# Tag Info

4

The easiest way is to notice that $f$ differs from the zero function (which is measurable) on a null-set. As null-sets are always measurable, $f$ is measurable. You can also do it explicitly: for $2^n\leq a$, $f^{-1}(a,\infty)=\{n,n+1,n+2,\ldots\}$. for $a=0$, $f^{-1}(a,\infty)=\mathbb Z$. for $a<0$, $f^{-1}(a,\infty)=\mathbb R$.

0

Proposition 1. $\|.\|$ is $\mathcal{C}^{\infty}(\mathbb{R}^n\setminus\{0\})$. Proof. $t\in\mathbb{R}_+^*\mapsto\sqrt{t}$ is $\mathcal{C}^{\infty}(\mathbb{R_+^*})$ and $\|\cdot\|:\mathbb{R}^n\setminus\{0\}\rightarrow\mathbb{R}_+^*$, it suffices to prove that $\|.\|^2$ is $\mathcal{C}^{\infty}(\mathbb{R}^n)$. One has: $$\forall x\in\mathbb{R}^n,\forall ... 0 \newcommand{\Reals}{\mathbf{R}}Just so this has an answer: Conceptually, M and N are isometric to a circle of fixed circumference, hence isometric to each other, and an isometry F:M \to N has the desired property. In more detail, let \ell = L[M] = L[N] be the common lengths of M and N. Pick a point m in M arbitrarily ... 0 Assuming you didn't know about the extended mean value theorem, I thought of this: Let b=a+h , and c\in(a,b)=(a,a+h), then$$\left[f(b)-(a)\right]g'(c)=\left[g(b)-g(a)\right]f'(c)\left[f(b)-(a)\right]\lim_{h\rightarrow0}\frac{g(c+h)-g(c)}{h}=\left[g(b)-g(a)\right]\lim_{h\rightarrow0}\frac{f(c+h)-f(c)}{h}$$... 1 The nontrivial part is to prove that the inverse image condition is sufficient. I am assuming that measurability of a function is defined with respect to the Borel sigma-algebra on the reals (for the larger Lebesgue sigma-algebra the statement is false). The Borel sigma-algebra is generated by open intervals so it is enough to prove that the inverse image of ... 1 Assuming we're dealing with the Borel \sigma-algebra on \mathbb{R}. You should how that \{[q_n,q_m): q_n,q_m \in Q\} generate the Borel \sigma-algebra. This is simple: Let a_n be a decreasing sequence of rational numbers which tends to -\infty , and b_n be an increasing sequence of rational numbers which converges to b, and b\in ... 0 Showing a) implies b) is straightforward. Given \epsilon we get a n_0 and then select an n with p(n)>n_0. For these n we have ||x_{p(n)}-x_{p(n+1)}||<\epsilon. For the reverse assume the sequence is not a Cachy sequence and lets construct an increasing sequence p(n) such that ||x_{p(n)}-x_{p(n+1)}||>\epsilon. There must exist an ... 0 Remark that if p is an (strictly) increasing function defined on N, p(n)>n. Suppose that x_n is a Cauchy sequence Let c>0 \exists N such that n,m>N implies that \|x_n-x_m\|<c, in particular \|x_{p(n)}-x_{p(n+1)}\|<c if n>N since p(n),p(n+1)>n>N. done. In the other hands, suppose that for every increasing ... 1 If I_{1} and I_{2} are sets (such as intervals of real numbers), define I_{1} + I_{2} to be their disjoint union and I_{1} \times I_{2} to be their Cartesian product. Since the Cartesian product distributes over disjoint unions,$$ (I_{1} + I_{2}) \times (I_{1} + I_{2}) = (I_{1} \times I_{1}) + (I_{1} \times I_{2}) + (I_{2} \times I_{1}) + (I_{2} ...

