New answers tagged

1

You have that there is $C>0$ and $\delta >0$ s.t. for all $t>0$, $$|f(t+h)-f(t)|\leq C|h|,\quad |h|\leq \delta .$$ Then if $u=t+h$ and thus $t=u-h$, we get $$|f(u)-f(u-h)|=|f(t+h)-f(t)|\leq C|h|.$$ Taking $h\to 0$ yields the wished result.


0

The continuity at $x = 0$ is really easy to show that I will skip the point. We now have to show that $$ \newcommand{\R}{\mathbb{R}} \forall a \in \R\setminus\{0\}:\forall \epsilon > 0:\exists \delta > 0:\forall x \in \R\setminus\{0\}: |x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon $$ when $n \in \mathbb{N}$ is odd, $$ f'(x) = \frac{1}{n}x^{\...


1

The proof has a clean idea, but it's a bit confusing the way you've written it - the assumptions are a bit scattered into the proof. I would start the proof as follows, where you lay out clearly what your goal is: We will show that, if $f:[a,b)\rightarrow (a,b)$ is continuous and injective, it cannot be surjective. This expresses, for the reader, exactly ...


1

You should specify what assumptions you use, for example that $f$ is bijective and continuous. Therefore, write "Assume that $f$ is bijective and continuous" in the beginning of the proof. It can also be good to tell the audience that you will use a "proof by contradiction". That will make it clear why you make that assumptions on $f$.


0

Substitute $ x $ for $ w + \frac{1}{3 w} $ 1 \begin{align} f(w + \frac{1}{3 w}) &= (w + \frac{1}{3 w})^3 - (w + \frac{1}{3 w}) + 3 \\ &= w^3 + \frac{1}{27 w^3} + w + \frac{1}{3 w} - w - \frac{1}{3 w} + 3 \\ &= w^3 + \frac{1}{27 w^3} + 3 \end{align} which reduces $ x^3 - x + 3 = 0 $ to the equation $$ w^3 + \frac{1}{27 w^3} + 3 = 0 $$ which is ...


0

For me it's a little easier to work with positive numbers. Obviously $f(x)>0$ if $x \ge 0, x \in \mathbb{Z}$, so let's assume $x<0, y:=-x, g(y) = f(x)=-y^3+y+3=3-(y^3-y), y \in \mathbb N.$ Now it's easy to show $y^3-y$ is monotone increasing with $y\in \mathbb N$ (via factoring or taking first derivative), so $g(y)$ is monotone decreasing and $g(1)=3&...


0

Hint) The definition of (open) ball: $$ B_r(c) =\{x \in [a,b)~|~d(x,c) < r \} $$ where $d$ is a metric given that $[a,c)$ is a metric space. That is, it is always true that $$ B_r(c) \subseteq [a,h) $$ for all $r >0 $ and all $c \in [a,c)$ where $h \in (a,c)$.


1

Let $O = (0, 2)$. Then, $f^{-1}[O] = (0, 1]$, which is not open. The strategy is that when you want to show the discontinuity by inverse image, let your open interval include the function value at the discontinuous point. In the example, we see that $f(1) = 1 \in O$.


3

The statement is incorrect (at the very least, it is extremely misleading) in the form it is currently phrased. Here are the correct statements: Let $f:\Bbb{R}^n\to \Bbb{R}^m$ be a function, and fix a point $p\in\Bbb{R}^n$. Then, $f$ is continuous at $p$ if and only if for every sequence $\{p_i\}_{i=1}^{\infty}$ in $\Bbb{R}^n$, $\lim\limits_{i\to \infty}p_i=...


1

Your proof is correct (with one inattentiveness in line -2: $f^{-1}(U) = M(a) \cap m(b)$). For a shorter proof see William Elliot's anwer. It sticks out that in the first part you work with the $\varepsilon$-$\delta$-definition of continuity, whereas in the second part you use a theorem about continuity: A function $f$ is not continuous if for some open ...


0

There are many types of continuity and most of them are subsumed under topology where we give the domain and codomain a certain topology and a fixed general notion of continuity always applies. Here the topologies are usually given by open sets, or equivalently by nets (which generalise sequences). One form of continuity that does not fall into this category ...


1

B is a subbase for a space S when { $\cap$F : F finite subset B } is a base for S. Theorem. If f:X -> Y, B is a subbase for Y, and for all U in B, f$^{-1}$(U) is open, then f is continuous. To apply this theorem to your problem, note that { (-$\infty$,x), (x,$\infty$) : x in R } is a subbase for R. The proof of the theorem is straight forward and ...


