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Jan
28
comment Order of Double Coset
@Sushil: The natural 1-1 and onto map is $axb\leftrightarrow axbx^{-1}$ for $a\in A$ and $b\in B$.
Jan
6
comment Annoying Polynomial Inequality
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
Dec
27
comment How to show that $f'(x)<2f(x)$
@Hans: Sorry, I don't know any other similar question, and after some failed attempts, I have to admit that I cannot explain my motivation of the "by (1)" step better than the post itself suggests.
Dec
25
comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear?
@VictorWang: You are welcome.
Dec
25
comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear?
@VictorWang: Sorry, I just read your comment because it is very long time since the last time I visited this website. Please note that since $\varphi(\Omega)\subset \Omega$, $\varphi^k$ is a well-defined holomorphic function on $\Omega$ for every $k$. Then from the choice of $r$ we know the radius of convergence of the Taylor expansion of $\varphi^k$ around $z_0$ is no less than $r$.
Dec
25
comment How to show that $f'(x)<2f(x)$
@Hans: Thank you for your upvote. I think the last time I visited this website is a couple of months ago, and I just read your comment. I am not sure how to explain the motivation of the step mentioned in your comment clearly. Are you still interested in this question?
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?