Wielandt theorem for unital normed algebras

Theorem. (Wielandt (1949)) Let $A$ be a normed unital algebra (with unit $e$). If $x,y\in A$ then $xy-yx\ne e$.

Proof.

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

$x^{n}y-y{x}^n$$=n{x}^{n-1}\ne 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 $x^n\ne 0$ and

$x^{n+1}y-y{x}^{n+1}$$=x^n(xy-yx)+(x^{n}y-y{x}^n)x$

$=x^{n}e+n{x}^{n-1}x=x^{n}e+n{x}^n=(n+e)x^n$

From this identity it follows that

$n\parallel x^{n-1}\parallel$$=\parallel x^{n}y-y{x}^n\parallel \le 2\parallel x^n\parallel \parallel y\parallel \le 2\parallel x^{n-1}\parallel \parallel x\parallel \parallel y\parallel$

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

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

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

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

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