1

$|TF_n(x)-TF(X)| \leq \int_a^h \left|\Phi(t,F_n(t))- \Phi(t,F(t)) \right| dt$. Given $\epsilon >0$ choose $\delta >0$ such that $|\Phi (t,x)-\Phi(t,y)| <\epsilon $ for $|x-y|<\delta$. This is possible because $\Phi$ is uniformly continuous. Now choose $n_o0$ such that $|F_n(t)-F(t)| <\delta$ for all $t$ if $n \geq n_0$. We now get $|TF_n(x)-...


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Use the hypothesis $\Phi$ is continous to go one step further in the integral


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You know that exists $a$ with $f'(a) = 1$. Similarly there is $b$ with $f'(b) = \frac{f(4) - f(1)}{4-1} = 8/3 > 2$. Now apply Darboux's theorem.


0

As $x$ increases from $2$ to $4$: the quantity $f(x)-f(1)$ varies continuously from $f(2)-f(1)=1$ to $f(4)-f(1)=8$; the quantity $x-1$ increases continuously from $2-1=1$ to $4-1=3$; and so the quantity $F(x) = \frac{f(x)-f(1)}{x-1}$ varies continuously from $F(2) = \frac{1}{1}=1$ to $F(4) = \frac{8}{3}$. Applying the Intermediate Value Theorem, since $F(2)...


0

Try the mean value theorem again but for other points: $$\frac{f(4)-f(2)}{4-2} = \frac{8-1}{4-2} = \frac 7 2$$ The mean value theorem for first interval $[1,2]$ gives that there must exist one point $\xi_1$ on that interval for which $f'(\xi_1)=1$. The mean value theorem for second interval $[2,4]$ gives that there must exist one point $\xi_2$ on that ...


0

It is easier to instead show that components of open sets are open, which is equivalent to local connectedness. Let $r\colon X\to A$ be the retraction and suppose $N$ is an open set in $A$ and $C$ is a component of $N$. Let $p\in N$ and $U = r^{-1}(N)$. By continuity, $U$ is open and by local connectedness, there is a connected open $V$ such that $p\in V\...


0

We can check if every $n$th root function is continuous in a slightly sneaky way by making use of a version of the inverse function theorem, which states the following: If $I\subseteq\mathbb{R}$ is an interval and $f:I\to\mathbb{R}$ is injective and continuous on $I$, then the function $g:I\to f(I)$ defined as $g(x)=f(x)$ is bijective and continuous. ...


1

The radical sign represents the principal or nonnegative square root. The $-b$ isn't being ignored: it's just negative (if $b$ is a positive real number, as you say) and thus it isn't $\sqrt{a}$ but is instead $-\sqrt{a}$. As a result, when you have the function $f(x)=\sqrt{4-x}$, your outputs are going to be nonnegative because the radical sign represents ...


1

Define $F(t)=\int_{a+t}^{b+t} f(x) dx = \int_{a}^{b} f(x+t) dx$. You know that $F$ is constant. So $\frac{d}{dt} F(t) = 0$. On the other hand \begin{align*} \frac{d}{dt} F(t) &= \frac{d}{dt} \int_{a}^{b} f(x+t) dx \newline &= \int_{a}^{b} \frac{d}{dt} f(x+t) dx \newline &= f(b+t)-f(a+t) \newline \end{align*} Finally $f(b+t)=f(a+t)$ $\forall ...


1

Note that your function can be written as a composition of continuous functions: $x^x = e^{\log(x^x)} = e^{x\log(x)}$. Therefore, it is a composition of functions $f(x) = e^x$ and $g(x) = x\log(x)$. I don't know how elementary can you be with your proof, but probably you can accept $e^x$ to be trivially continuous, as well as consider $g(x)$ as a product of ...


0

The problem in your first attempt is that, $y$ depends on $\varepsilon .$ Roughly speaking, you can regard this dependence as a one-to-one correspondence(N.B. this saying may not be logically correct because some $y$ may corresponds to many different $\varepsilon .$ But what I want to say is that, when you change $y$, $\varepsilon$ may be changed too.) So, ...


0

You can use the metric to prove continuity. \begin{align*} \lim_{x\to \alpha} d(f(x),f(\alpha)) &= \lim_{x\to \alpha} 2(|\cos x - \cos \alpha|+|\sin x - \sin \alpha|) \newline &=0 \end{align*} This last limit you need to use $\cos$, $\sin$ and absolute value continuity.


1

Suppose that $x_n$ converges to $x$. Let $\epsilon>0$, then the set $G_\epsilon=\{y\in M:\, f(y)>f(x)-\epsilon\}$ is open in $M$. Since, $x_n$ converges to $x$ there is an $m$ such that $x_n\in G_\epsilon$ for every $n\geq m$. In other words, $$f(x_n)>f(x)-\epsilon$$ for every $n\geq m$. This implies that $\liminf_{n}f(x_n)\geq f(x)-\epsilon$. As $\...


