What is the “continuity” as a term in computable analysis?


Background

I once implemented a datatype representing arbitrary real numbers in Haskell. It labels every real numbers by having a Cauchy sequence converging to it. That will let $ \mathbb{R}$ be in the usual topology. I also implemented addition, subtraction, multiplication, and division.

But my teacher said, "This doesn’t seem to be a good idea. Since comparison is undecidable here, this doesn’t look very practical. In particular, letting division by 0 to fall in an infinite loop doesn’t look good."

So I wanted my datatype to extend $ \mathbb{Q}$ . Since equality comparison of $ \mathbb{Q}$ is decidable, $ \mathbb{Q}$ is in discrete topology. That means a topology on $ \mathbb{R}$ must be finer than the discrete topology on $ \mathbb{Q}$ .

But, I think I found that, even if I could implement such datatype, it will be impractical.

Proof, step 1

Let $ \mathbb{R}$ be finer than $ \mathbb{Q}$ in discrete topology. Then $ \{0\}$ is open in $ \mathbb{R}$ . Assume $ + : \mathbb{R}^2 → \mathbb{R}$ is continuous. Then $ \{(x,-x): x \in \mathbb{R}\}$ is open in $ \mathbb{R}^2$ . Since $ \mathbb{R}^2$ is in product topology, $ \{(x,-x)\}$ is a basis element of $ \mathbb{R}^2$ for every $ x \in \mathbb{R}$ . It follows that $ \{x\}$ is a basis element of $ \mathbb{R}$ for every $ x \in \mathbb{R}$ . That is, $ \mathbb{R}$ is in discrete topology.

Proof, step 2

Since $ \mathbb{R}$ is in discrete topology, $ \mathbb{R}$ is computably equality comparable. This is a contradiction, so $ +$ is not continuous, and thus not computable.

Question

What is bugging me is the bolded text. It is well-known that every computable function is continuous (Weihrauch 2000, p. 6). Though the analytic definition and the topological definition of continuity coincide in functions from and to Euclidean spaces, $ \mathbb{R}$ above is not a Euclidean space. So I’m unsure whether my proof is correct. What is the definition of "continuity" in computable analysis?

Continuity of a differential of a Banach-valued holomorphic map

Originally posted on MSE.

Let $ U$ be an open set in $ \mathbb{C}^{n}$ let $ F$ be a Banach space (in my case even a dual Banach space), and let $ \varphi:U\to F$ be a holomorphic map. I seem to be able to prove that the differential map $ D\varphi:U\times\mathbb{C}^{n}\to F$ defined by $ $ D\varphi (z,v)= \lim\limits_{t\to 0}\frac{\varphi(z+tv)-\varphi(z)}{t}$ $ is holomorphic.

Is there a reference for this assertion? (Or at least for continuity)

I tried to look into some sources on infinite-dimensional holomorphicity and could not find such a statement, but some of those sources are rather complicated, and so it is likely I missed it.

Continuity of green functions

I have a technical question on a continuity of green function.

Setting

Let $ E$ be a locally compact separable metric space and $ m$ a locally finite measure on $ E$ .

Let $ X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in E})$ be a transient diffusion process. We assume that there exists a continuous function $ p_{t}(x,y):(0,\infty) \times E \times E \to [0,\infty)$ such that \begin{align*} E_{x}[f(X_t)]:=\int_{E}p_{t}(x,y)\,dm(y),\quad t>0,\ x,y \in E \end{align*}

My question

Then, under what conditions, can we show that $ g(x,y)=\int_{0}^{\infty}p_{t}(x,y)\,dt$ is a continuous function on $ E \times E$ off the diagonal.

Aim

I am interested in the case that $ X$ is a part process of a recurrent process. For example,

  • $ E=\bar{D} \setminus K$ , where $ D$ is a bounded smooth domain of $ \mathbb{R}^2$ and $ K \subset D$ is a closed ball,
  • $ X$ is a part process on $ \bar{D} \setminus K$ of a reflected Brownian motion on $ \bar{D}$ .

In this situation, $ X$ possesses a continuous heat kernel $ p_{t}(x,y)$ . However, I do not even know whether $ x \mapsto g(x,y)$ is locally bounded on $ \bar{D} \setminus \{y\}$ . I also think that upper bounds for $ p_{t}(x,y)$ is not known.

I think that the continuity of green functions is a fundamental problem, but is there a standard proof method?

Continuity of convolution on $\mathcal{D}’_+$

Let $ \mathcal{D}’_+:=\{T\in \mathcal{D}'(\mathbb{R}): \textrm{supp}(T)\subset [0,\infty)\}$ . Here $ \mathcal{D}'(\mathbb{R})$ is the usual space of distributions on $ \mathbb{R}$ , equipped with the weak$ \ast$ -topology induced by $ \mathcal{D}(\mathbb{R})$ , and $ \mathcal{D}_+’$ is given the subspace topology induced from $ \mathcal{D}'(\mathbb{R})$ .

Question: Is convolution $ \ast:\mathcal{D}’_+ \times \mathcal{D}’_+\rightarrow \mathcal{D}’_+$ separately continuous?

Hölder continuity of the Cantor function

The Cantor function (which I will define recursively below) is Hölder continuous and the exponent there is ln(2)/ln(3).

