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13m
revised Analytical solution to complex Heat Equation with Neumann boundary conditions and lateral heat loss
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20m
answered Analytical solution to complex Heat Equation with Neumann boundary conditions and lateral heat loss
5h
answered Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$.
13h
comment Show that an operator of rank $n$ can have at most $n$ nonzero eigenvalues.
If you have distinct non-zero eigenvectors $\{ \lambda_1,\cdots,\lambda_k \}$ with corresponding non-zero eigenvectors $\mathcal{E}=\{ x_1,\cdots,x_k\}$, can you find a way to show that $\mathcal{E}$ is a linearly independent set of vectors?
19h
answered Spectrum of double infinite shift using isometry to Fourier series
1d
comment Range of normal operator and its adjoint are equal
@Blazej : You're welcome.
1d
awarded  analysis
1d
answered prove that $u$ is equal a.e. to an absolutely continuous function
2d
answered Are the coefficients on the Fourier transform arbitrary?
2d
answered Range of normal operator and its adjoint are equal
2d
answered History of convolution
Jul
3
comment Operator theory curiosity
@Lukkio : So long as you have a deterministic algorithm, you have some kind of operator (maybe linear, maybe not) that takes data in some form $F_1$ to data in some form $F_2$ after application of the algorithm. Isolating some nice properties of the operator using properties of the algorithm is where you hope to gain traction.
Jul
3
comment How to evaluate this combination of sums and integrals?
(1) For fixed $m$, $\int_{0}^{1}\sin(n\pi x)\sin(m\pi x)dx$ is $0$ if $n \ne m$ and is $1/2$ if $n=m$. (2) You can show that $\{ \sqrt{2}\sin(n\pi x)\}_{n=1}^{\infty}$ is a complete orthonormal basis of $L^{2}[0,1]$, which can be deduced from the fact that $\{ \sqrt{2}e^{in\pi x} \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^{2}[-1,1]$.
Jul
3
comment Sot convergence of a sequence of operators implies uniform convergence
Have you considered the possibility that what you want to prove without the assumption of uniform convergence is false? For example, let $H=L^{2}[0,1]$ and let $A_{n}f = \chi_{[1-1/n,1]}f$. Then $\|A_{n}\|=1$ for all $n$ and $A_{n}f \rightarrow 0$ as $n\rightarrow \infty$ for all $f\in H$. However, what is the norm of $\|A_{n}-0\|$? (By the way, this comes from looking at a spectral measure.)
Jul
3
comment Sot convergence of a sequence of operators implies uniform convergence
You have a statement to prove: "If (...), prove ...". The condition (...) is the assumption of uniform convergence on the unit ball. So, the proof is that simple.
Jul
3
comment Orthonormal Basis of $L^2$
@Naji : Change the first sum to $\sum_{k=-\infty}^{\infty}$ and the finite sum to $\sum_{k=-n}^{n}$ or something like that.
Jul
3
comment Sot convergence of a sequence of operators implies uniform convergence
Given $\epsilon > 0$, there exists $N$ such that $\|A_{n}f-Af\| < \epsilon/2$ whenever $n \ge N$ and $\|f\| =1$. Therefore, $\|A_{n}-A\|=\sup_{\|f\|=1} \|A_{n}f-Af\| \le \epsilon/2 < \epsilon$ whenever $n \ge N$.
Jul
3
comment Operator theory curiosity
@Lukkio : Abstract operator that takes the data used in your algorithm and gives you new data after an application of the algorithm.
Jul
3
comment Sot convergence of a sequence of operators implies uniform convergence
You have assumed more than sot convergence. You have assumed uniform sot convergence on the unit disk. That's very different. Again, write out what the condition you stated actually means.
Jul
3
answered Question about the Fourier Inversion Formula