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seen May 23 at 11:41

Jun
3
awarded  Nice Answer
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 $-\frac{\pi}{2}$.
Apr
4
answered Vector spaces and intersections
Apr
3
revised What is $\sum_{n=0}^{\infty}|a_nz^n|^2=\frac{1}{2 \pi}\int_{-\pi}^{\pi}|f(ze^{it})|^2dt$ for?
corrected typo
Apr
1
awarded  Revival
Mar
25
awarded  Enlightened
Mar
25
awarded  Nice Answer
Mar
19
reviewed Approve suggested edit on Function $f$ from $[0,\infty)$ such that is limit at infinity equals zero and it's values greater than zero must decrease somewhere
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
10
revised A beautiful inequality for convex functions
reorganized
Mar
9
answered A beautiful inequality for convex functions
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.
Jan
9
awarded  Generalist
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.