Wielandt theorem for unital normed algebras

(Wielandt (1949)) Let $A$ be a normed unital algebra (with unit $e$). If $x,y\in A$ then $xy-yx\not=e$.

Proof.

Assume there are $x,y\in A$ such that $xy-yx=e$. Then for all $n\in\mathbb{N}$ we have

 $\displaystyle x^{n}y-yx^{n}$ $\displaystyle=nx^{n-1}\not=0$

We prove this by induction over $n\in\mathbb{N}$. It holds for $n=1$ by assumption. Assume it is valid for $n\in\mathbb{N}$. Then $x^{n}\not=0$ and

 $\displaystyle x^{n+1}y-yx^{n+1}$ $\displaystyle=x^{n}(xy-yx)+(x^{n}y-yx^{n})x$ $\displaystyle=x^{n}e+nx^{n-1}x=x^{n}e+nx^{n}=(n+e)x^{n}$

From this identity it follows that

 $\displaystyle n\|x^{n-1}\|$ $\displaystyle=\|x^{n}y-yx^{n}\|\leq 2\|x^{n}\|\|y\|\leq 2\|x^{n-1}\|\|x\|\|y\|$

It follows that $n\leq 2||x||||y||$ for all $n\in\mathbb{N}$ which is impossible. ∎

Corollary. The identity operator on a Hilbert space $\mathcal{H}$ cannot be expressed as a commutator of two bounded linear operators in $\mathcal{L}(\mathcal{H})$.

Remark. The above can be understood as a version of the uncertainty principle in one dimension. Let $H=L^{2}(\mathbb{R})$. Let $q\colon H\to H$ be $q(f)(x):=xf(x)$ with $D(q)=\{f\in L^{2}(\mathbb{R}):x\mapsto xf(x)\in L^{2}(\mathbb{R})\}$, the coordinate operator and $p\colon H\to H,p(f)(x)=-if^{\prime}(x)$ the momentum operator with $D(p):=\{f\in L^{2}(\mathbb{R}):f\ \text{absolutely \ continuous},f^{\prime}\in L% ^{2}(\mathbb{R})\}$. It follows that

 $\displaystyle pq-qp$ $\displaystyle=-i\mathrm{id}_{D}\ \text{on}\ D=D(q)\cap D(p)$

According to the corollary $D=L^{2}(\mathbb{R})$ can never be the case.

Title Wielandt theorem for unital normed algebras WielandtTheoremForUnitalNormedAlgebras 2013-03-22 19:01:22 2013-03-22 19:01:22 karstenb (16623) karstenb (16623) 6 karstenb (16623) Theorem msc 46H99