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comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
Cauchy integral formula.
comment Fourier Sequence Converges Uniformly Implies Almost Everywhere Pointwise Convergence
Yes, if the series converges uniformly to $g$, then $g$ must be continuous and, therefore, Riemann integrable. Uniform convergence also implies $L^2$ convergence. If you know that the Fourier series converges in $L^2$ to the original function $f$ (which you seem to know based on your proposed solution,) then you know that $f=g$ a.e. because $L^2$ limit functions are unique a.e..
comment Necessery and sufficient condition for existence of star cyclic vector for $M_\phi$
Do you realize you never defined $M_{\phi}$?
answered Approximate spectral decomposition
comment Find the solution of the Dirichlet problem in the half-plane y>0.
Do you mean as $|x|\rightarrow\infty$ instead of $|x|\rightarrow 0$?
answered Limit of sesquilinear forms is a sesquilinear form
comment Getting a Hermite polynomial expansion of Gaussian with given variance.
@Anirbit : Yes, but before it sounds too good, keep this in mind: uniform convergence on the plane is not enough to give you $L^2$ convergence. So when you want to put this into an integral and interchange limits, you have to start with functions that are compactly supported or are in $L^1$. There's something odd going on with the tails it seems, and you might expect that because of the increasing degree of polynomials involved in the sum.
comment Getting a Hermite polynomial expansion of Gaussian with given variance.
@Anirbit Actually, the Mehler kernel expansion converges uniformly to the bivariate Gaussian for $x,y\in\mathbb{R}$. The convergence is uniform on $\mathbb{R}\times\mathbb{R}$.
comment Getting a Hermite polynomial expansion of Gaussian with given variance.
@Anirbit : Uniform on compact subsets.
comment Operator Sum: Selfadjoint
@Freeze_S : You're welcome. Differentiation is a strange example. On $[0,1]$ there are these issues. On $\mathbb{R}$ it is selfadjoint. On $[0,\infty)$ it is symmetric with no selfadjoint extensions (f(0)=0) and you end up with a half-plane for the spectrum, which is why you get the Laplace transform.
answered Operator Sum: Selfadjoint
comment heat equation-uniqueness of solution
You are missing an initial condition $u(x,0)=f(x)$.
answered Monotone sequence of orthogonal projections on a complex Hilbert space
comment eigenvaue of Sturm Liouville problem
You're welcome @rosa .
answered Fourier transform of translation in $L^2$.
comment Evaluation of an integral associated with integral kernel of resolvent of Laplacian
Something is wrong in your first integral. Are you sure there isn't a negative in the exponent? Doesn't look very convergent to me.
comment Real Analysis, Folland problem 5.5.60 Hilbert Spaces
@Wolfgang : Your sketch of the proof looks correct to me. Is there a part of your plan that you're stuck on?
comment Continuity of functional calculus
Use the '@' symbol in front of the user name in order to ping them when you ask a question. Otherwise, they won't know you're asking them anything. This pings you because I'm posting under your question. If you post under someone's answer, then they will be pinged automatically, but they'll have to use @Kaushik to ping you back when they respond under their own answer.
comment Linear Algebra with functions
@user313448 Have you studied differential equations where they use the annihilator method? That's what I'm using. If not, you can do this with limits. Multiply by $e^{-3x}$ and let $x\rightarrow\infty$ in order to obtain $\gamma =0$. Then you can isolate $\beta=0$ and, finally, you isolate $\alpha=0$ with no limits.
answered Show $Tx=\sum^\infty _{n=1} \lambda_n \langle x , u_n \rangle v_n$ defines a bounded linear operator