# Tag Info

0

How about a bounded quasinilpotent operator $T \in \mathscr{L}(L^2[0,1])$ defined by $$(Tf)(x) = \int_{0}^{x}f(t)dt.$$ This operator has $\sigma(T)=\{0\}$. The resolvent cannot have a pole at $0$ because $T$ is not nilpotent of any order. So the resolvent has an essential singularity. You can solve for the resolvent by solving ...

8

Some points. 1) An oriented manifold $M$ which occurs as the boundary of an oriented manifold $W$ necessarily has $\chi(M)$ even. This means that not only are all of your surfaces not going to work, but neither is anything that's the boundary of a manifold. 2) Every odd-dimensional manifold has $\chi(M) = 0$ (even non-oriented ones). So you can't try that. ...

7

By the classification of surfaces, every closed oriented surface is a connected sum of tori, so you won't be able to find a $2$-dimensional example. By Poincare duality, any closed odd-dimensional manifold has Euler characteristic $0$, so the smallest dimension where you can find an example is $4$. To find a $4$-dimensional example, you can start with ...

1

Just pick a strictly negative function, let $f(x) = -1$ for all $x$. Then $$-1=\int_0^1 f\ge \int_0^{2/3}f+\int_{1/3}^1f = -{4\over 3}$$ Your positive function will not do the trick since $$\int_E1=m(E)$$ So if $\displaystyle\bigcup_{n=1}^\infty E_n=E$ $$\int_Ef\le \sum_{n=1}^\infty \int_{E_n}f \iff m(E)\le \sum_{n=1}^\infty m(E_n)$$ and the latter ...

0

As suggested by daw, take $u(x) = 1$. If you have some regularity of $\Omega$, then $$\int_{\mathbb R^n} u \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac\partial{\partial n} v \, \mathrm{d}s$$ for $v \in C_c^\infty(\mathbb R^n)$. The right hand side cannot be written as $\int_{\mathbb{R}^n} w \, v \, \mathrm{d}x$ for some $w \in ... 1 If$\Omega$is bounded, it suffices to take$u(x)\equiv 1$. Then$\hat u$cannot be in$W^{1,p}(\mathbb R^n)$for$p>n$since it is discontinuous. (Sobolev embedding) 1 In any infinite-dimensional Hilbert space$\mathcal H$, an orthonormal basis is not a basis in the algebraic sense. Suppose not: let$B$be such an orthonormal basis. Let$b_1, b_2, \ldots$be a sequence of distinct members of$B$and$x = \sum_{j=1}^\infty b_j/j$. If$x = \sum_{b \in B} c_b b$(where only finitely many$c_b \ne 0$), then$c_b = \langle b, ...

1

Values of Fabius function or it's first prime in dyadic rationals are rational, for example. Concernig non dyadic rational it seems to be an open question indeed.

26

Let $g \colon \mathbb{C} \rightarrow \mathbb{C}$ be a function that is discontinuous everywhere and bounded. Consider $f(z) := z^2 g(z)$. Then $f$ is continuous only at $z = 0$, and $f$ is differentiable at $z = 0$ as $$\lim_{z \to 0} \frac{f(z) - f(0)}{z} = \lim_{z \to 0} z g(z) = 0.$$

19

Consider $f(z)=|z|^2\cdot \mathbf 1_{\Bbb Q[i]}(z)$

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Your argument is this: There are $N:={12+5\choose5}$ different ways to put $12$ indistinguishable balls into $6$ boxes numbered from $1$ to $6$, and exactly one of these allocations has $2$ balls in each box. Therefore the probability of obtaining this outcome is given by $p={1\over N}$. But the random mechanism producing this allocation does not choose ...

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A source of radical but not prime: Let $K$ be a field, and let $S = K[x_1, . . . , x_n]$ be the polynomial ring in $n$ variables over $K$. The monomial prime ideals are all of the form $(x_{i_1}, . . . , x_{i_k} )$. But A monomial ideal $I$ is a radical ideal, iff $I$ is a squarefree monomial ideal (This is because: $\{\sqrt u : u ∈ G(I)\}$ is a set of ...

