Two-dimensional Divergence Theorem is equivalent to Green’s Theorem

The special planar version of Divergence Theorem is actually equivalent to Green’s Theorem, even though on first sight they look quite different.

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Outer Measure Zero implies Measurable

One quick way is to use Caratheodory’s Criterion:

Let \lambda^* denote the Lebesgue outer measure on \mathbb{R}^n, and let E\subseteq\mathbb{R}^n. Then E is Lebesgue measurable if and only if \lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^c) for every A\subseteq\mathbb{R}^n.

Suppose E is a set with outer measure zero, and A be any subset of \mathbb{R}^n.

Then \lambda^*(A\cap E)+\lambda^*(A\cap E^c)\leq\lambda^*(E)+\lambda^*(A)=\lambda^*(A) by the monotonicity of outer measure.

The other direction \lambda^*(A)\leq\lambda^*(A\cap E)+\lambda^*(A\cap E^c) follows by countable subadditivity of outer measure.


Functional Analysis List of Theorems

This is a list of miscellaneous theorems from Functional Analysis.

Hahn-Banach: Let p be a real-valued function on a normed linear space X with

(i) Positive homogeneity

(ii) Subadditivity

Let Y denote a linear subspace of X on which l(y)\leq p(y) for all y\in Y. Then l can be extended to all of X: l(x)\leq p(x) for all x\in X.

Geometric Hahn-Banach / Hyperplane Separation Theorem: Let K be a nonempty convex subset, K=int(K). Let y be a point outside K. Then there exists a linear functional l (depending on y) such that l(x)<c for all x\in K; l(y)=c.

Riesz Representation Theorem: Let l(x) be a linear functional on a Hilbert space H that is bounded: |l(x)|\leq c\|x\|. Then l(x)=\langle x,y\rangle for some unique y\in H.

Principle of Uniform Boundedness: A weakly convergent sequence \{x_n\} is uniformly bounded in norm.

Open Mapping Principle/Theorem:  Let X, U be Banach spaces. Let M:X\to U be a bounded linear map onto all of U. Then M maps open sets onto open sets.

Closed Graph Theorem: Let X,U be Banach spaces, M:X\to U be a closed linear map. Then M is continuous.

Weak* convergent sequence uniformly bounded

Theorem 11 (Lax Functional Analysis): A weak* convergent sequence \{u_n\} of points in a Banach space U=X' is uniformly bounded.

We will need a previous Theorem 3: X is a Banach space, \{l_v\} a collection of bounded linear functionals such that at every point x of X, |l_v(x)|\leq M(x) for all l_v. Then there is a constant c such that |l_v|\leq c for all l_v.

Sketch of proof:

Weak* convergence means \lim u_n(x)=u(x), thus there exists N such that for all n\geq N, we have |u_n(x)-u(x)|<1, which in turns means |u_n(x)|<1+|u(x)| via the triangle inequality. We have managed to bound the terms greater than equals to N.

For those terms less than N, we have |u_n(x)|\leq\|u_n\|\|x\|.

Thus, we may take M(x)=\max\{\|u_1\|\|x\|,\dots,\|u_{N-1}\|\|x\|,1+|u(x)|\}. The crucial thing is that M(x) depends only on x, not n.

Then, use Theorem 3, we can conclude that \|u_n\|\leq c for all n.

Hahn-Banach Theorem: Crown Jewel of Functional Analysis

Hahn-Banach Theorem is called the Crown Jewel of Functional Analysis, and has many different versions.

There is a Chinese quote “实变函数学十遍,泛函分析心犯寒”, which means one needs to study real function theory ten times before understanding, and the heart can go cold when studying functional analysis, which shows how deep is this subject.

The following is one version of Hahn-Banach Theorem that I find quite useful:

(Hahn-Banach, Version) If V is a normal vector space with linear subspace U (not necessarily closed) and if z is an element of V not in the closure of U, then there exists a continuous linear map \psi:V\to K with \psi(x)=0 for all x\in U, \psi(z)=1, and \|\psi\|=\text{dist}(z,U)^{-1}.

Brief sketch of proof: Define \phi:U+\text{span}\{z\}\to K, \phi(u+\lambda z)=\lambda, and use the Hahn-Banach (Extension version).

