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seen May 23 at 11:41

Apr
4
comment Vector spaces and intersections
@BeniBogosel: Thank you. Maybe I should mention that the example is a special case of the following observation. By choosing the basis $\{e_1,\dots, e_4\}$ appropriately, $V_1$, $V_2$, $V_3$ can always be of the form given in my answer and at the same time, $V_4$ is spanned by a pair of vectors $(e_1,e_3)A+(e_2,e_4)B$, where $A$, $B$ and $A-B$ are invertible $2\times 2$ matrices. Then $W$ exists iff $A^{-1}B$ has a real eigenvalue. In my answer, $A$ is identity and $B$ is rotation by $-\frac{\pi}{2}$.
Mar
10
comment A beautiful inequality for convex functions
@Julien: You are welcome!
Mar
10
comment A beautiful inequality for convex functions
@Julien: I edited my answer by reorganizing the argument. I hope it looks clearer now. In the current form, your question asks how I get $h_t$ in $(1)$. First I assume $h_t(x)=b\cdot V(x-a)$ for some constants $a$, $b$. Then from $h_t\ge f_t$ we know $b(t-a)=h_t(t)\ge f_t(t)=1$. To get $\int_0^1h_t \le 2\int_0^1f_t=2(1-t)$, I use a stronger inequality $\frac{b(1-a)^2}{2}=\int_a^1 h_t\le 2(1-t)$ instead. The inequalities determine a unique pair of $(a,b)=(2t-1,\frac{1}{1-t})$. There is probably more geometrical explanation, but maybe it needs more words due to the limitation of my English.
Mar
9
comment A beautiful inequality for convex functions
@Julien: I didn't show the motivation of the construction of $g$ in the linked answer, but the basic idea is quite similar to your approach, and the same idea can be applied to constructing $h$. Do I need to post an answer here to show all the details(probably it does not satisfy the "more intuitive" requirement in the earlier version of your question)?
Mar
9
comment A beautiful inequality for convex functions
@Julien: To see the history of all the editions of a post, you may simply click "edited +time" in the last line of the post.
Mar
9
comment A beautiful inequality for convex functions
@Julien: I almost forgot all the details in my answer to the linked question, and I just looked back at it briefly. It seems that the same argument there does not work, because it uses the fact that the supremum of convex functions is still convex, which fails to be true for infimum in general. However, I think the method in the remark of my answer there(or the first version of my answer) still works for constructing $h$, and this method shares the same basic idea with your own approach.
Mar
9
comment A beautiful inequality for convex functions
Here is a closely related question.
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
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: Thank you for being considerate of my mistakes.
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
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
4
comment Properties of a smooth bijection
For $M=N=\Bbb R$, $x\mapsto x^3$ is not of constant rank.
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 $\lim_{x\to\infty}\frac{f'(x+\theta(x))}{f'(x)}=1$.
Oct
26
comment Limit of $x_n/n$ as $n\to\infty$
@MartinArgerami: Thanks for your comment. I completely changed my argument after reading other answers.