The function is defined as the limit function of a sequence of functions, where $ f_{n}$ looks as below: \begin{cases} f_{n-1}(3x/2) & 0 \leq x\leq 1/3\ 0.5 & 1/3\leq x\leq 2/3 \ 0.5 + f_{n-1}((3x-2)/2) & 2/3\leq x \leq 1 \end{cases}

I’ve proved that $ f_{n}$ converges uniformly to the Cantor function and that $ ||f_{n} – f_{n-1}||_{\infty}$ $ \leq$ $ \frac{1}{2^n}$ , but I have no idea where to proceed from there. Any inputs will be appreciated.

A formal epsilon-delta proof for the Continuity Law for Composition

The continuity law for composition states, informally, that:

$ $ \text{IF } f \text{ is continuous at } g \ \text{ AND }\ g \text{ is continuous at } f(a) \text{ THEN } g(f(a)) \text{ is continuous at } a$ $

Or, using the limit definition of continuity:

$ $ \lim_{x \to a} f(x) = f(a) \wedge \lim_{x \to f(x)} g(x) = g(f(a)) \Longrightarrow \lim_{x \to a} g(f(x)) = g(f(a))$ $

Or, using the formal epsilon-delta definitions:

$ $ \forall \epsilon > 0, \exists \delta_1 > 0, \forall x \in \mathbb{R}, 0<|x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \epsilon \tag{1}\label{1}$ $

$ $ \forall \epsilon > 0, \exists \delta_2 > 0, \forall x \in \mathbb{R}, 0<|x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2}\label{2}$ $

$ $ \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R}, 0<|x – a| < \delta \Longrightarrow |g(f(x)) – g(f(a))| < \epsilon \tag{3}\label{3}$ $

$ $ \eqref{1} \wedge \eqref{2} \Longrightarrow \eqref{3}$ $


Proof:

We assume \eqref{1} and \eqref{2}. We want to show \eqref{3}.

Let $ \epsilon > 0$ .

We will use the same $ \epsilon$ in \eqref{2}, giving us:

$ $ \exists \delta_2 > 0, \forall x \in \mathbb{R}, 0<|x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2a}\label{2a} $ $

Let $ \delta_2 > 0$ such that the following is true:

$ $ \forall x \in \mathbb{R}, 0<|x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2b}\label{2b}$ $

We will substitute $ \delta_2$ for $ \epsilon$ in $ \eqref{1}$ to obtain:

$ $ \exists \delta_1 > 0, \forall x \in \mathbb{R}, 0<|x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1a}\label{1a} $ $

Let $ \delta_1 > 0$ such that the following is true:

$ $ \forall x \in \mathbb{R}, 0<|x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1b}\label{1b}$ $

Let $ x \in \mathbb{R}$ .

We will use the same $ x$ in \eqref{1b} to obtain:

$ $ 0 < |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1c}\label{1c}$ $

Let $ \delta = \delta_1$ .

We assume $ 0 < |x – a| < \delta = \delta_1$ .

By our assumption and \eqref{1c}, we know:

$ $ |f(x) – f(a)| < \delta_2 \tag{1d}\label{1d}$ $

We substitute $ f(x)$ for $ x$ in \eqref{2b} to obtain:

$ $ 0<|f(x) – f(a)| < \delta_2 \Longrightarrow |g(f(x)) – g(f(a))| < \epsilon \tag{2c}\label{2c}$ $


And here is where I’m stuck. All I need to show is $ 0 < |f(x) – f(a)| $ , and I will be able to complete the proof. But I don’t know how to show that. Any ideas?

Continuity of The Restriction Map Between Function Spaces

Let $ X,Y,Z$ be Hausdorff spaces and suppose that $ Z\subset X$ . Endow $ C(X,Y)$ and $ C(Z,Y)$ with the compact-open topologies and define the map $ \rho$ as \begin{align} \rho:&C(X,Y)\rightarrow C(Z,Y)\ &f\mapsto f|_Z. \end{align}

Is the map $ \rho$ continuous?

I see this “type of” operation used all the time in Sheaf theory but I never stopped to wonder if it was indeed continuous?

Uniform continuity of the function $f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $E=[1,+ \infty)$

I have the function $ f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $ E=[1,+ \infty)$ and I have to discuss the uniform continuity of f in E.

I’ve calculated the derivative $ y’$ and it tends to $ 0$ as $ x$ tends to $ \infty$

Can this fact be used to prove that $ f(x)$ is uniformly continuous in $ E$ ?

Characterising functions of bounded variation by their modulus of continuity

Given a a.e. finite measurable function $ \mathbb R^n \to \mathbb R$ , define the essential modulus of continuity, $ M(f): \ \mathbb R^n \times \mathbb R+ \to \mathbb R$ by

$ M(f) (x, e)$ $ =$ $ sup_{m(A) = 0} [d \geq 0| f[B_d (x) – A] \in [f(x) – e, f(x) + e]]$

It can be shown (details here: https://justpaste.it/37sij) that $ M$ is well defined a.e. in the product measure and does not depend on choice of representative.

Can we characterise functions of bounded variation by the behaviour of $ M$ ? That is, if $ S$ is the set of measurable functions, find some subset $ V$ of $ M(S)$ such that a function has bounded variation iff it is in $ V$ .