Closure is linear subspace

Let X be normed linear space, Y a subspace of X. The closure of Y, \bar{Y}, is a linear subspace of X.

We use the “sequential” equivalent definition of closure, rather than the one using open balls: \bar Y is the set of all limits of all convergent sequences of points in Y. Let z_1,z_2\in \bar{Y}, \alpha\in\mathbb{R}. There is a sequence (a_n) in Y such that a_n\to z_1. Similarly there is a sequence (b_n) in Y which converges to z_2.

Then (a_n+b_n) is a sequence in Y that converges to z_1+z_2. (\alpha a_n) is a sequence in Y that converges to \alpha z_1.

Norm of Cartesian Product

If we have two normed linear spaces Z and U, their Cartesian product Z\oplus U can also be normed, such as by setting |(z,u)|=|z|+|u|, |(z,u)|'=\max\{|z|,|u|\}, or |(z,u)|''=(|z|^2+|u|^2)^{1/2}. Note that we are following Lax’s Functional Analysis, where a norm is denoted as |\cdot |, rather than \|\cdot\| which is clearer but more cumbersome to write.

It is routine to check that all the above 3 are norms, satisfying the positivity, subadditivity, and homogeneity axioms. Minkowski’s inequality is useful to prove the subadditivity of the last norm.

We may check that all of the above 3 norms are equivalent. This follows from the inequalities \frac{1}{2}(|z|+|u|)\leq\max\{|z|,|u|\}\leq |z|+|u|, and M=(M^2)^{1/2}\leq (|z|^2+|u|^2)^{1/2}\leq (2M^2)^{1/2}=\sqrt 2 M, where M:=\max\{ |z|,|u|\}. In general, we have that all norms are equivalent in finite dimensional spaces.

{p(x)<1} is a Convex Set

This theorem can be considered a converse of a previous theorem.

Theorem: Let p denote a positive homogenous, subadditive function defined on a linear space X over the reals.

(i) The set of points x satisfying p(x)<1 is a convex subset of X, and 0 is an interior point of it.

(ii) The set of points x satisfying p(x)\leq 1 is a convex subset of X.

Proof: (i) Let K=\{x\in X\mid p(x)<1\}. Let x_1,x_2\in K. For 0\leq\alpha\leq 1,

\begin{aligned}p(\alpha x_1+(1-\alpha)x_2)&\leq \alpha p(x_1)+(1-\alpha)p(x_2)\\    &<\alpha+(1-\alpha)\\    &=1    \end{aligned}

Therefore K is convex. We also have p(0)=0\in K.

The proof of (ii) is similar.

Arzela-Ascoli Theorem and Applications

The Arzela-Ascoli Theorem is a rather formidable-sounding theorem that gives a necessary and sufficient condition for a sequence of real-valued continuous functions on a closed and bounded interval to have a uniformly convergent subsequence.

Statement: Let (f_n) be a uniformly bounded and equicontinuous sequence of real-valued continuous functions defined on a closed and bounded interval [a,b]. Then there exists a subsequence (f_{n_k}) that converges uniformly.

The converse of the Arzela-Ascoli Theorem is also true, in the sense that if every subsequence of (f_n) has a uniformly convergent subsequence, then (f_n) is uniformly bounded and equicontinuous.

Explanation of terms used: A sequence (f_n) of functions on [a,b] is uniformly bounded if there is a number M such that |f_n(x)|\leq M for all f_n and all x\in [a,b]. The sequence is equicontinous if, for all \epsilon>0, there exists \delta>0 such that |f_n(x)-f_n(y)|<\epsilon whenever |x-y|<\delta for all functions f_n in the sequence. The key point here is that a single \delta (depending solely on \epsilon) works for the entire family of functions.


Let g:[0,1]\times [0,1]\to [0,1] be a continuous function and let \{f_n\} be a sequence of functions such that f_n(x)=\begin{cases}0,&0\leq x\leq 1/n\\    \int_0^{x-\frac{1}{n}}g(t,f_n(t))\ dt,&1/n\leq x\leq 1\end{cases}

Prove that there exists a continuous function f:[0,1]\to\mathbb{R} such that f(x)=\int_0^x g(t,f(t))\ dt for all x\in [0,1].

