Category Archives: Analysis

Kuratowski Closure and Topological Spaces

Analysis,

This post was the result of conversations with Chris Gartland (from several years ago). In what follows we outline an alternate definition of a topological space due to Kuratowski.

We recall Kuratowski’s closure axioms for a set {X}. These are a set of axioms which can be used to define a topology on a set. Here, one replaces the notion of open (or closed) sets with a closure operation. The closure operation has the advantage of being defined on all subsets of {X}. For{A \subset X}, one may think of the closure of {A}, denoted {\overline{A}}, to be points that are “close to” {A}.

First we recall what is meant by a topological space. We let {\mathcal{P}(X)} be the power set of {X}, that is the set of all subsets of {X}.

{\ } Definition 1: Let {X} be a set and {T \subset \mathcal{P}(X)}. Then {(X,\mathcal{T})} is a topology with open sets {\mathcal{T}} if

  • (a) {\emptyset , X \in \mathcal{T}},
  • (b) if {A, B \in \mathcal{T}}, then {A \cap B \in \mathcal{T}},
  • (c) if {A_i \in \mathcal{T}} for all {i \in \mathcal{I}}, then {\cup_{i\in \mathcal{I}} A_i \in \mathcal{T}}.

Furthermore, we say

\displaystyle \mathcal{C} := \{X \setminus A : A \in \mathcal{T}\},

are the closed sets of {X}. {\spadesuit} {\ }

Note we could have defined (a), (b), (c) in terms of closed sets, utilizing DeMorgan’s law. We make use of this implicitly below. We now present the definition of Kuratowski closure.

{\ } Definition 2: Let {X} be a set and {\overline{\cdot} : \mathcal{P}(X) \rightarrow \mathcal{P}(X)}. We say {\overline{\cdot}} is a closure operation if

  • (i) {\overline{\emptyset} = \emptyset},
  • (ii) if {A \subset X}, then {A \subset \overline{A}},
  • (iii) if {D = \overline{A}}, then {\overline{D} = D},
  • (iv) if {A,B \subset X}, then {\overline{A\cup B} = \overline{A} \cup \overline{B}}. {\spadesuit}

{\ }

As mentioned above, one can informally think of {\overline{A}} as the set of points in {X} which are “close” to {A}. We now explain that these two definitions are equivalent. To define a closure operation from Definition 1, we set

\displaystyle \overline{A} := \bigcap_{A \subset C \ \text{closed}} C.\ \ \ \ \ (1)

Note that {\overline{A}} is closed by (c). Given a closure operation, we may define the closed sets to be

\displaystyle \mathcal{C} : = \{\overline{A} : A \subset X\}.\ \ \ \ \ (2)

{\ } Proposition 1: Let {(X, \mathcal{T})} be a topological space as in Definition 1. Then the operation defined in (1) is a closure operation in the sense of Definition 2. {\spadesuit} {\ }

Proof: Let

\displaystyle \mathcal{C} = \{X \setminus A : A \in \mathcal{T}\},

be the set of closed sets in {X}. Let {A , B \subset X}. By (1), we have {A \subset \overline{A}} and thus (ii) holds. As {X} is open, {\emptyset \in \mathcal{C}} and so (i) is satisfied. Note for any closed set, {C \in \mathcal{C}}, we have {\overline{C} = C} by (ii) and that {C} itself appears on the right hand side of (1). By (c), we have that

\displaystyle D = \bigcap_{A \subset C \ \text{closed}} C \in \mathcal{C},

and so {\overline{D} = D}. This proves (iii) and so it remains to show (iv). Note that {\overline{A} \cup \overline{B}} is a closed set containing {A \cup B}, by (b). Thus

\displaystyle \overline{A \cup B} \subset \overline{A} \cup \overline{B}.\ \ \ \ \ (3)

Now let {C \in \mathcal{C}} that contains {A \cup B}. Then {A, B \subset C} and so

\displaystyle A , B \subset \ \bigcap_{(A\cup B) \subset C \ \text{closed}} C.

This proves the reverse inclusion of (3) and thus of Proposition 1. {\spadesuit}

We now show that a closure axiom can be used to define a topology.