0

You appear to be saying that you do not know what a "limit" is. $\lim_{x\to a} f(x)= \lim_{x\to a+} f(x)$ and $\lim_{x\to a} f(x)= \lim_{x\to a-} f(x)$ where "$x\to a+$ means "using only numbers above a" and "$x\to a-$ means "using only numbers below a". If those two "one-sided" limits are not the same then ...


0

If you've not yet played around with what pairs delta and epsilon work, you might want to try something like this: https://www.desmos.com/calculator/iejhw8zhqd which can give you a feel for what happens as you shrink delta. The question you are asking is basically "if I narrow my x-window, does that necessarily shrink the y-window?" the answer ...


0

Nothing happens. Continuity says that if you are given $\epsilon > 0$ then you can find $\delta > 0$ such that the following statement is true: $(*)$ if $|x-x_0| < \delta$ then $|f(x)-f(x_0)| < \epsilon$. So, let's suppose that the $\epsilon$ has been chosen, and let's fix its value. And having done that, let's choose a value of $\delta$ which ...


0

Let $M =\sup \{f(x): x \geq 0\}$. If $M=0$ the the supremum is attained at $0$. Suppose $M\neq 0$. Then there is sequence $(x_n)$ such that $f(x_n) \to M$. If $x_n$ is unbounded we would a subsequence $(x_{n_k})$ tending to $\infty$ but then $M=\lim f(x_{n_k})=0$ a contradiction. Hence $(x_n)$ is bounded. This implies that $(x_n)$ has a subsequence ...


2

That is not correct. What you are supposed to prove is not that final expression. It is: for every $\varepsilon>0$, there is a $\delta>0$ such that$$(\forall x\in X):|x-x_0|<\delta\implies\bigl|g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)\bigr|<\varepsilon.\tag1$$Given $\varepsilon>0$, take $\delta'>0$ such that$$(\forall y\in Y):\bigl|y-f(x_0)\...


1

Since $f(|x|, u)$ is a composition of continuous functions (up to the boundary), then $f$ is continuous up to the boundary, whence $L^\infty(\overline \Omega)$ and therefore $L^p(\overline \Omega)$ for all $p \geq 1$. In particular, $f \in L^2(\Omega)$. Then, by Theorem 8.12 in Gilbarg-Trudinger, $u \in W^{2, 2} \Omega$. By the Agmon-Douglis-Nirenberg ...


0

Here is my proof for this. Let $k_1<\dots k_n\in\mathbb{R}$ such that $f$ is linear on $\left]k_i,k_{i+1}\right[$ for all $1\leqslant i\leqslant n-1$ and $f$ is linear on $\left]-\infty,k_1\right[$ and on $\left]k_n,+\infty\right[$. Let $c_i$ be the slope of $f$ on $\left[ k_i,k_{i+1}\right[$ for all $0\leqslant i\leqslant n$ with the convention $k_0=-\...


0

Let $f(x,y)=\frac{xy^{3}}{x^{2}+y^{6}}$ if $(x,y) \ne (0,0)$ and $f(0,0)=0.$ show that , for $x \ne 0$, we have $f(x,mx)=\frac{m^3x^2}{1+m^6x^4}$. Hence $f(x,mx) \to 0$ as $x \to 0.$ show that , for $y \ne 0$, we have $f(y^3,y)= 1/2.$. Hence $f(y^3,y) \to 1/2$ as $y \to 0.$


1

Consider the function $g: \Bbb R \to \Bbb R^2$ defined by $g(y) = (a,y)$. Then you can easily verify that $g$ is a continuous function. Now observe that $f(a,y) = f \circ g$ which is a composition of two consinuous function and therefore is continuous. Similarly, $f(x,b)$ can be shown to be a continuous real function of one real variable.


0

Well you have consider something that´s wrong. Imagine you have $A = (0, 1)$ and you take a succesion convergent to 0. Then de $d(0, A)=0$ but $0 \not\in A$. Lets consider this, instead: $d(x, A) = 0 \iff inf\lbrace||x-y||_2 \, \, y \in A\rbrace = 0 \iff \forall \varepsilon>0 \, \, \exists \, \, a_\varepsilon \in A\, \, || \, \, a_\varepsilon = d(x, y_\...


0

I think the function $F(x) = f(x - \lfloor x \rfloor ) + \lfloor x \rfloor$ is not a continuous lift. Let $S_1$ be $[0, 1]$ with $0, 1$ identified and let $f$ be rotation by $\frac 14$, $f(x) = x + \frac 14 \pmod 1.$ Let $F(x)$ be defined as above. Then $$\lim_{x \to \frac 34^-}F(x) = 1$$ while $$\lim_{x \to \frac 34^+}F(x) = 0.$$


2

In your specific example, the function is actually Lipschitz continuous. Such functions $f$ are $C^{\alpha}$ for each $0 < \alpha < 1$, and $\|f\|_{C^{\alpha}} \leq K_L$, where $K_L$ is any Lipschitz constant for $f$, for any $0 < \alpha < 1$. Another example of Lipschitz continuous functions include the restrictions of concave or convex ...


