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 2h answered Inverse of $I +T^*T$ 13h answered Laplace transform of the wave equation 14h accepted How to show Legendre Operator $L_{m}=-\frac{d}{dx}(1-x^{2})\frac{d}{dx}+\frac{m^{2}}{1-x^{2}}$ is Selfadjoint? 14h comment $\oint_{C}(A-\lambda I)^{-1}\,d\lambda=0$ implies interior of $C$ is in the resolvent. Thank you. I almost forgot about this question, and to mark your answer. 14h accepted $\oint_{C}(A-\lambda I)^{-1}\,d\lambda=0$ implies interior of $C$ is in the resolvent. 1d comment Laplace transforms to solve heat equation @copper.hat : Could be. I see how you're thinking now. 1d comment Laplace transforms to solve heat equation @copper.hat : The Laplace transform would be in the t variable, not in x. 2d comment Showing that an operator generates a contraction semigroup @user3482534 : $\Re(Dx,x) \le \|D\|\|x\|^{2}=\|D\|\Re(x,x) = \Re ((\|D\| I)x,x)$ 2d comment Exponential decay estimate involving $C_{0}$-semigroup and the principle eigenvalue @user3482534 : If you have a continuously differentiable function $F(t) \ge 0$ on $[0,\infty)$ and $F'(t) \le 0$, then $0 \le F(t) \le F(0)$ for all $t \ge 0$. 2d comment Showing that an operator generates a contraction semigroup @user3482534 : The real part of the inner product is bounded by the absolute value of the inner product, which, by Cauchy-Schwarz, is bounded by the products of norms of vectors. 2d comment What do $a_0$ ,$a_m$ and $b_m$ terms mean in the Fourier series formula? The Fourier series can represent the signals over a period of the Fourier series. After that, the signal being represented will not agree, unless the signal being represented has the same periodicity. You can represent a voice snippet over an interval $[0,T]$ and the Fourier series signal will repeat itself periodically after that. 2d comment What do $a_0$ ,$a_m$ and $b_m$ terms mean in the Fourier series formula? It's an offset, a center point about which everything vibrates. 2d comment How to use the Spectral Theorem to Derive $L^{2}(\mathbb{R})$ Fourier Transform Theory @GiuseppeNegro : your expression would be a special case where a function has one singularity at $x=0$. Stone's formula deals with the resolvent of a selfadjoint operator, which may have a continuum of singularities through the spectral measure. May25 comment What do $a_0$ ,$a_m$ and $b_m$ terms mean in the Fourier series formula? @pandu : an offset. May25 comment Proving identity of a $C_0$ semigroup @Leibniz1337 : An antiderivative is unique only up to a constant. I chose the constant so that one of the evaluation terms would vanish; I wanted the evaluation term at $s=t$ to vanish. It's a standard trick for deriving a truncated Taylor series. $\frac{d}{ds}(s-t) = \frac{d}{ds}s$. May25 comment solve integral equation using the theory of compact operator @user3382078 : The functions $1,\cos x,\cos 2x,\cdots$ are orthogonal. The only ones appearing on the right are $1,\cos 2x$. May24 answered Proving identity of a $C_0$ semigroup May24 comment Exponential decay estimate involving $C_{0}$-semigroup and the principle eigenvalue @user3482534 : $\inf\{ \lambda : \lambda\in\sigma(-\Delta)\}$. The inf is always a member of the spectrum because the spectrum is always closed. May23 comment Prob. 8, Sec. 3.5 in Erwin Kreyszig's Introductory Functoinal Anlaysis With Applications If you have a sequence $x_{n}=\sum_{k=1}^{m_{n}}a_{nk}e_k$ that converges to $x$, then $\|x-P_{m_{n}}x\| \le \|x-x_{n}\|$ by the previous remark, which means that $\|x-P_{l}x\| \le \|x-x_{n}\|$ for all $l \ge m_{n}$. That's enough to see that $\lim_{l}\|x-P_{l}x\|=0$. May23 comment Exponential decay estimate involving $C_{0}$-semigroup and the principle eigenvalue If $A$ is selfadjoint with minimum spectral element $\lambda_1$, then $(Ax,x) \ge \lambda_(x,x)$. In your case, $-\Delta$ has minimum spectral element $\lambda_1$. So $\lambda_1(f,f) \le (-\Delta f,f)$.