# Questions tagged [hilbert-spaces]

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

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• 7,514
1 vote
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### Every closed convex subset of a finite-dimensional subspace of real inner product space is Chebyshev.

In my post I was now able to prove the following theorem : If $K$ is a closed convex subset of an real inner product space $X$ that is contained in a complete subset of $X$, then $K$ is Chebyshev.. ...
22 views

### Spectrum of Compact Hermitian Operator Accumulating at 0 From Both Directions

I'm getting into spectral theory and I'm wondering if someone could give an example of a Hilbert space $H$ and a compact hermitian operator $T$ on $H$ whose spectrum $\sigma(T)$ has both a positive ...
• 493
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### Separable Hilbert spaces, total orthonormal sets and Schauder basis in Banach spaces

A Schauder basis for a Banach space $X$ is called the sequece $\{e_n\}$ where $x\in X$ can be expanded as: $$x=\sum^\infty_{k=1} a_ke_k$$ A total orthonormal basis in a Hilbert space $H$ is a Schauder ...
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### Differentiability (and other properties) of functions mapping into countable product of Hilbert spaces

I have Hilbert spaces $\{H_i\}$ indexed by the natural numbers, so a countable set. Let me define a function $$f\colon X \to H_1\times H_2 \times H_3 \times ...$$ by $$f(x) = (f_1(x), f_2(x), ...)$$ ...
• 73
1 vote
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• 373
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### Does the spectrum of $T(1+T^*T)^{-1}$ contain a compactification of that of $T$?

Let $T$ be a densely defined closed operator in the Hilbert space $\mathcal H$. Then $T^*T\ge0$, and $Q = T(1+T^*T)^{-1}$ has norm $\le1$ (Theorem 13.13 of Rudin's Functional Analysis). Define the ...
• 736
1 vote
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### There exists a set of complex numbers $\alpha_n:n\in\mathbb{Z^*}$ such that $x=\sum_{n\in\mathbb{Z^*}}\alpha_ne^{nx}$?

Note this is an infinite series! I have the same question for $1$, is it possible to have $1=\sum_{n\in\mathbb{Z^*}}\beta_ne^{nx}$ for some complex coeficients $\beta_n\in\mathbb{C}$? I am considering ...
• 141
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### Limits of sequences of nonlinear "bounded" operators

I have a sequence of nonlinear operators $T_n\colon X \to Y$ between two separable Hilbert spaces. The operators are bounded uniformly in that there is a constant $C$ with $$|T_n(x)| \leq C|x|.$$ I ...
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• 399
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### Limits of Compact Operators: a simple question about the diagonal trick.

Theorem. Let $H$ and $K$ be Hilbert spaces, and assume that $T_n\colon H \to K$ is compact for each $n \in \mathbb{N}$. If $T \in B(H,K)$ is such that $\lVert T − T_n\rVert \to 0$, then $T$ is compact....
• 1,070
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### Prove that a space is a Hilbert Space with inner product different to the usual inner product used

I'm working with following academic paper : Stability of the solutions of differential equations whose author is Bernard Beauzamy. I'm trying prove that the two next spaces $\mathcal{B}_2$ y $P_2$ are ...
• 123
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### "Every linear subspace of a closed space is closed"

I am wondering how to show something that my book writes about $L^2$ and Hilbert spaces. First I will talk about what my book does: It defines what a random variable in $L^2$ is. It shows that $L^2$ ...
• 1,476
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### For quadratic functions on arbitrary Hilbert spaces, does strict convexity still imply strongly convexity?

Let $\mathcal{H}$ be a separable real Hilbert space. Let $q:\mathcal{H} \to \mathbb{R}$ be a bounded quadratic function. By this, I mean $q(x)=(Qx,x)+(b,x)+c$ for some self-adjoint bounded linear ...
• 2,976
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### On the preimages under injective operators of a sequence of subspaces of a Hilbert space tending to 0

Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two separable Hilbert spaces and let $A_n:\mathcal{H}_1 \rightarrow \mathcal{H}_2$ be a sequence of uniformly bounded injective linear operators with a ...
• 515
1 vote
42 views

### Why weak convergence doesn't Imply strong convergence on $\infty$- dimensional Hilbert spaces.

To start off, I know this is wrong. Im hoping someone can explain to me where Im going wrong. I know that on a finite dimensional Hilbert space that weak convergence implies strong convergence. But I ...
18 views

### On strong convexity of the squared norm of an affine function.

Let $V,W$ be Hilbert spaces. Let $f:V \to W$ be a bounded affine function. That is, $f(x)=A(x)+b$ for some bounded linear function $A:V \to W$ and fixed $b \in W$. Since $V$ is a Hilbert space, the ...
• 2,976
216 views

### Operator norm of sum of tensor products

Let $A,B,C,D$ be bounded operators on a Hilbert space $\mathcal{H}$. I know that $$\|AB\| \leq \|A\otimes B\|$$ where $\otimes$ is the tensor product and $\|\cdot\|$ is the operator norm. I wonder ...
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### A is self-adjoint and $\|Ax\|=\|A\|$. Show x is an eigenvector for $A^2$.
Let $H$ be a Hilbert space and let $A$ be a bounded self-adjoint operator on $H$. Suppose that there exists a unit vector $x \in H$ such that $||Ax||=||A||$. i) Show that $x$ is an eigenvector of $A^2$...