3

If we allow $\mu$ to be an infinite measure, then at least precisely the same formulation will not work. Let $\mu = \infty \cdot \lambda$ be the Lebesgue measure "multiplied by $\infty$", i.e. $\mu(E) = 0$ if $\lambda(E) = 0$ and $\mu(E) = \infty$ otherwise. Then $L^\infty (\mu) = L^\infty (\lambda)$ is infinite dimensional, although $\mu$ only assumes two ...

0

Let $B=\{(x,y)\in [0,1]:\forall n (\{x_n,y_n\}\subset \{1,5\}) \}.$ $$\text { For each } n \text { let } B(n)=\{ \sum_{j=1}^{j=n}b_j10^{-j}:b_j\in \{1,5\} \}.$$ Note that $B_n$ has $2^n$ members.For each $x\in B(n)$ let $I(x,n) = (x-10^{-n},x+10^{-n}).$ $$\text { Let } J(n)=\cup_{x\in B(n)}I(x,n).$$ $$\text { For every } n \text { we have }A\subset ... 1 Given \epsilon > 0 and x, we want to show that there exists a \delta>0 such that |F(x)-F(y)|<\epsilon whenever |x-y|<\delta. Since [-x,x] is compact, the minimum is attained at some point(s) t_0\in[-x,x]. Suppose first that \pm x are not points of minimum, i.e. the minimum is attained strictly inside. Let \delta be such that ... 0 The Cantor set C has the same cardinal as R\backslash \{0\} and is Lebesgue-null. So let f:C\to R\backslash \{0\} be a surjection, and extend f to the domain [0,1] by letting f(x)=0 for x\in [0,1]\backslash C. 4 You have that$$\int_{[0,1]}fd\lambda=\int_{[0,0.1]}fd\lambda+\dots+\int_{[0.9,1]}fd\lambda.$$Note that f(x+0.1)=f(x) to get that$$\int_{[0,1]}fd\lambda=10\times\int_{[0,0.1]}fd\lambda.$$Now write$$\int_{[0,0.1]}fd\lambda=\int_{[0,0.01]}fd\lambda+\dots+\int_{[0.09,0.1]}fd\lambda=0.01\times[1+\dots+9]$$to finally get$$\int_{[0,1]}fd\lambda=10\times ...

3

The sequence $(a_n)_n$ is Cauchy: If $m>n>N$ then $$|a_m-a_n|\le\sum_{k=n}^{m-1}|a_{k+1}-a_k|\le \sum_{k=N}^\infty|a_k-a_{k+1}|$$ and the latter is $<\epsilon$ for $N$ large enough for this expresses exactly that $\sum_{k=1}^\infty|a_k-a_{k+1}|$ converges. Hence with $a:=\lim_{k\to\infty} a_k$, we have $\lim_{k\to\infty}|a_1-a_k|=|a_1-a|$.

2

$$|(a_1-a_r)-(a_1-a_s)|=|a_r-a_s|= |\sum_{k=r}^{s-1} a_{k+1}-a_k | \le \sum_{k=r}^{s-1} |a_{k+1}-a_k |$$ This is convergent.

2

Assume that $\lim_{n \to \infty} x_{n+m} = x$. Then, for every $\varepsilon > 0$, there exists $N_0$ such that if $n > N_0$ then $|x_{n+m} - x| < \varepsilon$. If we define $N := N_0 + m$ then we see that if $n > N$ then $|x_n - x| < \varepsilon$.

1

Notice that: $|\sum_{k=1}^{n} a_k-a_{k+1} | = |a_n - a_1 | \leq \sum_{k=1}^{n} |a_k-a_{k+1}|$ for each $n \in \mathbb{N}$

1

The $f_n$ are continuous functions on $\mathbb{R}$, but the poinwise limit of $f_n(x)$ is not, since the uniform convergence preserve the continuity you can conclude that $f_n$ can't converge uniformly on all $\mathbb{R}$.

