New answers tagged

0

Considering the $6n$ evenly spaces points on the circle, call a point exterior if it is one of the chosen points which splits the circle and interior if it is in the interior of one of the arcs. Let the dual splitting of the circle be the one generated by changing every interior point to exterior and vice versa. If no points are diametrically opposite, the ...


6

If $f$ is differentiable and $f(x_1) = f(x_2) = f(x_3) = 0$ with $x_1 < x_2 < x_3$ then you can apply the mean-value theorem (or Rolle's theorem) to both intervals $[x_1,x_2]$ and $[x_2, x_3]$. It follows that $f'$ has a root in each of the open intervals $(x_1, x_2)$ and $(x_2, x_3)$. That makes (at least) two roots of the derivative. In the same ...


1

The given function can also be written as $\int_0^{x}\frac {f(x-s)} {\sqrt s} ds$ (by the substitution $t=x-s$). Let $x_n \to x$. Then $I_{(0,x_n)}(s) \frac {f(x_n-s)} {\sqrt s}\to I_{(0,x)}(s) \frac {f(x-s)} {\sqrt s} $ for every $s$ except $s=x$. Also, $I_{(0,x_n)}(s) \frac {f(x_n-s)} {\sqrt s}$ is dominated by $\frac M {\sqrt s}$ where $M=\sup \{|f(s)...


2

Assume otherwise: On a circle of length $6n$, we choose $3n$ points such that they split the circle into $n$ arcs of length $1$, $n$ arcs of length $2$, $n$ arcs of length $3$; and no two points are diametrically opposite. Let's remove every arc of length $1$ from the circle, and also remove its opposite segment, which is the middle of an arc of length $...


4

Notice that taking the limit as $r \to 0$, like you did, only gets near zero through paths that are straight lines. This is not enough to ensure differentiability and this example is to show why. As pointed out in the comments, we have that $$ \frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta - \sin \theta} $$ is not bounded, when you approach zero through ...


3

Your mistake is in falsely assuming that $g(\theta) = \dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded. It has a discontinuity at $\theta = \pi/4$ so it is not bounded. Hence you cannot conclude $ \lim\limits_{r\to 0}f(r) g(\theta)=0$


1

For Dirichlet function, do you understand that $\mathbb{Q}$ is dense in $\mathbb{R}$? If so, you will find the function is discontinuous for every point just by $\epsilon-\delta$ approach. For the second function, you can just pick any $\epsilon$, then pick $\delta=\epsilon$, for $x\in(-\delta,\delta)$, if $x$ is rational, $f(x)=0<\epsilon$, if $x$ is ...


1

First, let us look at the $\epsilon-\delta$ definition of continuity. If $f$ is a real valued function, then it is said to be continuous at a point $x_0 \in \mathbb{R}$ if $$\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \forall x \in \mathbb{R} \text{ with } \left| x - x_0 \right| < \delta, \text{ we have } \left| f \left( x \right) - ...


1

Remember what epsilon-delta continuity is: anytime someone gives an epsilon, you have to respond with a delta that works if you want to prove your function is continuous. Sometimes the strategy can feel a little contrived, but imagining this sort of game can be a real help. When dealing with strange functions, think about the "special features" of your ...


4

Okay, a little late, but I just figured how to go about it myself so I'll post an answer for completeness: $|\frac{x^2 - \pi^2}{\sqrt{x^2 + 5} + \sqrt{\pi^2 + 5}}| <|\frac{x^2 - \pi^2}{x + \pi}| < |x - \pi| < \delta$ and choosing $\delta = \epsilon$ completes the proof!


2

You know the denominator is always greater than or equal to $\sqrt{\pi^2+5}$, so observe $\big|\frac{x^2-\pi^2}{\sqrt{x^2+5}+\sqrt{\pi^2+5}}\big|\le\big|\frac{x^2-\pi^2}{\sqrt{\pi^2+5^2}}\big|$. Then choose $\delta>0$ such that $|x^2-\pi^2|<\epsilon\sqrt{\pi^2+5^2}$.


0

I assume your $\overline y_n$ is given by $\overline y_n=(Y_1+\cdots Y_n)/n$. By the continuous mapping theorem (which can be considered Slutsky's theorem's big sister) if $\overline y_n$ converges in probability to a constant $c$, then $g(\overline y_n)$ converges to $c^2$. But without further info about the distribution of the $Y_i$ one does not know if ...


3

Suppose no such $x_n $ exists for some $n \in\mathbb N$. Then $f (x)\leq M-\frac1n $ for all $x\in [a,b] $, contradicting the assumption that $\sup A=M $.


