Curse of Knowledge

You may have known what it is...

Let us say you are a programmer who has been working in big companies for a decade. How does it feel when you want to help someone who starts studying programming from scratch? You may find it makes no sense that he or she cannot understand that, by copying several lines of code on the book, they has successfully made a programme printing "Hello, world!" on the screen. You know what I am talking about - the curse of knowledge.

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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 \(t\).

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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.

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(Kind of) Missing Content in Your Linear Algebra Class (Still on Progress)

I think it's quite often that, when you are learning mathematics beyond linear algebra, you are stuck at some linear algebra problems, but you haven't learnt that systematically before although you wish you had. In this blog post we will go through some content that is not universally taught but quite often used in further mathematics. But this blog post does not serve as a piece of textbook. If you find some interesting topics, you know what document you should read later, and study it later.

This post is still on progress, neither is it finished nor polished properly. For the coming days there will be new contents, untill this line is deleted. What I'm planning to add at this moment:

  • Transpose is not just about changing indices of its components.
  • Norm and topology in vector spaces
  • Representing groups using matrices
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Dedekind Domain and Properties in an Elementary Approach

You can find contents about Dedekind domain (or Dedekind ring) in almost all algebraic number theory books. But many properties can be proved inside ring theory. I hope you can find the solution you need in this post, and this post will not go further than elementary ring theory. With that being said, you are assumed to have enough knowledge of ring and ring of fractions (this post serves well), but not too much mathematics maturity is assumed (at the very least you are assumed to be familiar with terminologies in the linked post).\(\def\mb{\mathbb}\) \(\def\mfk{\mathfrak}\)


There are several ways to define Dedekind domain since there are several equivalent statements of it. We will start from the one based on ring of fractions. As a friendly reminder, \(\mb{Z}\) or any principal integral domain is already a Dedekind domain. In fact Dedekind domain may be viewed as a generalization of principal integral domain.

Let \(\mfk{o}\) be an integral domain (a.k.a. entire ring), and \(K\) be its quotient field. A Dedekind domain is an integral domain \(\mfk{o}\) such that the fractional ideals form a group under multiplication. Let's have a breakdown. By a fractional ideal \(\mfk{a}\) we mean a nontrivial additive subgroup of \(K\) such that

  • \(\mfk{o}\mfk{a}=\mfk{a}\),
  • there exists some nonzero element \(c \in \mfk{o}\) such that \(c\mfk{a} \subset \mfk{o}\).

What does the group look like? As you may guess, the unit element is \(\mfk{o}\). For a fractional ideal \(\mfk{a}\), we have the inverse to be another fractional ideal \(\mfk{b}\) such that \(\mfk{ab}=\mfk{ba}=\mfk{o}\). Note we regard \(\mfk{o}\) as a subring of \(K\). For \(a \in \mfk{o}\), we treat it as \(a/1 \in K\). This makes sense because the map \(i:a \mapsto a/1\) is injective. For the existence of \(c\), you may consider it as a restriction that the 'denominator' is bounded. Alternatively, we say that fractional ideal of \(K\) is a finitely generated \(\mfk{o}\)-submodule of \(K\). But in this post it is not assumed that you have learned module theory.

Let's take \(\mb{Z}\) as an example. The quotient field of \(\mb{Z}\) is \(\mb{Q}\). We have a fractional ideal \(P\) where all elements are of the type \(\frac{np}{2}\) with \(p\) prime and \(n \in \mb{Z}\). Then indeed we have \(\mb{Z}P=P\). On the other hand, take \(2 \in \mb{Z}\), we have \(2P \subset \mb{Z}\). For its inverse we can take a fractional ideal \(Q\) where all elements are of the type \(\frac{2n}{p}\). As proved in algebraic number theory, the ring of algebraic integers in a number field is a Dedekind domain.

Before we go on we need to clarify the definition of ideal multiplication. Let \(\mfk{a}\) and \(\mfk{b}\) be two ideals, we define \(\mfk{ab}\) to be the set of all sums \[ x_1y_1+\cdots+x_ny_n \] where \(x_i \in \mfk{a}\) and \(y_i \in \mfk{b}\). Here the number \(n\) means finite but is not fixed. Alternatively we cay say \(\mfk{ab}\) contains all finite sum of products of \(\mfk{a}\) and \(\mfk{b}\).


