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  • 0 posts edited
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  • 35 votes cast
Apr
23
asked estimate on the sum of Rademacher functions
Jan
27
asked asymptotic behavior of the two sequences defining exponential function
Jan
20
comment $\inf_{a\in\mathbb{C}}\|f-a\|_{L^{\infty}(I)}\le|I|\|f'\|_{L^{\infty}(I)}$
Good point. $|I|\|f'\|_{L^{\infty}(I)}$ is an upper bound for the length of the curve.
Jan
16
comment $\inf_{a\in\mathbb{C}}\|f-a\|_{L^{\infty}(I)}\le|I|\|f'\|_{L^{\infty}(I)}$
I see. So in fact we don't need the geometric pic. It is just fundamental theorem.
Jan
16
accepted $\inf_{a\in\mathbb{C}}\|f-a\|_{L^{\infty}(I)}\le|I|\|f'\|_{L^{\infty}(I)}$
Jan
12
asked $\inf_{a\in\mathbb{C}}\|f-a\|_{L^{\infty}(I)}\le|I|\|f'\|_{L^{\infty}(I)}$
Dec
28
comment On a property of $(Mf)^{\delta}$
Thank you. I thought "Normal Human" is a robot at first....
Dec
28
accepted On a property of $(Mf)^{\delta}$
Dec
26
comment average of maximal function is less than its infimum?
There are two typos: $Mf(x)=1/2^k$ if $x\in[1/2,1[$ and same mistake appears in the next line. Very nice construction. I have a conjecture for a modified statement: math.stackexchange.com/questions/1590031/…
Dec
26
revised On a property of $(Mf)^{\delta}$
edited title
Dec
26
asked On a property of $(Mf)^{\delta}$
Dec
23
comment average of maximal function is less than its infimum?
Thank you anyway. I think you understand my question very well. Your argument is very straightforward and already satisfactory, whereas Thiele used some commonly used standard estimates, which can be hard to follow for beginners. Both of you make use of the property of $J$ in the last step. The difference is the initial setup.
Dec
23
comment average of maximal function is less than its infimum?
Thiele wants to show exactly the same inequality you mentioned in your answer. He achieved this by three steps, each with an equality. Firstly, the LHS can be controlled by inf over $I$ of Maximal function. See math.stackexchange.com/questions/1066910/… for details. Secondly, enlarge the range of inf to ancestor $J$, but pay a price of $2^k$ (this is where I don't understand). Lastly, use the property of $J$ (you also used this property in your proof) to bound inf by a constant.
Dec
23
comment average of maximal function is less than its infimum?
Thanks! Can you understand Thiele's argument? The display in the bottom of page 25 consists of three inequalities and I can understand all but the middle one, which is this question.
Dec
17
awarded  Nice Question
Dec
10
comment average of maximal function is less than its infimum?
@MattRosenzweig ams.org/bookstore?fn=20&arg1=cbmsseries&ikey=CBMS-105 bottom of page 25
Dec
8
accepted write abc-ABC in terms of linear combination of products
Dec
8
asked write abc-ABC in terms of linear combination of products
Nov
23
comment average of maximal function is less than its infimum?
@MattRosenzweig The statement is true, but the lemma I posed is my conjecture.
Nov
17
asked average of maximal function is less than its infimum?