1

$\ell^1$ is a metric space and its closure equals to the set of all elements that have distance $0$ from $A$. On one side, any element $g$ of $$B:=\{f\in\ell^1:\,\,\forall\,\,n\,\,|f(n)|\leq \frac{1}{2^n}\}$$ has distance $0$ from $A$, because it either in $A$ or a limit of elements in $A$. On the other side, $\ell^1\setminus B$ is open (being equal to ...

1

The integral bound $$\int_0^t |x(s)| ds \le t^4$$ don't say much when $t$ is large. In particular, the set $$L = \{ f\in C[0,1] : f(x) = 0 \text{ on } [0,1/2], \ \|f\|_\infty \le 1\}$$ is a subset of $K$. This $L$ is as big as $\{ f\in C[2/3,1]: \|f\|_\infty \le 1\}$, so $L$ cannot be compact and so is $K$.

1

The uniform norm is bounded by the D norm but not vice versa. Consider a sequence of sine waves scaled so that the frequency increases very rapidly and the amplitude converges slowly to 0.

2

$\cos x = 1 -\frac12x^2+\frac{1}{24}x^4+\cdots$ $e^{-\frac12x^2} = 1 -\frac12x^2+\frac{1}{8}x^4+\cdots$ So if $\varepsilon$ is small and $|x| < \varepsilon$, they differ by at most $\approx \frac{1}{12}\varepsilon^4$.

0

They have the same value, slope and curvature at $x=0$ (verify this by writing the first few terms of the Taylor expansion - in fact the third derivatives are zero, too)