3

The whole argument is actually contained in the proof. Since $M - \frac{1}{n}$ is not an upper bound, there must be an element larger than that, say $f(x_n)$. On the other hand, $f(x_n) \leq M$ because $M$ is an upper bound.


3

I wasn't able to get access to Scott's paper, so I instead adapted Mandelkern's proof of TT below. One direction of UT is easy, so we only have to prove that if $A \subset X$ is such that any two completely separated sets in $A$ are completely separated in $X$, then $A$ is $C^*$-embedded in $X$ -- it suffices to show that any continuous $f : A \to [0, 1]$ ...


1

By definition you look at continuity at points in domain hence $x$ approaches some constant and not $\infty$ hence we have $xy^2\sin(\frac1 y) \to 0$ as $y\to 0$ paired with the fact that composition and product of continuous functions is continuous you get that the function is continuous.


2

Càdlàg functions are everywhere right-continuous, and have left limits everywhere. Let $$ A = \{ t \mid \text{$y$ is continuous at $t$}\} $$ For all $t \in A$ we have $y(t-) = y(t)$, and consequently $x(t) = y(t)$. In Cardinality of set of discontinuities of cadlag functions it is shown that càdlàg functions have at most countably many discontinuities, ...


1

other Let $$x_n=1+\frac 1n, y_n=2+\frac 1n.$$ we have $$f(x_n,y_n)=4+2(\frac 3n +\frac{1}{n^2}),$$ $$(x_n,y_n)\to (1,2),$$ $$f(x_n,y_n)\to 4$$ and $$f(1,2)=0\ne 4$$ then $f$ is not ...


3

In problems like these it is useful to approach the point in question via different directions. In many cases, if there is a discontinuity, it will emerge in this way. Here, for example, if we look at the line $y=2x$, and take a sequence of points along this line tending to the point $(1,2)$, we find that the value of $f(x,y)$ along this line is $2x(2x)=4x^2$...


0

You mean « invertible » instead of « inventively », right? If so, consider the mapping $\psi=Id: \mathscr{S} \rightarrow \mathscr{S}$, where $\mathscr{S}$ is the space of Schwartz functions on $\mathbb{R}$ (smooth functions of which all derivatives vanish at infinity faster than any polynomial growth). The starting space is endowed with the norm $\|f\|_s=...


1

Hint: using $\sin(A-B)=\sin \, A \cos\, B- \cos \, A \sin \, B%$ and the formulas $\cos^{2}(x)=\frac {1+\cos(2x)} 2$, $\sin^{2}(x)=\frac {1-\cos(2x)} 2$ you can see that if a solution exists it must be of the form $f(x)=x+A\sin (4\pi x)+B \cos (4 \pi x)$. [ This is a Fourier series expansion where $\cos (2\pi nx)$ and $\sin (2\pi mx)$ terms are non-zero only ...


2

No. More generally, for any regular space $X$, the continuous maps $\mathbb{R}_K\to X$ and $\mathbb{R}_u\to X$ are the same. To prove this, suppose $f:\mathbb{R}_K\to X$ is continuous and let $U\subseteq X$ be open. Since $f^{-1}(U)$ is open in $\mathbb{R}_K$, it contains an open interval around all its points except for possibly $0$, so we just have to ...


4

First of all, your definition of "conjugate" is highly unnatural because it requires $\varphi:\mathbb{R}\to\mathbb{R}$ instead of $\varphi:X\to X$. Note also that by your definition, multiplication on $\mathbb{R}^+$ is not conjugate to addition on $\mathbb{R}$, since you have only defined what it means for two operations on the same set to be conjugate. ...


4

Consider the function $$\color{#df0000}{\boxed{f : \Bbb R^+ \times \Bbb R^+ \to \Bbb R^+, \qquad f(x, y) := x + y + 1}} .$$


0

The definition of limit does not make sense in isolated points because in that case any real number would be the limit of the function at that point. Recall the definition of limit: Let $f: A \longrightarrow \mathbb{R}$ be a real function and $a$ be an adherent point of A. We say $f$ has limit $\lambda$ as $x$ approaches $a$ if for every real $\varepsilon ...


3

What I proved in that answer was that, if $f$ was discontinuous at some point $x$, then there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of $X$ converging to $x$ such that the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ is not convergent. But we are assuming that no such sequence exists.


0

How you organize the proof depends on how your definitions are given. As some other answers have noted, if you define a continuous function as one under which the preimage of every closed set is closed, then the proof is immediate, but that's not the definition that you're using, which makes the proof a little more involved. So let me first clarify the ...


