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Since you seem really interested in the inequality $\rho(A)\le\|A\|$ ($\rho$ the spectral radius), here is a simple and elegant proof. In the 2nd edition of my book Matrices (Springer Verlag GTM216), …

The answer may depend upon which functional space you are dealing with. But since you insist upon the symbol and the Fourier transform, let me assume that you have $L^2({\mathbb R}^d)$ in mind. Becaus …

I presume that $V$ is real valued. Then the equation on the sphere is $L\psi=E\psi$, an eigenvalue equation for the self-adjoint operator $L$, with compact resolvant. Therefore the eigenvalues are rea …

If $A,A^T=\ell^p\rightarrow\ell^p$, then the adjoints $A^T,A$ map $\ell^{p'}$ into itself. By interpolation (Riesz-Thorin), they map $\ell^2$ into itself. It will be often the case that the spectrum o …

This is not the right way to think about operator norms. Instead, you can say that if $T:L^2\rightarrow L^2$ and $T:L^\infty\rightarrow L^1$ (as you consider) are bounded, then $T:L^p\rightarrow L^{p' …

I don't see which kind of condition you are looking for, as there are a lot of pairs $T,S$ such that $TS=\lambda ST$ and $\lambda\ne1$, even in finite dimension. Such pairs are said to $\lambda$-commu …

If $uV$ is real valued, then the convex set of complex functions $\psi$ with $\|\psi\|_{L^2}\le A$ ($A$ a given constant) is invariant under the flow. It is hard to find something else, because the S …

The most elementary case of Weyl's inequality says that, if $\lambda_i(S)$ denote the $i$th eigenvaue of the Hermitian matrix $S$ (increasing order), then $\lambda_i(S)+\lambda_1(T)\le\lambda_i(S+T)\l …

Here is a hint. Let me begin with a formal calculus. The Baker-Campbell-Hausdorff formula tells you that $$e^{-Y}e^{-X}e^{X+Y}\sim e^{\frac12[Y,X]},$$ where $[\cdot,\cdot]$ is the commutator. Applying …

The answer is negative, and this happens as soon as $n=2$. The question is whether the composition $X\mapsto L:=L_{X^2}$ is globally Lipschitz over ${\bf SPD}_n$. Let $x_j\in{\mathbb R}^n$ denote the …

The answer is No. Here is a counter-example: $$X=\begin{pmatrix} 9 & 3 \\ 3 & 1 \end{pmatrix},\qquad Y=\begin{pmatrix} 1 & 3 \\ 3 & 9 \end{pmatrix}.$$

Miscellaneous results. If $A$ is strictly upper triangular, then $x\cdot\nabla$ consists only is terms $x_j\partial_k$ with $j<k$. The action of $L$ over homogenous polynomials of degree $d$ is descr …

I don't have the answer to your question, but I can give you the following: $$\|\sqrt A-\sqrt B\|_\infty\le\sqrt{\|A-B\|_\infty\,}\,,$$ where the $\infty$-Schatten norm is nothing but the operator nor …