0

The word "integration" is used in two different meanings: (i) It can mean finding an antiderivative, or primitive, $F$ of a given function $f$. In this sense integration is just the inverse of differentiation, and there is no theorem whatsoever involved. (ii) Integration can mean the mental process of capturing "the area under the curve $y=f(x)$ for $a\leq ... 0 The convergence can't be uniform since the$f_n$are even not uniformly bounded. It's easy to compute the$\sup_{(0,1)}|f_n-f|$and see the it doesn't converge to$0$. For$[1/4,1/2]$the convergence is indeed uniform. It's easy to compute$\sup_{[1/4,1/2]}|f_n-f|$and see that it converge to$0$. 0 $$f(x)=\int_0^x K(x,y)f(y)\,dy+g(x)$$ Taking derivatives both sides with respect to$x$we get $$f'(x)=K(x,x)f(x)+g'(x)$$ $$f'(x)-K(x,x)f(x)=g'(x)$$ Let$\mu(x)=\exp(\int -K(x,x)\,dx)$and multiply both sides by it $$\mu(x)f'(x)-K(x,x)\mu(x)f(x)=\mu(x)g'(x)$$ Notice that$-K(x,x)\mu(x)=\mu'(x)$so: $$\mu(x)f'(x)+\mu'(x)f(x)=\mu(x)g'(x)$$ ... 1 Assume that$f$is continuous. Let$\epsilon > 0$and consider $$G_k := \{x \in [0,1] : |f_k(x) - f(x)| \leq \epsilon/2\}$$ and set $$F_n = \cap_{k=n}^{\infty} G_k = \{x \in [0,1] : \sup_{k\geq n}|f_k(x) - f(x)| \leq \epsilon/2 \}$$ Then$F_nis closed and by pointwise convergence $$[0,1] = \bigcup_{n=1}^{\infty} F_n$$ So by Baire Category, ... 1 It's true. I'll show transitivity. Define the relation between sequences of closed intervals as: \begin{align} ([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \iff & \forall n\in\Bbb N: \\ &\quad a_n \le d_n, \text{ and }\tag{i} \\ &\quad c_n\le b_n.\tag{ii} \end{align} A sequence of nested intervals is a sequence with the properties ... 1 There cannot be such a sequence.\chi_{\Bbb Q}$is a double limit of continuous functions, cf. the Dirichlet function. Thus, it's a Baire class 2 function. As the article states, it can't be a Baire class 1 function (single pointwise limit of continuous functions) because such functions have a meager set of discontinuities, unlike$\chi_{\Bbb Q}$. For a ... 0 In a similar spirit, you could also consider an open interval$(a_{m}, b_{m})$. Then take intervals of the form$(c_{m}, d_{m})$with$c_{m}, d_{m} \in \mathbb{Q}$and each of length at most$2^{-m}(1 - \alpha)$. Then the set$E = [0, 1] \backslash \cup_{m \in \mathbb{N}}(c_{m}, d_{m})$is compact as it is a closed set in a compact set and its interior is ... 0 The first interval you remove ($\gamma_{11}$) has measure$\frac{\beta}{2}$. The next two ($\gamma_{21}$and$\gamma_{22}$), as you have written them, each have measure$\frac{\beta}{4}$. But then their combined measure$\frac{\beta}{2}$. This construction cannot be correct, for then you remove intervals of combined measure$\frac{\beta}{2}$at each step. ... 0 It is not enough to assume that$g$satisfies $$g(x)=\frac{1}{4}\left(g(\tfrac{x}{2})+g(\tfrac{x+1}{2})\right),$$ and then check that$g(x)=g(x+1)$. This is the converse of what you are being asked to prove, and it is not true in general. 1 Hint: note that the function is periodic of period$1$, so it is sufficient (why?) to study it in the interval$[0, 1]$. If your$g$was unbounded, it would have to be unbounded in the interval$[0, 1]$. But what do you know about continous functions on closed, bounded intervals? 2 Here is a direct proof (as suggested by @AndréNicolas in the comments) that if$a_1,a_2,a_3\cdots\to g$then$a_1,g,a_2,g,a_3,g\cdots\to g$. Note that the latter sequence is$b_1,b_2,b_3\cdots$, where$b_{2n}=g$and$b_{2n-1}=a_n$. So take$\varepsilon>0$, then there is$N$such that if$n>N$then$|a_n-g|<\varepsilon$. Let$K=2N-1$. Then if ... 0 If$(e_n)_n \notin \ell^1$, show that$A $is not even bounded in$\ell^1$and thus not totally bounded. For the other direction, show that$A $is compact by showing that it is sequentially compact. Indeed, for a sequence$(x_n^{(m)})_n \in A $, use the usual diagonal argument to construct a subsequence$(x_n^{(m_k)})_n $such that the limit$x_n =\lim_k ...

1

According to my Differential Geometry professor, it means that the closure of $V_{\alpha}$ is contained in $U_{\alpha}$. According to Silvia Ghinassi and other sources, it generally means that the closure of $V_{\alpha}$ is a compact subset of $U_{\alpha}$, in which case the notation $V_{\alpha}\Subset U_{\alpha}$ is read "$V_{\alpha}$ is compactly ...

3

Being "continuous on the whole interval" means, by definition, being continuous at every point of the interval. So, if your function is not continuous at just one point, it won't be continuous on the whole interval. It will still be continuous on the interval $(0,1]$, for instance.

0

I don't know if the following example is nontrivial enough for you, but I'll let you decide that. Let $\psi$ be a Schwartz function such that $\widehat{\psi}$ has support in the annulus $\{2^{-1/2}\leq|\xi|\leq 2^{1/2}\}$. Consider the symbol $a$ given by $$a(x,\xi)=\sum_{j=1}^{\infty}a_{j}(x)\widehat{\psi}(2^{-j}\xi), \tag{1}$$ where the $a_{j}$ are ...