{\ } Proposition 2: Let {\overline{\cdot} : \mathcal{P}(X) \rightarrow \mathcal{P}(X)} be a closure axiom as in Definition 2. Then the sets defined in (2) form the closed sets of a topology in the sense of Definition 1. {\spadesuit} {\ }

Proof: We implicitly make use of DeMorgan’s law to transition bewtween closed and open sets. By {(i)}, we have {\emptyset \in \mathcal{C}} and by (ii), we have {X \in \mathcal{C}}. Thus {(a)} is established. Let {A, B \in \mathcal{C}}. By (iii), (iv) and (2), we have

\displaystyle A \cup B = \overline{A \cup B} \in \mathcal{C},

which establishes (b). Let {C_i \in \mathcal{C}} for all {i \in \mathcal{I}} some index set. Then we have

\displaystyle \overline{\cap_{i \in \mathcal{I}} C_i}= \cap_{i \in \mathcal{I}} C_i.

Indeed, the backwards inclusion follows from (ii). To see the forward direction, it is enough to show for any {i \in \mathcal{I}},

\displaystyle \overline{\cap_{i \in \mathcal{I}} C_i} \subset \overline{C_i} = C_i.

The second equality is (iii). The first subset inequality follows from (iv) as if {A \subset B} then

\displaystyle \overline{A} \subset \overline{A} \cup \overline{B} = \overline{A \cup B} = \overline{B}.

We apply this with {A = \cup_{i\in \mathcal{I}} C_i} and {B = C_i}. {\spadesuit}

Recall if {(X , \mathcal{T}_X)} and {(Y , \mathcal{T}_Y)} are topological spaces, then

\displaystyle f : X \rightarrow Y,

is said to be continuous if the pre-image of every open set in {Y} is open in {X}, that is

\displaystyle f^{-1}(B) \in \mathcal{T}_X,\ \ \ \ \ (4)

for all open {Y}. As {f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)}, (5) is equivalent to the analogous definition for closed sets. It turns out that we may equivalently define a map to be continuous if

\displaystyle f(\overline{A}) \subset \overline{f(A)}.\ \ \ \ \ (5)

This can be informally interpreted as points that are close to {A} are mapped to points that are close to {f(A)}. We now prove this.

{\ } Proposition 3: Let {f : X \rightarrow Y} be a function. Then {f} is continuous in the sense of (4) if and only if it is continuous in the sense of (5). {\spadesuit} {\ }

Proof: We start with the forward direction. Let {A \subset X}. Then by (4), {f^{-1}(\overline{f(A)})} is closed. As {f(A) \subset \overline{f(A)}}, we have

\displaystyle A \subset f^{-1}(\overline{f(A)}).

Since the right hand side is closed, by (ii) and (iii), we have

\displaystyle \overline{A} \subset f^{-1}(\overline{f(A)}).

Applying {f} to both sides establishes (5).

Now we show the reverse implication. Let {B} be a closed set. Then by (5),

\displaystyle f(\overline{f^{-1}(B)}) \subset \overline{f(f^{-1}(B))} =\overline{B} = B .

As any element which is mapped to {B} by {f} must lie in {f^{-1}(B)}, we have

\displaystyle \overline{f^{-1}(B)} \subset f^{-1}(B).

By (ii), equality holds and thus {f^{-1}(B)} is closed. {\spadesuit}

Hölder’s Inequality

Analysis,

Let {(\Omega,\mu)} be a {\sigma}–finite measure space and {f,g : \Omega \rightarrow \mathbb{C}} be measurable functions. Hölder’s inequality asserts for {p,q \in [1, \infty]}, {p^{-1} + q^{-1} = 1} one has

\displaystyle \int |fg| \leq ||f||_{p} ||g||_{q}.

One standard proof is to integrate the AM-GM inequality

\displaystyle \frac{|f(x)||g(x)|}{||f||_{p} ||g||_q} \leq \frac{|f(x)|^p}{p||f||_p^p} + \frac{|g(x)|}{q ||g||_q^q}.

Here, I provide an alternative proof. First consider the case where {g = \chi_E} is the characteristic function of a set {E} of finite measure. Let

\displaystyle f_{<M} = f(x) 1_{f(x) < M} , \ \ \ \ f_{\geq M}(x) = f(x) 1_{f(x) \geq M} .