1

I would split up the analysis into cases. Case 1: If $y=0$, then $f(x,y)=0$ by definition. Case 2: If $x=0$ and $y\neq 0$, then $$\Big|f(x,y)\Big|=\lim_{(x,y) \to (0,y)} \Big|x\sin\frac{1}{y}\Big|\leq \lim_{(x,y) \to (0,y)}\Big|x\Big|$$ since $|\sin\frac{1}{y}|\leq 1$. In this case, $f(x,y)\to 0$ since $x=0$. Case 3: If $x\neq 0$ and $y\neq 0$, then $$...


1

It is true that The sum of a finite number of continuous functions is a continuous function. The product of a finite number of continuous functions is a continuous function. To prove these, assume that $f(x)$ and $g(x)$ are continuous functions at a point $a$. Then, $$\lim_{x\to a}f(x)=f(a)$$ $$\lim_{x\to a}g(x)=g(a)$$ So, $$\lim_{x\to a}f(x)\cdot g(x)=...


2

Here's another approach. Since $\exp x:=\sum_{n\ge0}\frac{x^n}{n!}$ converges for all $x$ by the ratio test, and you can show $\exp x\exp h=\exp (x+h)$ with the binomial theorem, we need only show continuity at $0$, since if $|h|<\delta\implies|\exp h-1|<\frac{\epsilon}{\exp x}$ then $|h|<\delta\implies|\exp(x+h)-\exp x|<\epsilon$. By the ...


1

Another prove different from the prove of Hamam can be to prove that $e^x$ is a derivable function, and so it must be also continuos: $\lim_{x\to x_0}\frac{e^x-e^{x_0}}{x-x_0}=$ $\lim_{x\to x_0}e^{x_0}\frac{e^{x-x_0}-1}{x-x_0}=e^{x_0}\lim_{y\to 0}\frac{e^y-1}{y}=e^{x_0}$ so $e^x$ is derivable in each point and it must be continuos in each point. Your ...


1

$$e^x=\sum_{n=0}^{+\infty}\frac{x^n}{n!}$$ the Radius of convergence is $$R=\lim_{n\to+\infty}\frac{(n+1)!}{n!}=+\infty$$ thus the sum of the power series is infinitely differentiable, and of course continuous at $$(-R,R)=(-\infty,+\infty).$$ but, for example $\sum_{n=0}^{\infty}x^n$ is not continuous at $x=1$.


5

Infinitely many continuous functions added together doesn't imply the result is also continuous. Take the Fourier series of a square wave function for example. All terms are continuous but the result is not. $$f(x)=\frac{4}{\pi}\sum_{n=1,3,5,...}{\frac{1}{n}\sin(\frac{n\pi x}{L})}$$


0

It's easier to use sequences: Fix $0\neq x\in \mathbb R$. There are sequences $(y_n)$ and $(z_n)$ such that $y_n\to 0$ and $z_n\to 0$ and such that $\sin(1/y_n)=1$ and $\sin(1/z_n)=-1$ (why?). Then, $\alpha_n=(x,y_n)\to (x,0)$ and $\beta_n=(x,z_n)\to (x,0)$ but $f(\alpha_n)\to x$ and $f(\beta_n)\to -x$ so $f$ is not continuous at $(x,0).$ The case $x=0$ is ...


0

As mentioned by @5xum, this is not true as it's written. Either you need further assumptions on $f$, or you want to prove a different result. The reason I say this is that this kind of problem is usually tackled through the Intermediate Value Theorem. Since polynomials are continuous functions, given $a<b$ such that $f(a)\neq f(b)$ (without loss of ...


1

I believe there might be a typo in your question. We can prove that there exists $c \in (a, b)$ such that $$1975 f(a) + \color{red}{29} f(b) = 2004 f(c)$$ Indeed, consider the function $$g(x) = 1975f(a) + 29f(b) - 2004 f(x)$$ Suppose that $f(a) < f(b)$. Then: $$\begin{align*} g(a) & = 1975 f(a) + 29 f(b) - 2004 f(a) = 29 (f(b) - f(a)) > 0 \\ g(b) &...


2

The claim is false. For example, take $f(x)=x^2$, and take $a=-1, b=\sqrt{\frac{30}{297}}$. (Note here that $b<1$, it will be important later) Then, $1975 f(a)+297 f(b)=2004f(c)$ translates to $$c^2=\frac{1975 \cdot 1 + 297\cdot\frac{30}{297}}{2004} = \frac{1975 + 30}{2004} = \frac{2005}{2004} > 1.$$ This means that $|c|>1$, but $c\in (a,b)\...


