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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


7h
comment Existence of a semigroup of bounded operators which is not $C_0$
Try to find a non-zero discontinuous function $E : \mathbb{R}\rightarrow\mathbb{R}$ such that $E(x+y)=E(x)E(y)$ for all $x,y\in\mathbb{R}$. Then you'll have what you want on $X=\mathbb{R}^{1}$.
7h
answered Linear operator in $\ell^2$
13h
comment Differential Equation - Water dripping from roof
@Simon : If you multiply $y'+ky$ by $e^{kt}$, then you get $\frac{d}{dt}(e^{kt}y(t))=e^{kt}(y'+ky)$.
22h
comment Differential Equation - Water dripping from roof
@Simon : I have added more detail now.
22h
revised Differential Equation - Water dripping from roof
added 256 characters in body
1d
answered Differential Equation - Water dripping from roof
1d
comment Existance of inverse of an operator
What if you take $A=0$? Do you still find $(T(t_{0})-I)$ to be invertible?
1d
comment Formal definition of “node” with respect to eigenvalues and functional analysis.
@MeLoco : use the 'at' ping if you want someone to get your message. Selfadjoint guarantees eigenfunction expansions (integrals and/or sums.) Assuming limit-point case at $\pm \infty$: If you want to work on $(-\infty,+\infty)$, then you need $f, Lf\in L^{2}$ and that's all. If you split this into $[a,\infty)$, then the only condition you need other than $f, Lf \in L^{2}$ is $A f(a)+Bf'(a)=0$ where $A$, $B$ are real and not both $0$. Both cases give you selfadjoint operators.
1d
comment Formal definition of “node” with respect to eigenvalues and functional analysis.
@MeLoco : If you assume limit circle case at both $\pm \infty$, then I don't think you'll have to split the interval. In that case, then I believe that all the eigenfunctions (even for real $\lambda$) should be in $L^{2}$. But you probably want the limit point case at both endpoints so that the operators are essentially selfadjoint on compactly supported functions; in that case there are no guarantees that any eigenfunction with real eigenvalue is ever in $L^{2}$ at $+\infty$ or at $-\infty$.
1d
revised Characterisation of one-dimensional Sobolev space
added 52 characters in body
1d
answered Characterisation of one-dimensional Sobolev space
2d
comment Formal definition of “node” with respect to eigenvalues and functional analysis.
@MeLoco : Okay, so you are working on $\mathbb{R}$. You need the limit point case at $\pm\infty$ in order to not require conditions (limit circle requires conditions.) So I assume that's what you mean. And $q$ has a very special form, which is why you can find an eigenfunction with eigenvalue $0$.
Aug
19
comment Formal definition of “node” with respect to eigenvalues and functional analysis.
Something is missing in your statement. What does "=" mean in your quote, and what is $\varphi'$ in $L\varphi'=0$. In a Physics context, "nodes" may mean zeros of the eigenfunction. You haven't specified the interval on which you are working, the endpoint conditions, or the nature of $q$. That leaves a lot in doubt.
Aug
19
accepted Finite-Dimensional Subspaces Invariant under Differentiation
Aug
19
answered Finite-Dimensional Subspaces Invariant under Differentiation
Aug
18
comment Is it possible to use complex logarithm to integrate $1/(z+i)$ along a path?
@helplessKirk : Continuous and piecewise smooth is enough to make the argument go through.
Aug
18
answered Eigenvalues of symmetric elliptic operators
Aug
16
answered Is it possible to use complex logarithm to integrate $1/(z+i)$ along a path?
Aug
16
comment Proving unitary inequivalence
The elements of the Spectral Theorem are unique. If $A$ is a densely-defined selfadjoint operator, then there exists a unique representation $A=\int \lambda dE(\lambda)$, where $E$ is a Borel orthogonal projection-valued measure satisfying $E(\mathbb{R})=I$, $E(\emptyset) = 0$, $E(S)E(T)=E(S\cap T)$.
Aug
16
answered Poles of Fourier transform