The idea is to use Arzela-Ascoli Theorem. Hence, we need to show that (f_n) is uniformly bounded and equicontinuous.

We have

\begin{aligned}|f_n(x)|&\leq |\int_0^{x-\frac{1}{n}} 1\ dt|\\    &=|x-\frac{1}{n}|\\    &\leq |x|+|\frac{1}{n}|\\    &\leq 1+1\\    &=2    \end{aligned}

This shows that the sequence is uniformly bounded.

If 0\leq x\leq 1/n,

\begin{aligned}|f_n(x)-f_n(y)|&=|0-f_n(y)|\\    &=|\int_0^{y-\frac{1}{n}} g(t,f_n(t))\ dt|\\    &\leq |\int_0^{y-\frac{1}{n}} 1\ dt|\\    &=|y-\frac{1}{n}|\\    &\leq |y-x|    \end{aligned}

Similarly if 0\leq y\leq 1/n, |f_n(x)-f_n(y)|\leq |x-y|.

If 1/n\leq x\leq 1 and 1/n\leq y\leq 1,

\begin{aligned}|f_n(x)-f_n(y)|&=|\int_0^{x-1/n} g(t,f_n(t))\ dt-\int_0^{y-1/n}g(t,f_n(t))\ dt|\\    &=|\int_{y-1/n}^{x-1/n}g(t,f_n(t))\ dt|\\    &\leq |\int_{y-1/n}^{x-1/n} 1\ dt|\\    &=|(x-1/n)-(y-1/n)|\\    &=|x-y|    \end{aligned}

Therefore we may choose \delta=\epsilon, then whenever |x-y|<\delta, |f_n(x)-f_n(y)|\leq |x-y|<\epsilon. Thus the sequence is indeed equicontinuous.

By Arzela-Ascoli Theorem, there exists a subsequence (f_{n_k}) that is uniformly convergent.

f_{n_k}(x)\to f(x)=\int_0^x g(t,f(t))\ dt.

By the Uniform Limit Theorem, f:[0,1]\to\mathbb{R} is continuous since each f_n is continuous.

Function of Bounded Variation that is not continuous

This is a basic example of a function of bounded variation on [0,1] but not continuous on [0,1].

The key Theorem regarding functions of bounded variation is Jordan’s Theorem: A function is of bounded variation on the closed bounded interval [a,b] iff it is the difference of two increasing functions on [a,b].

Consider g(x)=\begin{cases}0&\text{if}\ 0\leq x<1\\  1&\text{if}\ x=1  \end{cases}

h(x)\equiv 0

Both g and h are increasing functions on [0,1]. Thus by Jordan’s Theorem, f(x)=g(x)-h(x)=g(x) is a function of bounded variation, but it is certainly not continuous on [0,1]!

Interpolation Technique in Analysis

Question: Let f belong to both L^{p_1} and L^{p_2}, with 1\leq p_1<p_2<\infty. Show that f\in L^p for all p_1\leq p\leq p_2.

There is a pretty neat trick to do this question, known as the “interpolation technique”. The proof is as follows.

For p_1<p<p_2, there exists 0<\alpha<1 such that \displaystyle\boxed{p=\alpha p_1+(1-\alpha)p_2}. This is the key “interpolation step”. Once we have this, everything flows smoothly with the help of Holder’s inequality.

\displaystyle\begin{aligned}    \int |f|^p\ d\mu&=\int (|f|^{\alpha p_1}\cdot |f|^{(1-\alpha)p_2})\ d\mu\\    &\leq\||f|^{\alpha p_1}\|_\frac{1}{\alpha}\||f|^{(1-\alpha)p_2}\|_\frac{1}{1-\alpha}\\    &=(\int |f|^{p_1}\ d\mu)^\alpha\cdot (\int |f|^{p_2}\ d\mu)^{1-\alpha}\\    &<\infty    \end{aligned}

Thus f\in L^p.

Note that the magical thing about the interpolation technique is that p=\frac{1}{\alpha} and q=\frac{1}{1-\alpha} are Holder conjugates, since \frac{1}{p}+\frac{1}{q}=1 is easily verified.

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