0

You will use it follows. hint First, you prove that $$f(n)=nf(1)$$ $$f(\frac pq)=\frac pq f(1)$$ and for $x$ real, you use the fact that $$x=\lim_{n\to+\infty}\frac{p_n}{q_n}$$ and by continuity $$f(x)=\lim_{n\to+\infty}f(\frac{p_n}{q_n})$$


0

Lower hemicontinuity Let $u_0 \in U$ and $V$ be an open of $\mathbb{R}^n$ intersecting $h^{-1}_{u_0}(\{0\})$. Let then $x_0 \in \mathbb{R}^n$ such that $h(u_0,x_0) = 0$ and $(u_0, x_0) \in V$. We will prove that there exists $W$ a neighborhood of $u_0$ such that $h^{-1}_u \cap V \neq \emptyset$ for all $u \in W$. For clarity, we define the following ...


0

If the assumption is that $f$ is continuous at $x_0$ (i.e., not in the whole $X$), then $f^{-1}(V)$ is NOT NECESSARILY an open subset of $X$. Example $$ f(x)=\left\{\begin{array}{ccc} 0 & \text{if $x$ is irrational} \\ x & \text{if $x$ is rational} \end{array}\right. $$ Then $f$ is continuous only at $x=0$, and $$ f^{-1}\big((-1,1)\big)=(-1,1)\cap \...


2

Not necessarily. Take $f\colon\mathbb R\longrightarrow\mathbb R$ defined by$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in\mathbb Q\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is continuous at $0$. But $f^{-1}\left(\left(-1,1\right)\right)=(\mathbb R\...


5

$f(x_n)-f(y_n) \to 0$ whenever $x_n-y_n \to 0$.


2

HINT: use uniform continuity. Fix $x_0$ and let $\epsilon > 0$. Choose $t_0 \in \left[-x, x\right]$ such that $f(x_0) = \psi(x_0, t_0)$. Since $\psi$ is continuous, there exists a $\delta > 0$ such that $$|\psi(x, t) - \psi(x_0, t_0)| < \epsilon$$ for all $(x, t)$ in the closed ball of radius $\delta$ (let's say with respect to the norm $\|\cdot\|...


0

I suppose there will be mechanisms to relate discrete and continuous mathematics, but if your question is destined to the absence of cycles in the Collatz conjecture, you could try to put your question in the graph theory, for example, is it a forest, or equivalently, is a disjoint union of trees? Is a directed tree to the case where the edges are all ...


9

If you take a constant function, i.e., pick some $a\in\mathbb{R}$ and set $f(x) = a$ for all $x$, then $f(\mathbb{Q}) = \{a\}$ $f(\mathbb{R}\setminus \mathbb{Q}) = \{a\}$ and so, by the condition on $f$, you need both $\{a\}\subseteq \mathbb{R}\setminus \mathbb{Q}$ and $\{a\}\subseteq \mathbb{Q}$. That is, you need $a\in\mathbb{Q}\cap (\mathbb{R}\setminus \...


1

This community wiki solution is intended to clear the question from the unanswered queue. Yes, your proof is correct.


2

I don't think this works, since the fact that your sequence $y'_i$ has the property that $f(y'_i)\to f(a)$ doesn't seem to tell you anything about $f(y_i)$. I think the right way to consider subsequences. If $f(y_i)\not\to f(a)$, show that there must be some subsequence $y_{i_j}$ satisfying $f(y_{i_j})\to b$ for some $b\neq f(a)$ (possibly $b=\infty$ or $b=...


3

The function $$f(x,y,z)=x+e^yz$$ is not Lipschitz. Consider $$f(1,y_1,1)-f(1,y_2,1)= e^{y_1 } -e^{y_2}$$ There is no constant $k$ which makes $$ | e^{y_1 } -e^{y_2}|<k|y_1-y_2|$$ for all $y_1$ and $y_2$ Thus the function is not Lipschitz.


1

Hint: You can write $$x_2+e^{y_{2}} z_2-x_1-e^{y_1}z_1=x_2-x_1+e^{y_2} z_2-e^{y_1}z_2+e^{y_1} z_2-z_1e^{y_2}=x_2-x_1+(e^{y_2}-e^{y_1})z_2+e^{y_1}(z_2-z_1)$$


2

If $f(x,y,z)$ were Lipschitz then $g(x) =f(0,y,1)=e^y$ should be Lipschitz but it is not true since the inequality $$|e^y -1|\leq M|y|$$ is not true for any $M$ for all $y\in\mathbb{R}$


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