1

You have developed the partial fractions into a power series or Laurent series at $z_0 =0$ and $z_0 = \infty$, which unfortunately does not help to get the Laurent series at $z_0 = i$. Instead you have proceed as follows: For $|z-i| < 2$ you have $$\frac{1}{z+i} = \frac{1}{(z-i) + 2i} = \frac{1}{2i(\frac{z-i}{2i} + 1)} = -\frac i2 \frac{1}{1 - \frac ... 2 That works if you first prove (by induction) that all a_n are positive. That will let you apply the ordering axiom as you stated. For an example of why you need them positive,$$5 \geq 4-1 \geq -1$$but it is not true that$$-5 \geq -4$$Strictly speaking, you are appealing to a slightly stronger induction principle than the one which is usually ... 1 If I understand your question correctly, it is the following: Let \epsilon > 0 be given. For what \alpha, \beta \in \mathbb C does the following hold:$$ \left|\frac{\alpha+z_0}{\beta z_1+z_0}\right|<\epsilon \quad \text{ for all } z_0, z_1 \in \mathbb C \text{ with } |z_0|<\epsilon \text{ and } |z_1|<\epsilon \, . $$First let ... 1 HINT:$$\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n}\right)\right)^{-n}=\lim_{n\to\infty}\left(-1+\left(1+\frac{1}{n}\right)^n-\frac{1}{n}\right)^{-n}=\lim_{n\to\infty}\exp\left(\ln\left(\left(-1+\left(1+\frac{1}{n}\right)^n-\frac{1}{n}\right)^{-n}\right)\right)=$$... 2 Hint.$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x=\exp(x)$$From here on you can easily conclude the result as you mentioned in the comments of your post. 0 You need to use something like the Generalized Dirichlet Test (proven in this answer). This answer shows that \left(1+\frac1n\right)^n is an increasing sequence. Therefore,$$ \frac{\cfrac{(n+1)^{n+1}}{e^{n+1}(n+1)!}}{\cfrac{n^n}{e^nn!}}=\frac{\left(1+\frac1n\right)^n}e\lt1 $$and so \dfrac{n^n}{e^nn!} is a decreasing sequence. When ... 0 Apply ratio test:$$\left| {\frac{z \cdot n}{{n+1}}}\right| $$Converges when |z|<1, diverges when |z|\le1 4 Let$$\sup_{n\in \mathbb N} \| f'_n\|_2 = D <\infty.$$Using the Fundamental Theorem of Calculus, if f = f_n,$$|f(x)| = |f(x) - f(0)| = \left|\int_0^x f'(s) ds \right| \le \sqrt x \|f'\|_2 \le D.$$So \{f_n\} has a uniform C^0 bound. Similarly,$$|f(x) - f(y)| \le \sqrt{|x-y|} \|f'\|_2 \le \sqrt{|x-y|}D.$$Thus the family \{f_n\} is ... 0 Here is a sketch of how I would approach the problem. It isn't complete, but hopefully it has the main ideas. First suppose the argument of u is real, i.e. \theta=0:$$u(r)=\frac{1}{2\pi}\int_0^{2\pi}\frac{R^2-r^2}{R^2-2Rr\cos(\phi)+r^2}f(\phi)d\phi.$$Using the triangle and reverse-triangle inequalities we can show that |u(r)| is bounded, and so we ... 0 Assuming no Asymptotic discontinuities if f(x) is Strictly increasing Then f'(x)>0. Letting$$g(x)=f^{-1}(x)$$we have$$f(g(x))=x$$differentiating Both sides$$f'(g(x))g'(x)=1$$\implies$$g'(x)=\frac{1}{f'(g(x))}$$and since f' is Positive g'\gt 0, hence f^{-1} is Increasing. 1 Yes. To see this, h(x):=g(x)-f(x)>0 for all x \in [a,b]. Additionally, h(x) is continuous since g(x) and f(x) are continuous. It follows that \int_a^b h(x)dx>0, so$$0>\int_a^b h(x) dx = \int_a^b (g(x)-f(x))dx = \int_a^b g(x)dx - \int_a^b f(x)dx. Thus $\int_a^b f(x)dx< \int_a^b g(x)dx$. To see why $\int_a^b h(x)>0$: given \$c ...

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