(Proposition 1) A Dedekind domain \(\mfk{o}\) is Noetherian.

By Noetherian ring we mean that every ideal in a ring is finitely generated. Precisely, we will prove that for every ideal \(\mfk{a} \subset \mfk{o}\) there are \(a_1,a_2,\cdots,a_n \in \mfk{a}\) such that, for every \(r \in \mfk{a}\), we have an expression \[ r = c_1a_1 + c_2a_2 + \cdots + c_na_n \qquad c_1,c_2,\cdots,c_n \in \mfk{o}. \] Also note that any ideal \(\mfk{a} \subset \mfk{o}\) can be viewed as a fractional ideal.

Proof. Since \(\mfk{a}\) is an ideal of \(\mfk{o}\), let \(K\) be the quotient field of \(\mfk{o}\), we see since \(\mfk{oa}=\mfk{a}\), we may also view \(\mfk{a}\) as a fractional ideal. Since \(\mfk{o}\) is a Dedekind domain, and fractional ideals of \(\mfk{a}\) is a group, there is an fractional ideal \(\mfk{b}\) such that \(\mfk{ab}=\mfk{ba}=\mfk{o}\). Since \(1 \in \mfk{o}\), we may say that there exists some \(a_1,a_2,\cdots, a_n \in \mfk{a}\) and \(b_1,b_2,\cdots,b_n \in \mfk{o}\) such that \(\sum_{i = 1 }^{n}a_ib_i=1\). For any \(r \in \mfk{a}\), we have an expression \[ r = rb_1a_1+rb_2a_2+\cdots+rb_na_n. \] On the other hand, any element of the form \(c_1a_1+c_2a_2+\cdots+c_na_n\), by definition, is an element of \(\mfk{a}\). \(\blacksquare\)

From now on, the inverse of an fractional ideal \(\mfk{a}\) will be written like \(\mfk{a}^{-1}\).

(Proposition 2) For ideals \(\mfk{a},\mfk{b} \subset \mfk{o}\), \(\mfk{b}\subset\mfk{a}\) if and only if there exists some \(\mfk{c}\) such that \(\mfk{ac}=\mfk{b}\) (or we simply say \(\mfk{a}|\mfk{b}\))

Proof. If \(\mfk{b}=\mfk{ac}\), simply note that \(\mfk{ac} \subset \mfk{a} \cap \mfk{c} \subset \mfk{a}\). For the converse, suppose that \(a \supset \mfk{b}\), then \(\mfk{c}=\mfk{a}^{-1}\mfk{b}\) is an ideal of \(\mfk{o}\) since \(\mfk{c}=\mfk{a}^{-1}\mfk{b} \subset \mfk{a}^{-1}\mfk{a}=\mfk{o}\), hence we may write \(\mfk{b}=\mfk{a}\mfk{c}\). \(\blacksquare\)

(Proposition 3) If \(\mfk{a}\) is an ideal of \(\mfk{o}\), then there are prime ideals \(\mfk{p}_1,\mfk{p}_2,\cdots,\mfk{p}_n\) such that \[ \mfk{a}=\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_n. \]

Proof. For this problem we use a classical technique: contradiction on maximality. Suppose this is not true, let \(\mfk{A}\) be the set of ideals of \(\mfk{o}\) that cannot be written as the product of prime ideals. By assumption \(\mfk{U}\) is nonempty. Since as we have proved, \(\mfk{o}\) is Noetherian, we can pick an maximal element \(\mfk{a}\) of \(\mfk{A}\) with respect to inclusion. If \(\mfk{a}\) is maximal, then since all maximal ideals are prime, \(\mfk{a}\) itself is prime as well. If \(\mfk{a}\) is properly contained in an ideal \(\mfk{m}\), then we write \(\mfk{a}=\mfk{m}\mfk{m}^{-1}\mfk{a}\). We have \(\mfk{m}^{-1}\mfk{a} \supsetneq \mfk{a}\) since if not, we have \(\mfk{a}=\mfk{ma}\), which implies \(\mfk{m}=\mfk{o}\). But by maximality, \(\mfk{m}^{-1}\mfk{a}\not\in\mfk{U}\), hence it can be written as a product of prime ideals. But \(\mfk{m}\) is prime as well, we have a prime factorization for \(\mfk{a}\), contradicting the definition of \(\mfk{U}\).

