Questions tagged [hilbert-spaces]

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

Let $H$ a vector space over the field $\mathbb C$, and $\langle \cdot,\cdot\rangle\colon H\times H\to \mathbb{C}$ a map which satisfies

  1. $\langle x,x\rangle =0\Longrightarrow x=0$ and $\langle x,x\rangle\geqslant 0$ for all $x\in H$,
  2. $(\forall x,y\in H):\langle x,y\rangle=\overline{\langle y,x\rangle}$,
  3. $(\forall x_1,x_2,y\in H)(\forall\alpha_1,\alpha_2\in\mathbb C):\langle \alpha_1 x_1+\alpha_2 x_2,y\rangle=\alpha_1\langle x_1,y\rangle+\alpha_2\langle x_2,y\rangle$.

The map $\lVert\cdot\rVert\colon H\to\mathbb R_+$, defined by $\lVert x\rVert =\langle x,x\rangle^{\frac 12}$ is a norm.

If $(H,\lVert \cdot\rVert)$ is complete, then $H$ is called a Hilbert space.

Example: The space $H$ of all sequences $x_0,x_1,x_2,\ldots$ of complex numbers such that $\sum_{n=0}^\infty|x_n|^2<\infty$, with the inner product $$\bigl\langle(x_0,x_1,x_2,\ldots),(y_0,y_1,y_2,\ldots)\bigr\rangle =\sum_{n=0}^{+\infty}x_n\overline{y_n}$$is a Hilbert space.

8254 questions
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orthogonal subspaces in a Hilbert space

Is it true that if $A,B$ are closed subsets of a Hilbert space $H$, such that $A\perp B$, we have $A+B+(A\cup B)^{\perp} =H$ ? What if $A,B$ are closed subspaces ?$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
user38178
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Hilbert Space, showing a sequence in Cauchy

Suppose $X$ is a Hilbert space, $M\subset X$ is a closed subspace and $y\notin M$. Let $d = \inf\{ \|x-y\|:x\in M\}$ show that if $\{x_n\}_{1}^{\infty}$ and $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ then $\{x_n\}$ is Cauchy. Attempted proof -…
Wolfy
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The special case of the Riemann lebesgue lemma

I'm trying to prove the following Let $A$ be a measurable subset of $[0,2\pi]$ $$\lim_{n\to \infty} \int_A e^{inx} \, dx=0$$ There is a hint "this is the special case of the Riemann Lebesgue lemma" But I have no idea how to approach this problem…
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not understanding a step in a proof

Hi: I'm reading some introductory notes on hilbert spaces and there is a step in a proof that I don't follow. I will put the exact statement below. If someone could explain how it is obtained, it's appreciated. Note that commans between two terms…
mark leeds
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Hilbert space projection theorem: how to finish my proof?

The Hilbert space projetion theorem is the following theorem: Let $H$ be a Hilbert space and $C$ any closed convex subset. Then for $h \in H$ there exists a unique $c_0 \in C$ such that $\|h-c_0\| = \inf_{c\in C}\|h-c\|$. Please could someone show…
user167889
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Continuous spectrum of $R+L$, where $R$ and $L$ are the right and left shift of sequences in $l^2$

Consider the Hilbert space $l^2$ and the Left and Right-shift operator \begin{align*} L(x_1,x_2,\cdots) &= (x_2,x_3,\cdots)\\ R(x_1,x_2,\cdots) &= (0,x_1,x_2,\cdots ) \end{align*} I'm trying to find out the continuous spectrum of $R+L$, as I've…
fabiana
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A bilinear form whose corresponding quadratic form has a larger norm

I know that given a bilinear form $B$ and it's corresponding quadratic form $Q$, assuming that $B$ is complex and bounded, that $\|B\| \le 2\|Q \|$ However, I've failed to manufacture an example of a bilinear form $B$ where this inequality is…
Shai Deshe
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Let $H$ be a Hilbert space, $A$ is unitary and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$?

Let $H$ be a Hilbert space, and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$? What I have is if $x\in S^{\perp}$ then $x\perp A(A^*A^*Ax)$ then $(x,A^*Ax)=(Ax,Ax)=0$, so $x$ is in $\operatorname{Null}(A)$, is there a more obvious…
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is there a convex bounded subset A of H such that A is not norm closed and A∩L is norm closed for every finite dimensional subspace L of H

"Given an infinite dimensional Hilbert space H. Show that there is a convex bounded subset A of H such that A is not norm closed and A∩L is norm closed for every finite dimensional subspace L of H." This question was my exam question. I cannot give…
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Basis of intersections of $L^p$ spaces

I keep confusing myself about a subspace basis and I can only find intelligible material discussing the finite, linear algebra, case. It is known that the Hilbert space $L^2(X)$ has a basis, for example given by Fourier modes when $X=[-L,L]$. …
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$\|T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|$ for complex Hilbert spaces

Let $H$ be a hilbert space over the complex numbers. Let $T:H\to H$ be a linear bounded operator prove that $$\| T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|.$$ Obviously $\sup_{\| f\|=1} |\langle Tf,f\rangle| \le \| T\|$ by C-S inequality. But how…
Agumon
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Prove that if $X$ is a Hilbert space, then $B(X)$ is not a Hilbert space

I`m having a homework question that goes like this: X is a Hilbert space, a complete inner product space, show that B(X) is not a Hilbert space. I`m quite stuck and I would love to understand this problem! My attempt so far: Let $T \in B(X)$ be a…
Pim
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Showing that $R(T)=R(T^*)$ for a normal operator $T$

For a normal operator $T$ acting on a Hilbert space it is easy to show that the kernel of $T$ coincides with the kernel of the adjoint $T^*$. Thus the norm-closures of the ranges $R(T)$ and $R(T^*)$ coincide. Using the spectral theorem for normal…
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product of Hilbert spaces

Let $H$ be an infinite dimensional Hilbert space. claim: $H\times H$ with the norm $\|(x,y)\|=\|x\|+\|y\|$ is an Hilbert space. I can't find a counterexample..
sky90
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Orthogonal complement properties in Hilbert spaces

We have a Hilbert space $H$ and $x_0\in H$. We need to show that if we let $V$ be a closed subspace of $H$, then $$ \min\{\|x-x_0\|\,:\,x\in V\}=\max\{|\langle y,x_0\rangle|\,:\,y\in V^\perp,\,\|y\|=1\}. $$ I have a good intuitive idea of why this…
Laars Helenius
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