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 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?