Then

\displaystyle \int |f| \chi_E = \int |f_{<M} |\chi_E + \int |f_{\geq M} |\chi_E \leq M \mu(E) + \frac{1}{M^{p-1}} \int |f|^p.

By choosing the optimal value for {M= \frac{||f||_p}{\mu(E)^{\frac{1}{p}}}}, we obtain

\displaystyle \int |f| \chi_E \leq 2 \mu(E)^{1-\frac{1}{p}} ||f||_p.\ \ \ \ \ (1)

Remark: Note that if {|f|} is bounded away from {\lambda}, we may improve the above argument. For instance, suppose {f(x) \notin [M^{-1} \lambda , M^{\frac{1}{p-1}} \lambda]} for some {M\geq 1} and for all {x \in \Omega}. Then

\displaystyle \int |f| \chi_E \leq 2 M^{-1} \mu(E)^{1-\frac{1}{p}} ||f||_p.

♠

We can get rid of the constant 2 in (1) by using the tensor product trick. That is, apply (1) to

\displaystyle F: \Omega^n \rightarrow \mathbb{C} , \ \ \ \ \ \ \ F(x_1 , \ldots , x_n) = f(x_1) \cdots f(x_n).

It follows that

\displaystyle \left(\int |f| \chi_E \right)^n = \int |F|\chi_{E^n} \leq \mu(E^n)^{1-\frac{1}{p}} ||F||_p = 2 \left(\mu(E)^{1-\frac{1}{p}} ||f||_p\right)^n.

Taking {n^{\rm th}} roots and letting {n \rightarrow \infty} establishes Hölder’s inequality in the case {g = \chi_E}.

Remark: In many applications, (1) is enough. Nevertheless, extending to simple functions is an easy matter and Hölder’s inequality follows by the density of simple functions in {L^p}. ♠

Getting a lower bound for an L1 norm using higher moments

Analysis,

I would like to discuss a principle that came up in this recent talk of Adam Harper as well as my own research with Burak Erdogan.

Let f : \Omega \to \mathbb{C} be a function on some measure space with measure \mu (for instance \Omega = [0,1] or a finite set).

Often one is interested in finding lower bounds for the L^1 norm of f, that is \int_{\Omega} |f| d\mu, but has no way to directly estimate it. As a toy example, we can consider g_N: \mathbb{R} / \mathbb{Z} \to \mathbb{C} via g_N(x) = \sum_{1 \leq n \leq N} e(x n^2). Estimating the L^1 norm directly seems hard.

But sometimes, we are able to estimate higher L^p norms of f. This is useful to our original problem, since an application of Holder’s inequality reveals that a lower bound on the L^2 norm and an upper bound on the L^4 norm gives a lower bound on the L^1. To see this, note \int |f|^2 \leq (\int |f|^4 )^{1/3} (\int |f|)^{2/3}.

We can apply this idea to our original example. Parseval’s identity gives that \int |g_N|^2 = N, while orthogonality and the divisor bound give that \int |g_N|^4 \lesssim_{\epsilon} N^{2 + \epsilon}. This gives \int |f| \gtrsim_{\epsilon} N^{1/2 - \epsilon} and is expected by the heuristic that a typical exponential sum should be about square root of the length of the sum.

My intuition is the following. Suppose the measure space is a probability space. Then \int |f| \leq (\int |f|^2)^{1/2}. We are basically trying to reverse this inequality. Equality holds when f is constant, that is f is not too concentrated. The upper bound on the L^4 norm of f implies that indeed f is not too concentrated.

We mention that there is nothing too special about the exponents 2 and 4 chosen for the above discussion (although they are convenient for the specific example I chose).

Convergence near the boundary

Analysis

DiscApproxId

In the above pdf, I consider in a unified manner the question of convergence of a power series on the boundary of their radius of convergence, and of Dirichlet series on the boundary of their abscissa of convergence. I end with an application to L(1,\chi), where \chi is the nontrivial multiplicative character modulo 4.

This was motivated by these notes of Keith Conrad and a discussion with Kyle Pratt.