# Tag Info

4

This is not possible for all $A\subseteq\mathbb{R}$ of measure $0$. Define $\mathbb{D}[f]$ to be the set of non-differentiable points of $f$. The following result is due to Z. Zahorski, exhibited in this $1964$ paper: Theorem. Given $X\subseteq \mathbb{R}$, there exists $f$ such that $\mathbb{D}[f]=X$ if and only if $X=\mathcal{A}\cup\mathcal{B}$ ...

3

This is just a special case of a general fact about topology: If $X$ is compact, $Y$ is Hausdorff, and $f:X\to Y$ is a continuous bijection then $f^{-1}$ is continuous: Since $f$ is a bijection, the inverse image of $S$ under $f^{-1}$ is just $f(S)$. So you have to show that $f(S)$ is open for every open set $S\subset X$. Since $f$ is a bijection this is ...

3

Reality Check: What is the integration variable for the inner integral doing in the outer bounds of the outer integral? $$\displaystyle\int_0^\infty\int_{-pt}^0 \textsf{stuff}\operatorname d x\operatorname d t \neq \int_{\color{red}{-pt}}^0 \int_0^\infty \textsf{stuff}\operatorname d t\operatorname d x$$ What we have is ...

3

Real analysis is hard. It took mathematicians centuries to figure out careful correct logical ways to describe things like the difference between continuity and uniform continuity, in order to avoid the things that look like paradoxes to you. I think this really is both the present and the future of mathematics. If you find it alien even after struggling to ...

3

Using the condition, for all $\epsilon >0$, there is $N=N_\epsilon$ large so that $$|f(x+1) - f(x) -L|< \epsilon$$ for all $x\ge N$. Then for each $M = 1, 2, 3, \cdots$, by induction on $M$ we have $$| f(x+ M) - f(x) -ML|< M\epsilon$$ if $x\ge N$. As $f$ is bounded on bounded intervals, there is $K >0$ so that $|f(x) | \le K$ when $x\in ... 2 Hint: Assume$(x_n)$is C-convergent to$x$. Fix$n > N$. Then for all$\epsilon > 0$we know that$0 \leq d(x_n,x) < \epsilon$. So$d(x_n,x) \in \cap_{\epsilon > 0} [0,\epsilon) = \{0\}$, which is to say that$x_n = x$. Thus a sequence$(x_n)$is C-convergent if and only if there exists$x \in \mathbb{R}$and$N$such that$x_n = x$for all$n ...

2

Hint: Let $X$ be the reals in $(0,1)$ with the usual metric, and $Y$ the reals with the usual metric. Let $f(x)=\tan(\pi x/2)$.

2

Hint: This is an alternative to your approach that's more straightforward. For all $x \in C[0,1]$ (hence, for all $x \in K$) $$|f(x)| \leq \|x\|_\infty\|y\|_\infty.$$ (Why? Show this if you don't have a theorem that proves this already.) Hence $$\|x_n\|_\infty \to 0\text{ as }n \to \infty \implies |f(x_n)| \to 0\text{ as }n\to\infty,$$ as well.

2

$n_0$ is an upper bound, meaning that every element of $A$ is less than it. $n_0-1$ is not an upper bound, meaning some element of $A$ is greater than it. Thus this number falls between the two. $0<\frac{1}{y-x}<N$ holds by using 0.21 to find an $n$ such that $n\leq\frac{1}{y-x}<n+1$ and setting $N=n+1$. $n\leq Nx<n+1$ is exactly what 0.21 ...

2

(this proof takes a little longer, but gives a neat bound on the rate of convergence) If you want a universal bound of the form $a_n\le M$, the recursion tells you that you're going to need $M\le M^2+\frac{M}{6}+\frac{1}{9}$, which works out to be equivalent to $(6M-1)(3M-2)\le0$. The lowest bound this allows us to take is $M=\frac{1}{6}$, i.e. that ...

2

If you want an easier way to find what value the limit converges to, consider that by the definition of convergence, as $n\to\infty$, $a_{n+1} = a_n$. Thus, let's let $a_{n+1} = a_n = L$. Substituting this in, we get: $$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$$ $$L = \frac{6(1+L)}{7+L}$$ $$L = \frac{6 + 6L}{7+L}$$ $$7L + L^2 = 6 + 6L$$ $$L ^2 + L - 6 = 0$$ ...

2

Convergence: Note that $$a_{n+1}=6-\frac{36}{7+a_n}$$ The function $\displaystyle f(x) = 6 - \frac{36}{7+x}$ has derivative $\displaystyle f'(x) = \frac{36}{(7+x)^2}.$ So we have $0 \leq f'(x) <1$ so the sequence converges by the Contraction Mapping Theorem.

2

The last bit is confused. You know that $\sup(Y)$ exists and is an upper bound for $X$ as well, as $X \subseteq Y$. So $\sup(X)$ exists. But $\sup(X)$ is the smallest of all upperbounds of $X$, and $\sup(Y)$ is one of the possible upper bounds. So..

1

Note that $$Gr(f) = \{ (x, f(x)) : x\in A\}$$ and $$A\times f(A) = \{ (x, f(y)): x, y \in A\}$$ So $A\times f(A)$ is strictly bigger in general. To show that $Gr(f)$ is comapct, you might check directly that $Gr(A)$ is sequential compact: let $(x_n, f(x_n)) \in Gr(f)$ be a sequence. Then compactness of $A$ implies that $x_n \to y \in A$ (taking ...

1

The graph is a closed subset of the compact set $A\times f(A)$ so it is compact. done.

