4

Hint: In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and $f(x, y)=\frac{2xy}{x^2 + y^2}=\frac{2r^2\cos\theta\sin\theta}{r^2}=2\cos\theta\sin\theta$. When $r \to 0$, the limit does not exist. Why?


3

My Question: Can we tell that in $S^{n-1}(n\ge3)$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ contains a set homeomorphic to $S^{n-2}$? No. Consider any closed subset $A\subseteq S^{n-1}$ such that $A$ disconnects the sphere into two nonempty open components $U_1,U_2$. Since $S^{n-1}$ is perfectly normal then we can construct $f:S^{n-1}\to [-1,1]$ such that $f^...


3

Yes $f: \Bbb R \to \Bbb R$ defined by: $$f(x) = \begin{cases} 1 & x \in \Bbb Q\\ 0 & x \notin \Bbb Q\\ \end{cases}$$ is everywhere discontinuous but constant (so continuous) on both dense sets $\Bbb Q$ and $\Bbb R\setminus \Bbb Q$.


3

For the continuity there is no problem, as it is a local feature. More precisely, let $s \in S$. For every $V$ neighborhood of $f(s)$ there exists $U$ neighborhood of $s$ inside $B_{\epsilon_s}(s)$ such that $f(U) \subset V$. Then $U$ is also a neighborhood of $s$ in $S$, so $f : S\to Y$ is continuous. For every $s \in S$ let $\epsilon_s$ as in the ...


2

hint $ f $ is continuous at $ 0 $ because $$(\forall x\in \Bbb R)\;\;|f(x)|=|x|$$ and $$\lim_{x\to 0}f(x)=0=f(0)$$ Let $ a\ne 0$ rational. Take, for $ n $ great enough, $$a_n=a+\frac{\pi}{n}\notin \Bbb Q$$ then $$\lim_{n\to+\infty}f(a_n)=\lim_{n\to+\infty}(-a_n)=-a\ne f(a)$$ Let $ b $ be an irrational number. Take $$b_n=\frac{\lfloor 10^nb\rfloor}{10^n}\in \...


2

Here's a counterexample . . . Let metric spaces $X,Y$ be given by \begin{align*} X&=\mathbb{Z}\,{\large{\setminus}}\{0\}\\[4pt] Y&=\{0\}\cup\Bigl\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}n=2,3,4,...\Bigr\}\\[4pt] \end{align*} with metrics inherited from the standard metric on $\mathbb{R}$. Note that the topology on $X$ is discrete, so every subset ...


1

For $F=\emptyset$ we can take $g: x \to \Bbb R$ to be a constant function with value $1$, so that $g^{-1}[\{0\}] = \emptyset = F$. So we can assume that $F \neq \emptyset$. Let $f(x) = \inf \{d(x,y): y \in F\}$, which is well-defined as the infimum of a non-emptys set that is bounded below (by $0$) in $\Bbb R$. Then it's not too hard to see that $$\forall x,...


1

By the uniform Cauchy criterion, there exists $N(\epsilon)$ such that for all $m > n \geqslant N(\epsilon)$ and all $x \in (a,b)$, we have $|f_m(x) - f_n(x) | < \epsilon$. By continuity, this implies that $|f_m(a) - f_n(a)| = \lim_{x \to a+}|f_m(x) - f_n(x)| \leqslant \epsilon$, and similarly as $x \to b-$.


1

Edit: My first post only showed that $f$ cannot be continuous by declaring $f(0,0) = 0$. Now I adjusted the proof to show that $f$ is not continuous at $(0,0)$. Let us suppose for the sake of contradiction that $f$ is continuous at $(0,0)$ and $\lim_{(x,y)\rightarrow(0,0)}f(x,y) = L$. Note that $f(x,x) = \frac{2x^2}{x^2+x^2} = 1$ and $f(2x,x) = \frac{4x^2}{...


1

Suppose that the limit exists and it is equal to $L$. For any $\varepsilon>0$, we can find $\delta>0$ so that $|(x,y)|<\delta$ implies that $|f(x,y)-L|<\varepsilon$. Fix $\varepsilon>0$, but arbitrary. Find such $\delta>0$, so for $|x|<\delta$ and $y=0$ we have that $|(x,y)|<\delta$, thus $|f(x,y)-L|<\varepsilon$, but since $y=0$ ...


1

Hint: So, (and I am completely using @EricWofsey's answer linked above), you could let $f_n(x)=\begin{cases}x,x\in[n,n+1]\\n,x\lt n\\n+1,x\gt n+1\end{cases}$


1

$\mathbf{ Claim:}$ The set $ D $ is not close in $ I $. To see this we proceed by contradiction, in particular we will show that if $ D $ is close then there exists an extension. Suppose $ D $ close and non-empty. Let $ a = \min\{ x \in D \} $ and $ d = \max \{ x \in D \} $. We extend $ f $ in $ \mathbb{R} \setminus [a,b] $ by setting $ \tilde{f}(x) = f(a) $...


1

Let $f_n(t): \frac{1}{n} \sin (nt).$ Then $||f_n||_{\infty} \le \frac{1}{n}$ for all $n$. Hence $$ ||f_n -0||_{\infty} \to 0.$$ But we have $(h(f_n))(t)= \cos (nt)$ and therefore $(h(f_n)))$ does not converge to $(h(0))$ in the norm $||\cdot||_\infty$


1

Continuity of $h$ means given $f$ and $\epsilon >0$ there exist $\delta >0$ such that $\|g-f\|_{\infty} <\delta$ implies $\|h(f)-h(g)\|_{\infty} <\epsilon$. In your example $\|h-f\|_{\infty} <\delta$ is not satisfied for any $\delta \in (0,1)$ so the example says nothing about continuity of $h$. [BTW, you should avoid using $h$ with two ...


1

You can just define: \begin{equation} f(x) = \begin{cases} 1 & \text{if $\ x \in D $}\\ 0 & \text{if $\ x \in \mathbb{R} \setminus D$} \end{cases} \end{equation} It is a discontinuous function at each point of $D$ and it is also continuous at each point of $\mathbb{R} \setminus D$ assuming $D$ is a finite set or an infinite set without any ...


1

Using Leibnitz integral rule: $xf(x) = \dfrac{x}{x^4+1}$ Let $x \in \mathbb{R} - \{ 0\}.$ Now we can divide the above expression by $x$ as $x\ne0.$ $f(x) = \dfrac{1}{x^4+1}$ which is clearly less than one as denominator is always greater than numerator. Now lets include zero into the domain... Substituting zero into the original equation we have: $0f(0) = \...


1

$$ a = \max \{ x \in [0, 1] \mid f(x) \le 0 \} $$ is well-defined since $f(0) = 0$ and $f$ is continuous. Then $f(x) \le 0$ on $[0, a]$ (since $f$ is convex), and $f(x) > 0$ on $(a, 1]$. The function $F(x) = \int_0^x f(t) \, dt$ is decreasing on $[0, a]$ and increasing on $[a, 1]$ with $F(0) = F(1) = 0$. Now we have $$ \begin{align} \int_0^1 |f(x)| \, dx &...


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