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 8h comment Normal Operators: Draft @Freeze_S : If you want a new construction, please post a new problem. 1d comment Normal Operators: Draft If $\{ x_{n} \}$ converges to $x$, and $\{ N(I+N^{\star}N)^{-1}x_n \}$ converges to $y$, then $\{ z_n=(I+N^{\star}N)^{-1}x_n\}$ converges to $(I+N^{\star}N)^{-1}x$ (why?) and $\{ Nz_n \}$ converges to $y$. Because $N$ is closed, then $y = N\lim_{n}z_n = N(I+N^{\star}N)^{-1}x$. Apr18 comment Normal Operators: Draft @Freeze_S : I added a different argument, but forgot to ping you. Apr18 comment What is the Fourier basis for all 2 dimensional functions? Technically, the $sin$ and $cos$ functions are not an orthonormal basis because they're not in $L^{2}(\mathbb{R})$. That's why I chose the example I did. (Plus, the set of Hermite functions is convenient for dealing with the Fourier transform because they are eigenfunctions of the Fourier transform on $L^{2}(\mathbb{R})$! The Fourier transform has 4 eigenvalues: $1,i,-1,i$ and this basis diagonalizes the Fourier transform.) Any true orthonormal basis of $L^{2}(\mathbb{R})$ must be countably infinite, and not uncountable infinite. Any cardinality can be a dimension for an orthonormal basis. Apr18 answered How to construct a continuous function that is (mean) convergent to a given square integrable function Apr18 revised Showing the range of Fourier transform on $L^1(\mathbb{R})$ is in $L^1(\mathbb{R})$ for a particular scenario added 7 characters in body Apr18 comment Finding the norm of a linear functional @Functionalanalysis : I forgot to ping you in the last remark. Apr18 comment What is the Fourier basis for all 2 dimensional functions? For example, the set of functions $\{ h_{n}=c_{n}e^{x^{2}/2}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\}_{n=0}^{\infty}$ is a complete orthonormal basis of $L^{2}(\mathbb{R})$, whether you are dealing with real or complex functions, provided the $c_n$ are chosen so that $\|h_{n}\|=1$ for all $n \ge 0$. (These are the Hermite functions.) Then $\{ h_{n}(x)h_{m}(y) \}_{n,m \ge 0}$ is a complete orthonormal basis of $L^{2}(\mathbb{R}^{2})$, which is the set of jointly measurable, square integrable functions on $\mathbb{R}^{2}$. This is a "tensor product" of bases. Apr18 comment Finding the norm of a linear functional A function $\chi_{S}$ is the characteristic function of the set $S$; it is $1$ for $x \in S$ and is $0$ otherwise. Every continuous linear functional $\Phi$ on $C[a,b]$ has the form $\Phi(f)=\int_{a}^{b}f(t)dg(t)$, where $g$ is a function of bounded variation; $g$ is unique when some normalization is required such as $g(a)=0$ and $g(t+0)=g(t)$ for $a \le t < b$. Then $\|\Phi\|=\|g\|$, where $\|g\|$ is the variation of the function of bounded variation $g$. If $g(t)=\int_{a}^{t}h(t)dt$, then it is well know that the variation of $g$ is $\|g\|=\int_{a}^{b}|h(t)|dt$. Apr18 revised Normal Operators: Draft added 1579 characters in body Apr18 comment Showing the range of Fourier transform on $L^1(\mathbb{R})$ is in $L^1(\mathbb{R})$ for a particular scenario @RakeshBalhara : I assume you straightened out the details? The Gaussian convolution with $f$ at $0$ converges to $f(0)$ if you only require that $f \in L^1$ and is continuous at $0$. And the Fourier transform identity holds very generally. If you have spotted a potential problem, I would be happy to look into it. Apr18 comment Showing the range of Fourier transform on $L^1(\mathbb{R})$ is in $L^1(\mathbb{R})$ for a particular scenario @Harish : The first expression is the Fourier transform solution of the heat equation $u_{t}=u_{xx}$ obtained by transforming this equation in $x$, and assuming initial condition $u(x,0)=f(x)$. That's how you get $e^{-ts^{2}}$ into the picture. Then, for $f,g \in L^{1}$, you always have $\int \hat{f}(s)g(s)ds = \int f(y)\hat{g}(y)dx$, which is a good identity to remember. All you have to do is apply with $g(s)=e^{-ts^{2}}e^{-isx}$. Properties of the heat equation give you $u(t,x)\rightarrow f(x)$ as $t\downarrow 0$ if $f$ is continuous. Apr18 answered Showing the range of Fourier transform on $L^1(\mathbb{R})$ is in $L^1(\mathbb{R})$ for a particular scenario Apr18 comment What is the Fourier basis for all 2 dimensional functions? What spaces, what norms? If you're working on $L^{2}(\mathbb{R}^{2})$, then you can normally mutiply an orthonormal basis in one variable by an orthonormal basis basis in the other. Apr18 answered Normal Operators: Draft Apr18 comment I need help understanding the proof of Lemma 2.4-1 from Kreyszig's Functional Analysis. I posted a different proof here: math.stackexchange.com/questions/1043581/… Apr18 answered A rather abstract strongly continuous semigroup Apr18 comment A question regarding Parseval's identity. Your aguments look sound. More generally: If $S$ is an orthonormal subset of an inner product space $X$, and if $M$ is linear span of elements of $S$, then the closure $M^{c}$ includes $x$ iff Parseval's equality holds for $x$ with respect to $S$. The proof works because you have something for $\sum_{s \in S}(x,s)s$ to converge to. (If $\sum_{s\in S}|(x,s)|^{2} < \|x\|^{2}$, then you can't necessarily form $\sum_{s\in S}(x,s)s$ if $X$ is not complete.) Apr16 answered Deflection of String Apr16 answered About the self-adjoint extension of an operator.