1

If we assume that $f_n(x)$ tends to $f(x)$, then, intuitively we'll have $\frac{f_1(x) + f_2(x) + ... +f_n(x)}{n} = \frac{f_1(x) + ... + f_{N-1}(x)}{n} + \frac{f_N(x) + ... + f_n(x)}{n}$, where you can get $|f_N(x) -f(x)| < \epsilon$ for sufficiently large $N$. Then, as $n$ becomes large, and since $N$ is finite (for some $\epsilon$) the first part will ...

1

Yes. Suppose $\lim_{x\to+\infty}f(x)=L$. That is, for every $\epsilon>0$ there exists $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Let $x_n\to+\infty$, that is for every $M>0$ exists $n_0\in\Bbb N$ such that $n>n_0$ implies $x_n>M$. I claim that $f(x_n)\to L$. Indeed, let $\epsilon>0$ be given. Then pick $x_0$ such that ...

1

(i) if $\phi_1(x)=\phi(x)$ except on a set of measure zero $A$, then $g(x,\phi(x))=g(x,\phi_1(x))$ except on that same set of measure zero. Thus two functions which are members of the same a.e. equivalence class in $L^r$ give two functions equivalent in $L^s$. (ii) holds because $x\mapsto g(x,\phi(x))$ is a real function of $x$ defined for $x\in \Omega$ ...

1

The convergence implies an inequality of the form $$\frac{l}{2} \leq \frac{|a_n|}{|b_n|} \leq \frac{3l}{2}$$ valid for all sufficiently large $n$. Simply pick $\epsilon = \frac{l}{2} > 0$. This will allow you to use one series to dominate the other as needed.

1

Let $f(z)=\sum_{n=3}^{\infty}\left(1-\frac{1}{n^2}\right)^{-n^3}z^n$ be a given power series. Using the Root Test, the radius of convergence $R$ is given by \begin{align} R&=\frac{1}{\lim_{n\to \infty}\left(\left(1-\frac{1}{n^2}\right)^{-n^3}\right)^{1/n}}\\\\ &=\frac{1}{\lim_{n\to \infty}\left(1-\frac{1}{n^2}\right)^{-n^2}}\\\\ &=1/e ... 1 Hint: show that f(C) is closed for every C closed in A. 1 Differentiating we get y'=1-4\int_0^ty(x)dx-4ty+4ty=1-4\int_0^ty(x)dx. $$Differentiating again, we obtain$$ y''=-4y, $$which has the solutions y(t)=a\cos(2t)+b\sin(2t). Substituting in the original equation gives$$ a\cos(2t)+b\sin(2t)=t[1-2a\sin(2t)+2\cos(2t)-2]+4\int_0^t s(a\cos(2s)+b\sin(2s))ds. $$Solving for a and b, you finally compute ... 1 Hint: Make the system of intervals disjoint and show that equality holds. Then, realize that the length of what you've removed must have been positive. To make the system disjoint, you may let$$E_0 := I_0,E_1 := I_1 - I_0,E_2 := I_2 - (I_0 \cup I_1),$$that is for k > 1$$E_k := I_k - (I_1 \cup \cdots \cup I_{k-1}).$$For the ... 1 Rewrite$$ f(x,y) = \frac{x^{a_1}y^{b_1} + x^{a_2}y^{b_2} \cdots}{x^{c_1}y^{d_1} + x^{c_2}y^{d_2} \cdots} $$as$$ f(x,y) = \frac{x^{a_1} x^{b_1}(y/x)^{b_1} + x^{a_2}x^{b_2}(y/x)^{b_2} \cdots}{x^{c_1}x^{d_1}(y/x)^{d_1} + x^{c_2}x^{d_2}(y/x)^{d_2} \cdots} $$by multiplying monomials upstairs and downstairs by a suitable power of x, and then simplify ... 1 Hint: Say that you have a sequence x_n\to x such that x_n(0) \in [-3,4] and |x_n(t)-x_n(s)| \leq d |t^2-s^2| for all t and s (you mean in [0,1]), for each n. What happens in these two properties if n\to\infty? 1 Suppose \;\{x_n\}_{n=1}^\infty\subset K\; is such that \;x_n\xrightarrow[n\to\infty]{}x\; . We must show that \;x\in K\;. By continuity:$$x(0)=\lim_{n\to\infty}x_n(0)\in[-3,4]\;,\;\;\text{as this last is a closed interval}$$Also, again by continuity, for all \;t,s\in[0,1]\; , we have ... 1 C-convergence is much stronger than convergence. The sentence for every \epsilon>0 and every n>N, d(x_n,x)<\epsilon is the same as for every n>N, x_n=x. 1 This is true only if \lvert z\rvert <1. The reason is quite simple and relies on a well known identity from high school:$$1-z^n=(1-z)(1+z+z^2+\dots+z^{n-1})$$which we can rewrite as$$\frac1{1-z}=1+z+z^2+\dots+z^{n-1}+\frac{z^n}{1-z}.$$From this we deduce that$$\biggl\lvert\frac1{1-z}-(1+z+z^2+\dots+z^{n-1})\biggr\rvert=\frac{\bigl\lvert ...

1

You need to define a metric on the domain of F. You seems using $||_{\infty}$ on the domain of $F$. You actually used a strong assumption on $\psi$. You need $\psi$ is uniform continuous on $[0,1] \times \mathbb{R}$

1

The proof here, works for $K=[a,b]\subset \mathbb R^1$, but I think it can be modified for $K\subset \mathbb R^n$, because Arzela-Ascoli theorem can be generalized for any metric compact space $K$. Let us denote with $|.|_\infty$ the supremum norm in $C^0(K)$. You have that $\{f_n^{(k-1)}\}$ is equicontinuous, since $\{f_n^{(k)}\}$ is uniformly bounded: ...

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