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I figured out what is wrong in my example. Of course $D$ as I defined it is not dense. I mixed up two spaces and didn't saw it later on. Thanks for any comment - they helped alot to rethink my example!

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Here is a counterexample. Let $X=Y=H$ be an infinite-dimensional Hilbert space with orthonormal basis $\{e_j\}$. Let $T_k$ be the projection onto the span of $\{e_1,\ldots,e_k\}$. Let $D$ be the span of $\{e_1,e_2,\ldots\}$. For each $x\in D$, for $k$ big enough $T_kx=x$, so the sequence $\{T_kx\}$ is Cauchy. As we have $T_kx\to x$, we have $T_k\to I$ ...

1

Consider $X=S^1\vee S^3$ and its double cover $X_2$ i.e, attach two copy of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) =\mathbb{Z} = \pi_1(X_2)$. And covering map induced isomorphism in $\pi_n$ for all $n\geq 2$. But they are not homotopically equivalent/ they have different homology groups since their Eular Characteristics are ...

0

Splitting fields can be taken with respect to any family of polynomials over a fixed field $K$. Fields $E$ that are splitting fields of families of polynomials over a field $K$ are precisely those fields such that $E/K$ is normal. Galois extensions are defined to be normal and separable. If the family is finite, say $f_1,\ldots,f_s$, then $E$ is the ...

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The review of Perlman and Chaudhuri, Reversing the Stein effect, Statist. Sci. 27 (2012), no. 1, 135–143, MR2953500, says (in part), "The authors make their point in a rather unorthodox way using a made-up Star Trek episode. The starship Enterprise is lost in space and unable to move. To be rescued it is necessary to send a probe close enough to the ...

1

This is moderately common in Theoretical Computer Science. A lot of problems and algorithms in that field are posed and discussed in terms of mini stories, even in technical papers. In his celebrated paper introducing what is known as the "Arthur-Merlin Protocols", Laszlo Babai writes as if telling a fable about Camelot. Leslie Lamport, the founder of the ...

2

One piece that comes to mind is Gromov's "Metric structures for Riemannian and Non-riemannian spaces." Everything from the numbering - a Gromov hallmark - to some of the colorful yet sometimes remarkably illuminating language - a wonderful example is 1.25.1/2, "If one feels disgusted by the spineless flexibility of arc-wise isometric maps, ..." - is rather ...

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The averaging operation, defined by $$a\oplus b= \frac{a+b}2$$ is commutative but not associative.

2

For outer measure, if $A \subseteq B$ and $m^*(A)<\infty$, then $m^*(B\setminus A) \geq m^*(B)-m^*(A)$ because $A \cup (B\setminus A)=B$, so $m^*(B) \leq m^*(A)+m^*(B\setminus A)$ by the property of outer measure that you mentioned.

0

By contradiction, suppose $\cap_{n\in N}A_n=\phi.$ For each $x\in A_1$ let $f(x)$ be the least $n\in N$ such that $x\not \in A_n.$ Then $B=\{f(x):x\in A_1\}$ is a finite subset of $N,$ so there exists $m\in N$ such that $\forall n\in B\;(m\geq n).$ For such an $m,$ we have $\forall x\in A_1\;(x\not \in A_m)$ because $\forall x\in A_1\;(x\not \in ... 0 So I'll claim$\bigcap_{n = 1}^{\infty} A_{n} = A_{1}$, which is finite. Recall that$X \subseteq Y$means that if$x \in X$, then$y \in Y$. So let$x \in A_{1}$, then$A_{1} \subseteq A_{2} \Rightarrow x \in A_{2}, A_{2} \subseteq A_{3} \Rightarrow x \in A_{3}$, and on, so$x \in A_{n}$for all$n$. But if$y \not \in A_{1}$, then$y \not \in \cap_{n = ...

