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seen Oct 8 at 8:44

Dec
3
comment Operators on $C([0,1])$ that is compact or not.
@user62138: You may use a similar argument as part b) in my answer. If $(f_n)$ is bounded in $L^p$ for some $1<p<\infty$, Hölder's inequality implies that $F_n$ is equicontinous.
Dec
3
comment Operators on $C([0,1])$ that is compact or not.
@user62138: I think it's compact iff $p\ne 1,\infty$.
Nov
25
comment How to prove the inequality between mathematical expectations?
@Siméon: I am the same user as "Landscape". I changed my user name one month ago, so I didn't notice your message until I come back to this post now. Unfortunately, I have nothing new about either zyx's approach or any other intuitive way in solving the original problem. I wasn't so self-confident that I could solve it by myself within a short time, so I even didn't think about it deeply. However, I am still interested, so please let me know if you succeed in finding a more intuitive way to solve the original problem, and then I can post a question and/or set a bounty for you to answer.
Nov
14
comment Trace of $A$ if $A =A^{-1}$
The condition $A\ne\pm I$ didn't show up when I was answering the question. With the additional assumption $A\ne\pm I$, the trace of $A$ must be $0$.
Nov
14
revised Contour integration of $\int_{-\infty}^{\infty} \frac{1-b+x^2}{(1-b+x^2)^2 + 4bx^2}dx = \pi$
added 24 characters in body
Nov
14
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: Thank you for being considerate of my mistakes.
Nov
12
revised Show that some $C^\infty$ real function is analytic
added 26 characters in body
Nov
7
comment Uniform convergence in $\mathbb{R}^2$
@Jack: Thank you for your reminding. As a response, I deleted that equality. Is it better now?
Nov
7
revised Uniform convergence in $\mathbb{R}^2$
added 3 characters in body
Nov
7
revised Uniform convergence in $\mathbb{R}^2$
added remark
Nov
7
comment Uniform convergence of constant speed $C^1$ curves with the same endpoints
$\sigma_n'$ may not be uniformly convergent. For example, let $a=(0,0)$, $b=(0,1)$, $L_n=\frac{n+1}{n}$, and $\sigma_n(t)=\frac{1}{2\pi n}(\cos 2(n+1)\pi t -1, \sin 2(n+1)\pi t)$, $t\in[0,\frac{1}{n+1}]$ and $\sigma_n(t)=(0,\frac{(n+1)t-1}{n})$, $t\in[\frac{1}{n+1},1]$. By the way, do you have any problem with my answer in the linked post?
Nov
7
comment Uniform convergence in $\mathbb{R}^2$
@Jack: Please see here.
Nov
6
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: Moreover, when I checked my answer, I found I had given a wrong link to Gauss–Bonnet theorem. I cannot believe how that could happen! Anyway, the link is corrected now. I sincerely apologize for these mistakes.
Nov
6
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: When I looked back at this post incidentally, I found that the remark $\gamma'(\Bbb R)=S^1$ in the last version of my answer is incorrect in general, so I removed it. (The statement is correct when $\gamma$ is a simple closed curve and when I made the remark, I overlooked the self-intersecting case.) Here is a counter-example of the remark for self-intersecting case. Let $\gamma$ be the arc-length reparametrization of the curve $t\mapsto (2\sin t, \sin 2t)$. Then it's easy to check that $(0,1)\notin \gamma'(\Bbb R)$.
Nov
6
revised For a closed plane curve, showing some inequalities.
removed an incorrect remark and corrected a link
Nov
4
comment Properties of a smooth bijection
For $M=N=\Bbb R$, $x\mapsto x^3$ is not of constant rank.
Nov
4
awarded  Revival
Nov
1
revised prove $f(x)$ has at least $2n$ roots
deleted 160 characters in body
Nov
1
answered prove $f(x)$ has at least $2n$ roots
Oct
27
comment Limit of $x_n/n$ as $n\to\infty$
@sundaycat: To take limit in $(5)$, I need assumption (ii) to apply $(1)$, where a reasonable requirement is $\theta_n$ cannot be too large. Without any assumption, I only know $0<\theta_n<\delta_n$. To control the upper bound of $\theta_n$, I need assumption $(1)$ to obtain $\delta_n\le\delta$ when $n$ is large. For example, consider $f(x)=x^p$($p\ge 1$) and an arbitrary $\theta:(0,\infty)\to (0,\infty)$. Then you need some condition on $\theta$ to get $\lim_{x\to\infty}\frac{f'(x+\theta(x))}{f'(x)}=1$.