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Trial And Error.

The picture is a portrait of a famous Mathematician who was often criticized for not being rigorous enough in his new inventions.

"Anyone who has never made a mistake has never tried anything new." - Albert Einstein


5h
comment Misunderstanding about Laplace operator
$A$ is bounded with dense range; it is injective, but it is not surjective. That means that the inverse $\Delta$ has dense domain and is surjective. Both are closed, but there is no contradiction because the inverse is only densely-defined and, so, is not defined on the full complete space.
5h
comment Periodic Laplace operator non closed in $ C^2(0,L)$
@E.Hilbert : Directly check with Calculus.
1d
answered proving Orthonormal basis
1d
answered Periodic Laplace operator non closed in $ C^2(0,L)$
2d
answered Intuition $C^*$-identity
Aug
29
answered Riesz Functional Calculus vs. Holomorphic Functional Calculus
Aug
29
comment Positive Linear Transformations: What good for?
@Freeze_S : $A$ is semibounded below by $m$ iff $((A-mI)x,x) \ge 0$. The Friedrichs extension came out of Friedrichs resolving a question of John von Neumann in the affirmative: If $A$ is a symmetric and semibounded below by $m$, is there a selfadjoint extension with the same lower bound? von Neumann could show that it could be done with any bound arbitrarily close, but not the same. And this is useful for Quantum, which is where von Neumann was working at the time.
Aug
29
answered Positive Linear Transformations: What good for?
Aug
29
revised Condition under which a locally convex topological vector space becomes a normed linear space
added 15 characters in body
Aug
29
answered Condition under which a locally convex topological vector space becomes a normed linear space
Aug
28
answered Selfadjoint Operator: Empty Spectrum
Aug
28
comment Selfadjoint Operator: Empty Spectrum
@Freeze_S : I added information for you about numerical range, but I assume the spectral theorem.
Aug
28
revised Selfadjoint Operator: Empty Spectrum
added 1218 characters in body
Aug
28
answered Selfadjoint Operator: Empty Spectrum
Aug
27
answered Norm of functional associated to vector $p$-norm
Aug
26
comment Banach Spaces: Totally Bounded vs. Bounded
@Freeze_S : Adding to what JHance had to say, a set $K$ in a metric space is compact iff $K$ is complete and totally bounded.
Aug
26
comment Banach Spaces: Totally Bounded vs. Bounded
@Freeze_S : Yes, the Riesz lemma shows that the unit ball of an infinite-dimensional normed space is not totally bounded, even though it is clearly bounded.
Aug
26
answered Banach Spaces: Totally Bounded vs. Bounded
Aug
26
comment When is $V=U\oplus U^{\perp}$?
@Bombyxmori : You are welcome.
Aug
26
comment When is $V=U\oplus U^{\perp}$?
I like your criteria. Where did you learn of it?