Next we show uniqueness up to permutation. If \[ \mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k=\mfk{q}_1\mfk{q}_2\cdots\mfk{q}_j, \] since \(\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k\subset\mfk{p}_1\) and \(\mfk{p}_1\) is prime, we may assume that \(\mfk{q}_1 \subset \mfk{p}_1\). By the property of fractional ideal we have \(\mfk{q}_1=\mfk{p}_1\mfk{r}_1\) for some fractional ideal \(\mfk{r}_1\). However we also have \(\mfk{q}_1 \subset \mfk{r}_1\). Since \(\mfk{q}_1\) is prime, we either have \(\mfk{q}_1 \supset \mfk{p}_1\) or \(\mfk{q}_1 \supset \mfk{r}_1\). In the former case we get \(\mfk{p}_1=\mfk{q}_1\), and we finish the proof by continuing inductively. In the latter case we have \(\mfk{r}_1=\mfk{q}_1=\mfk{p}_1\mfk{q}_1\), which shows that \(\mfk{p}_1=\mfk{o}\), which is impossible. \(\blacksquare\)

(Proposition 4) Every nontrivial prime ideal \(\mfk{p}\) is maximal.

Proof. Let \(\mfk{m}\) be an maximal ideal containing \(\mfk{p}\). By proposition 2 we have some \(\mfk{c}\) such that \(\mfk{p}=\mfk{mc}\). If \(\mfk{m} \neq \mfk{p}\), then \(\mfk{c} \neq \mfk{o}\), and we may write \(\mfk{c}=\mfk{p}_1\cdots\mfk{p}_n\), hence \(\mfk{p}=\mfk{m}\mfk{p}_1\cdots\mfk{p}_n\), which is a prime factorisation, contradicting the fact that \(\mfk{p}\) has a unique prime factorisation, which is \(\mfk{p}\) itself. Hence any maximal ideal containing \(\mfk{p}\) is \(\mfk{p}\) itself. \(\blacksquare\)

(Proposition 5) Suppose the Dedekind domain \(\mfk{o}\) only contains one prime (and maximal) ideal \(\mfk{p}\), let \(t \in \mfk{p}\) and \(t \not\in \mfk{p}^2\), then \(\mfk{p}\) is generated by \(t\).

Proof. Let \(\mfk{t}\) be the ideal generated by \(t\). By proposition 3 we have a factorisation \[ \mfk{t}=\mfk{p}^n \] for some \(n\) since \(\mfk{o}\) contains only one prime ideal. According to proposition 2, if \(n \geq 3\), we write \(\mfk{p}^n=\mfk{p}^2\mfk{p}^{n-2}\), we see \(\mfk{p}^2 \supset \mfk{p}^n\). But this is impossible since if so we have \(t \in \mfk{p}^n \subset \mfk{p}^2\) contradicting our assumption. Hence \(0<n<3\). But If \(n=2\) we have \(t \in \mfk{p}^2\) which is also not possible. So \(\mfk{t}=\mfk{p}\) provided that such \(t\) exists.

For the existence of \(t\), note if not, then for all \(t \in \mfk{p}\) we have \(t \in \mfk{p}^2\), hence \(\mfk{p} \subset \mfk{p}^2\). On the other hand we already have \(\mfk{p}^2 = \mfk{p}\mfk{p}\), which implies that \(\mfk{p}^2 \subset \mfk{p}\) (proposition 2), hence \(\mfk{p}^2=\mfk{p}\), contradicting proposition 3. Hence such \(t\) exists and our proof is finished. \(\blacksquare\)

Characterisation of Dedekind domain

In fact there is another equivalent definition of Dedekind domain:

A domain \(\mfk{o}\) is Dedekind if and only if

  • \(\mfk{o}\) is Noetherian.
  • \(\mfk{o}\) is integrally closed.
  • \(\mfk{o}\)​ has Krull dimension \(1\)​ (i.e. every non-zero prime ideals are maximal).

This is equivalent to say that faction ideals form a group and is frequently used by mathematicians as well. But we need some more advanced techniques to establish the equivalence. Presumably there will be a post about this in the future.

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 \[ \begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned} \]