# Tag Info

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### Representation of isometry of Hilbert spaces with orthonormal basis

I agree with you. I didn't check your counterexample, but a simpler one is$$\Bbb C\to\Bbb C,\;z\mapsto\overline z.$$ See also Mazur-Ulam-like theorem for complex Hilbert spaces.
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### Function orthogonal to the square of any eigenfunction

Here is an argument that’s similar in ideas to my comments, and that proves the result for a large class of $V$: Let $\psi$ be the Gaussian density function appearing in all $u_j$. (Recall that $u_j$ ...
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### How can I show that $\|\,|A|\,|A^*|\,\|=\|A^2\|$?

We need the following identities: $\|B^* B\| = \|B B^*\| = \|B\|^2$. Using them several times we get: \begin{align*} \| \;|A|\; |A^*|\; \|^2 &= \| \; |A| \; |A^*|\; |A^*| \; |A| \;\|\\ & = \| \...
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1 vote
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### quotient space in inner product space

The assumptions in question are not inconsistent. The following is a concrete example. Let $H = l^2$, $W = \{e_0, \sum_{n=1}^\infty \frac{1}{n}e_n\}^\perp$. Then $W \subset H$ is closed and of co-...
• 10.6k
1 vote
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### Show that $\exp(D_s)$ converges strongly on $L^2$ to $T_1$ as $s \to 0$.

Taking the Fourier transform of $D_s(f)$ and letting $s\to 0$ will probably not work directly because $D_s(f)$ does not converge for all $f\in L^2$ as $s\to0$ (basically because not every such $f$ is ...
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### The direct sum of two closed subspace is closed? (Hilbert space)

Just want to point out that by adding an orthogonal condition, the statement is true: The direct sum of two orthogonal closed subspaces is closed. Let $M, N$ be orthogonal closed subspaces of a ...
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### Compact + absolute convergence of eigenvalues $\Rightarrow$ trace class?

Let $T: \ell^2\to \ell^2$ be defined by $$T(x_0,x_1,\ldots )=(0,a_0x_0,a_1x_1,\ldots )$$ where $a_n\to 0$ and $\sum |a_n|^2=\infty.$ Then $T$ is compact. Moreover $T$ does not admit any eigenvalues ...
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• 10.6k

### Show that function is positive definite

One way to factor the entire expression in pairs is by forming the standard inner product of the vector $$\frac1{\sqrt{2}}\,\begin{pmatrix} (1+i)\,z_2\\ (1-i)\,z_1 + 2z_2\end{pmatrix}$$ with itself. ...
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