# Tagged Questions

For question involving Hilbert spaces, that is, complete normed spaces whose norm comes from an inner product.

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### Let $C$ be a convex closed nonempty subset of a Hilbert space $H$. Show that there is a unique element in $C$ with the minimum norm.

Let $H$ be a Hilbert space and $C \subset H$ be a convex, closed and nonempty subset of $H$. Prove that there exists a unique $x_0\in C$ with minimum norm among the elements of $C$. I don't know ...
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### Sequence in hilbert space, mutually orthogonal vectors

Let $y_1,y_2,\cdots$ be a sequence in a Hilbert space. Let $V_n$ be the linear span of $\{y_1,\cdots,y_n\}$. Assume that $||y_{n+1}||\leq ||y-y_{n+1}||$ for all $y\in V_n$ for $n=1,2,3,\cdots$. Show ...
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### positive operator, projection on Hilbert,$Q|T|Q \ge |QTQ|?$

Let $T$ be an operator on a Hilbert space $H$. And $Q$ be a projection. Whether $$Q|T|Q \ge |QTQ|?$$ Obviously, if $T$ is positive, then $Q|T|Q = |QTQ|$. Also, there are some $T$ such that $QTQ=0$ ...
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### Compact Operator with Infinite rank Doesn't have a Closed Image

Let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space. Claim: A compact operator $T$ which has infinite-rank has an image that isn't closed. I'm trying to prove this claim but I'm ...
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### Several questions about the proof of this lemma

I am reading this paper. I have Several questions about the proof of the lemma 3.3. How to Prove 1,2,3,4?
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### Prove that $(A^\bot)^\bot = \overline{\operatorname{span}A}$

I proved that $(A^\bot)^\bot \supset \overline{\operatorname{span}A}$. To prove the reverse inclusion, let $x \in (A^\bot)^\bot$, then $\langle x, y\rangle = 0$ for all $y \in A^\bot$. How can I ...
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### Parameterizing the set of subquotients of a Hilbert space

If you want to parameterize the set of subspaces of a finite-dimensional Hilbert space $V$, and naturally induce a norm derived from $V$ on the resulting moduli space, the classical approach is to ...
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We were recalling some basic stuff about Hilbert spaces in class, and the professor gave the remark: If $H$ is a Hilbert space and $V$ is a finite dimensional vector subspace of $H$, then for all $f\... 1answer 42 views ### Why is it true that the multiplication operator in a reproducing kernel Hilbert space is always continuous? In my functional analysis I was met with this seemingly trivial theorem on RKHS If$ \mathbb{H} $is a reproducing Kernel Hilbert Space and we have a multiplier$ \phi $meaning it satisfies$ \...
We know that if $A$ is a self-adjoint unbounded operator on a Hilbert space $(H;\left<.,.\right>)$ then $\sigma(A) \subset \mathbb R$. Now, how it can be shown that if $A$ is more positive i.e. ...