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Let $m$ be the minimum of the sizes of the nonempty finite sets $A_1,A_2,\dots$. Since they are nonempty, each $|A_k|\ge 1$, hence $m\ge 1$. Now, $m$ is the minimum, so one of the sets $A_k$ has in effect $m$ elements. But then all $A_j$ with $j\ge k$ also has $m$ elements. So $A_k$ is repeating until the end, and this will be the intersection as well.

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Answer based on previous formulation of question: Consider a planar system (i.e. for $x(t)$ and $y(t)$), and write it in polar coordinates. Suppose that system has the form \begin{align} \dot{r} &= f(r),\\ \dot{\theta} &= 1. \end{align} Then, there exists a limit cycle, taking the form of a circle, for every root of $f(r)$. This way, you can ...

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$$f(x)=\frac{1}{x}$$ on $[1,\infty [$. Notice that on $(a,b)$, $$L^2(a,b)\subset L^1(a,b).$$ Therefore, such a result would be impossible on bounded set.

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Yes, for instance if you define $f : \Bbb R \to \Bbb R$ by $f(x)=0$ if $x<1$ and $f(x)=1/x$ if $x≥1$, then $f \in L^2(\Bbb R)$ but $f \not \in L^1(\Bbb R)$.

2

Simplify your example by taking $f(x) =x$ for $x\in E$ and $f(x)=-x$ for $x\in E^c$.

4

The converse is false. Take the Sierpinski space $X = \{0,1\}$ with topology given by $\tau = \{ \varnothing, \{0\}, \{0,1\} \}$. This topology is first-countable but it is not Hausdorff, hence not metrizable.

4

The long line is locally homeomorphic to $\mathbb R$ (and so first countable) but is not metrizable. Less exotically, the lower limit topology on $\mathbb R$ is also first countable but not metrizable.

2

If you just want to have an actual example, here it is: Set $n=3$, $x_1=x_2=x_3=1 + 2^{-52}.$ Then you have (the first values are the hexadecimal binary64 values) x1 = x2 = x3 = 3FF0000000000001 = 1.00000000000000E+0000 x1^2 = x2^2 = x3^2 = 3FF0000000000002 = 1.00000000000000E+0000 a = 3*(x1^2 + x2^2 + x3^2) = 4022000000000002 = ...

5

Let $K$ be any field and $P(x,y)\in K[x,y]$ be a polynomial and $e\in K$ be such that $P(x,e)=x$, $P(e,y)=y$, and $P(P(x,y),z)=P(x,P(y,z))$. (Note: We require that these be equalities of polynomials, not just of functions, which makes a difference if $K$ is finite.) Then $P$ is conjugate to either addition or multiplication by an affine map $K\to K$. ...

2

Theorem: If $X$ is countably compact and sequential (without any additional separation property and thus not necessarily $T_1$) then $X$ is sequentially compact. I found a proof in T. P. Kremsater, "Sequential Space Methods" (Master of Arts thesis). In the $T_1$ case the singleton sets $\{ x \}$ are closed. In the non-$T_1$ case consider instead the ...

1

In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$. Now $G=\cap F$ is closed and $G^c=S\backslash G$ is open. If $V$ is a family of open sets with $\cup V\supset G,$ then $V'=V\cup \{G^c\}$ is an open cover ...

3

In Dieudonné's Elements of Analysis, volume 1, you'll find an exercise which states that if $f$ is $C^1$ and $f_{xy}$ exists and is continuous on an open set $U$, then $f_{yx}$ exists and equals $f_{xy}$ on $U$. (So, of course, $f_{xy}$ is continuous as well.) Here's a sketch of the proof. As in the usual proof that a $C^2$ function has equal mixed partial ...

1

Apart from a counter example you can also try to prove this statement as follows: Let x be of the form $2k$, where $k\in \mathbb Z$ and y be of the form $2p+1$, where $p\in \mathbb Z$. Then their sum will of the form $2(p+k)+1$=$2K+1$, where $K \in \mathbb Z$. This is definitely odd.

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