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 Apr12 comment inserting absolute value in Hilbert transform and a discrete version of Hilbert transform Ah okay, I just copied what I had cited in a paper, I forgot they didn't supply the proof. I don't have access to it at the moment, but do they cite a paper by Marsden and Moreka (I probably have spelled that wrong, and there should be a third author, either Richards or Riemenschneider)? The proof might be in there. If not I will look again on Monday, I know I have seen it before. Apr11 revised inserting absolute value in Hilbert transform and a discrete version of Hilbert transform deleted 9 characters in body Apr10 comment inserting absolute value in Hilbert transform and a discrete version of Hilbert transform I neglected to mention, if it is not clear, you can do a change of variables in the Hilbert transform to look at is as $p.v.\int\frac{f(y)}{x-y}dy$, and same for the discrete version. Apr10 revised inserting absolute value in Hilbert transform and a discrete version of Hilbert transform added 87 characters in body Apr10 answered inserting absolute value in Hilbert transform and a discrete version of Hilbert transform Mar2 comment M bounded if and only if $\sup\{|\langle x,y \rangle|:y\in M \}<\infty$ Also I think the UBP is used to show one direction of your claim that a set is bounded iff it is weakly bounded. Mar2 comment M bounded if and only if $\sup\{|\langle x,y \rangle|:y\in M \}<\infty$ Do you mean for every $x\in M$, the supremum in your first line is finite? Mar2 comment How to prove inequality in positives? What is the inequality here? I see an equality. Feb27 comment Big error in basis of tensor product space I am not exactly sure what you are asking. But one thing you might make note of is that the tensor product of any ONBs is an ONB for the tensor product space. Nov17 comment Inverse of a particular operator What space are you acting the operator on? It is also likely to be unbounded. If you are looking at $C^\infty$ functions on $[0,1]$ for example, the sequence of functions $\sin(nx)$ should give a sequence whose uniform norm is unbounded when applying the operator. Nov17 asked $p$-stable Random Variables for $p>2$? Nov17 awarded Nice Question Oct16 comment Find a subset of the real numbers Yes, I understand that they are not open, I was trying to give a hint rather than the solution. But the problem asks for an open and dense subset of $\mathbb{R}$. No subset of the rationals would be open in $\mathbb{R}$. Oct16 comment Find a subset of the real numbers A good starting question for you: Are the rationals open in the real numbers? If not, then you cannot use a subset of them. Oct16 answered Decoding / reverse engineering math content Jul2 awarded Curious Apr28 revised Proof of integral inequality Poster did not know Latex. Apr28 suggested approved edit on Proof of integral inequality Apr21 comment Limit of a Cosine Sequence I chose the other for being very elementary, but I like this argument, it is a nice one to have in the back of my mind. Apr21 accepted Limit of a Cosine Sequence