# 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?

## Continuity of the restriction of a function to a set

Let $$X$$ be a topological space and $$f$$ a real-valued function defined on $$X$$. Let $$S\subset X$$ and suppose $$f$$ is continuous with respect to the induced topology on $$S$$ (opens are in the form $$S\cap V$$, where $$V$$ is an open of $$X$$). Is this the same as saying that the restriction of $$f$$ to $$S$$ is 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 $$V$$.