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Choose $N\in \mathbb{N}$ so that $\sum_{n\geqslant N} \|f_n-e_n\|^2 < 1$. We will show that $(f_k)_{k\geqslant N}$ is a basis for $\{e_k\colon k<N\}^\perp$, which will be enough. Let $f\in \{e_k\colon k<N\}^\perp$ and suppose that $(f,f_{n})=0$ for all $n\geqslant N$. I claim that $f=0$. Assume not. Then $$ \|f\|^{2}=\sum_{n\geqslant ...


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Answer. $X^\perp=Y=\big\{\{x_n\}\in\ell^2\,\,\,\text{where}\,\,\, x_{2n}=-x_{2n-1}/n\big\}$. a. Show that, if $x\in X$ and $y\in Y$, then $(x,y)=0$. b. Write any element of $\ell^2$ as a sum of elements of $X$ and $Y$.


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$i(A+1)$ is a symmetric tri-diagonal matrix on each of the subspaces spanned by the even and odd coordinates. These are equivalent to Jacobi matrices (real symmetric with positive off-diagonal elements). It is a classical result that these define essentially self-adjoint operators if the off-diagonal elements grow no more rapidly than $n$. By Theorem 2.7 of ...


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Hint 1: If $H$ is separable: Let $(e_i)$ be an orthonormal basis for $H$. Then for $x\in H$ and $N<\infty$ the function $y\mapsto \sum_{i=1}^N\langle y-x,e_i\rangle^2$ is weakly continuous, and so $\sigma(\tau^*)$-measurable. Also the limit $y\mapsto\|y-x\|^2=\lim_N \sum_{i=1}^N\langle y-x,e_i\rangle^2$ is $\sigma(\tau^*)$-measurable. Thus every ...



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