Question

Let M be a sequence of the Hilbert space x . prove that M is dense in x iff ( Y⊥ M ⟹ y=0 )

Answer 1

Answer to Question #44670 - Math - Functional Analysis

Let $M$ be a sequence of the Hilbert space $X.$ Prove that $M$ is dense in X iff

$(Y\bot M)\Rightarrow y=0$

Consider that $M$ is dense in $X$. If it is, then $\overline{M}=X$. Then, $M\bot Y$ means that $(m,y)=0\forall m\in M\forall y\in Y$. As $\overline{M}=X$, then $\forall x\in X\exists m_{k}:m_{k}\rightarrow x$. Therefore, if we consider a sequence

$(m_{k},y),k\in\mathbb{N},$

then, because of the fact that dot product is a continuous function of one multiplier,

$(m_{k},y)\rightarrow(x,y)\quad\forall x\in X(k\rightarrow\infty).$

Therefore,

$(x,y)=0\quad\forall x\in X\forall y\in Y$

As $Y\subset X$, this equality can hold only in case when $y=0$, because $x$ changes through all possible values in $X$

www.AssignmentExpert.com