambda } \right)\overline K \in \overline K$ for every . Therefore  is a convex set.

So what I am going to tell you is how correct the solution is? If no, what’s the problem, otherwise, what’s the main point? I will show you a little bit latter. I think I should go for sleep, it’s late now 😦