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.

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If $K : D(K)\to Y $ is a continuous one-to-one operator and C is a subspace, $D(K)$ is a compact set, then the inverse operator is continuous.

If $K : D(K)\to Y $ is a continuous one-to-one operator and $C$ is a subspace and $D(K)$ is a compact set, then the inverse operator $(K\mid C)^{-1}$ is continuous? Here, $K\mid C$ denotes the restriction of $K$ to $C$ and $D(K)$ is the domain of…
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