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visits member for 2 years, 6 months
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Aug
20
asked Proving a function is not integrable
Aug
18
accepted Sequence of indicator functions is weakly convergent.
Aug
17
asked Sequence of indicator functions is weakly convergent.
Aug
14
accepted Riccati type equation
Aug
6
asked Riccati type equation
Aug
4
accepted Prove that the solution tends to $0$ as $t$ goes to infinity
Aug
4
comment Prove that the solution tends to $0$ as $t$ goes to infinity
Excellent.. thank you did..
Aug
4
comment Prove that the solution tends to $0$ as $t$ goes to infinity
I don't get your point about symmetry.. can you please provide more details?
Aug
4
comment Prove that the solution tends to $0$ as $t$ goes to infinity
I don't know.. form your proof $<0$ seems to suffice indeed.. i copied the text as i found it..
Aug
4
comment Prove that the solution tends to $0$ as $t$ goes to infinity
Where did you used that the matrix is simmetric? Was it just a redundant information?
Aug
4
asked Prove that the solution tends to $0$ as $t$ goes to infinity
Aug
4
accepted Prove that this is a Banach space
Jul
28
asked Prove that this is a Banach space
Jul
21
accepted Prove that this function is measurable
Jul
21
comment Prove that this function is measurable
Can you please provide more details in the first point of your argument? Still I cannot follow you completely even though you are convincing me you are right... Moreover, I can't see what is $j$ in the formula.. Of course later i will accept your answer.. Thank you Davide
Jul
21
asked Prove that this function is measurable
Jun
11
accepted $A$ has measure $0$
Jun
11
comment $A$ has measure $0$
you mean... $\sin(n_kx)\to\pm\frac{1}{\sqrt2}$?
Jun
11
comment $A$ has measure $0$
but what if, for example, $n_k=4k+1$ and $x=\frac\pi2$? In this case the limit exists and it is constantly $1$ so I cannot claim that if the limit exists then it is $0$. I don't get your point sorry...
Jun
11
asked $A$ has measure $0$