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Jan 19, 2018 at 17:12 comment added user118240 Thank you for your comments. I completely agree with it: just like all the sines $\{\sin i\pi x\}_{i=1}^\infty$ are complete in $L^2(0,1)$.
Jan 18, 2018 at 15:04 comment added Math604 you can also think of the simpler case with the Dirichelt boundary condition eigenfunctions; so they are zero on the boundary. Yet any $L^2(\Omega)$ function can be written as an infinite sum.
Jan 16, 2018 at 11:51 answer added Alexandre timeline score: 4
Jan 15, 2018 at 13:11 comment added user118240 Thanks for the comment. I would also expect so but it is a result proved on page 37 of the note (in the link), and I do not find any problem with the proof.
Jan 14, 2018 at 18:35 review Close votes
Jan 15, 2018 at 12:47
Jan 14, 2018 at 18:06 history asked user118240 CC BY-SA 3.0