0

You can not assume $f(x)=x$. You need to prove this. Hints: Let $g(x)=f(x)-x$. Can you see that $g$ is continuous? Use Intermediate value theorem. Good luck!


0

For the forward implication, note that as $d(.,A)$ is a continuous function and ${0}$ is a closed set then $d(.,A)^{-1} ({0})$ is closed. $d(.,A)^{-1} ({0})$ also certainly contains A, so we must have that it contains $\bar{A}$ by definition of closure. For the reverse implication, note that $\bar{A}=A\cup A'$ where $A'$ is the set of limit points of A. If $...


0

Hint: The open ball of radius $\frac 1n$ around $x$ intersects $E$


1

When $\Psi \equiv 0$ the function $F_{\beta} $ is continuous for any $\beta$, so we cannot prove that $F_{\beta} $ can never be continuous for $\beta=1$. But we can given an example to show that $F_{\beta} $ need not be continuous when $\beta=1$. For this take $d=1$ and take $\Psi$ such that the function is equal to the identity function in some interval ...


2

You are right when you state But everything would've made sense if he did the following: .... Note that your approach requires the general assumption that $x$ is chosen such that $$(0^*) \quad |x| \ge 1, 2n|a_{n-1}|,...,2n|a_0| .$$ Spivak's proof is nevertheless correct although it has little shortcomings. In fact, $(1)$ is true only for $k \ge 2$, thus ...


2

No, it's not, because the sequence $$\begin{gather} 0.09, \\ 0.099, \\ 0.0999, \\ 0.09999, \\ \vdots \end{gather}$$ converges to $0.1$, while the sequence $$\begin{align} f(0.09) &= 0.0009, \\ f(0.099) &= 0.000909, \\ f(0.0999) &= 0.00090909, \\ &\,\,\,\vdots \end{align}$$ converges to $0.00\overline{09} \neq 0.01 = f(0.1)$.


1

$f$ being continuous on $(a,b)$ does not mean that $f$ is continuous at $a$, because $a \notin (a,b)$. $f$ is not even defined at $a$, so there is no "$f(a)$" for those two limits to converge to. The whole point of this theorem is that when $f$ is uniformly continuous on $(a,b)$ you can extend the domain of $f$ to include $a$ in such a way that $f$ ...


1

NO. Consider the function $$ f(x)=\left\{ \begin{array}{ccc} \sin(1/x) & \text{if} & x<0, \\ 0 & \text{if} & x\ge 0. \end{array} \right. $$ Then $f$ is right continuous everywhere, but has no left limit at $x=0$.


0

It's useful to think of the "nice pullbacks for open sets" definition of continuity: $f:X \to Y $ is continuous if for every open set $U \subset Y$, $f^{-1}(U)$ is an open set in the $X$ (where here $f^{-1}$ denotes the set inverse, i.e., $$ f^{-1}(U) = \{x \in X \mid f(x) \in Y \}. $$ Now suppose that $F : X \to F(X) \subset Y$ is continuous. Then ...


2

Since you included the general-topology tag, I’ll point out that this is a special case of a much more general result that has a short, simple proof. Theorem. Let $f$ and $g$ be continuous functions from a space $X$ to a Hausdorff space $Y$, and let $D$ be a dense subset of $X$ such that $f\upharpoonright D=g\upharpoonright D$; then $f=g$. Proof. If not, ...


1

Indeed, $f$ is a closed map. Take a closed subset $F\subseteq \mathbb S^1\times\mathbb S^1$. Since $\mathbb S^1\times \mathbb S^1$ is compact, it follows that $F$ is compact. Then, by continuity, $f(F)$ is compact. Finally, since $f(F)$ is compact in a Hausdorff space (because $\mathbb S^1\times\mathbb S^1$ is a Hausdorff space), $f(F)$ is closed.


1

Every closed subset of a compact metric space is compact. And a continuous map maps compact sets onto compact sets. And, finally, every compact subset of a metric space is closed. So, yes, your map is compact.


1

In fact you can simplify your proof a lot by denoting $h=f-g$. $h$ is continuous as the difference of two continuous maps. You then just have to prove that if a continuous map $h$ vanishes on a dense set $X_\alpha$, then $h$ vanishes on $\mathbb R$. Which is almost immediate. Take $a \in \mathbb R$. As $X_\alpha$ is dense, it exists a sequence $\{a_n\}$ of $...


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