A Property on Unital Normed Algebra

Question

I want to prove that: For any unital normed algebra, there do not exist two elements $x_1$ and $x_2$ so that $x_1x_2-x_2x_1=e$, where $e$ is the unit in that algebra.

I notice that the algebra has to be non-commutative (or otherwise the conclusion is trivial) and that one can utilize some multiplicative linear functional (whose existence is dubious).

Answer 1

Here's a beautiful proof courtesy of Rudin. We will use the fact that if there exists $x,y$ such that $xy-yx=1$ in a unital normed algebra, then for all $n$ $$xy^n-y^nx=ny^{n-1}$$

If this is true, then $$n||y^{n-1}||=||ny^{n-1}||=||xy^n-y^nx||\le 2||x||\ ||y^n||\le 2||x||\ ||y||\ ||y^{n-1}||$$ which implies that $$n\le 2||x||\ ||y||$$

But since $n$ is arbitrary, choose $n$ large enough to violate this bound and we have a contradiction.

All that remains is to prove the fact mentioned above. It follows via induction. The $n=1$ case is the assumption of our problem. Let it hold for some $n=j$. Then for $n=j+1$

\begin{align*}xy^{j+1}-y^{j+1}x&=xy^{j+1}-y^jxy+y^jxy-y^{j+1}x\\ &=(xy^j-y^jx)y+y^j(xy-yx)\\&=jy^{j-1}y+y^j1\\&=jy^j+y^j\\&=(j+1)y^j \end{align*}

and the proof is complete.

Answer 2

We may assume the algebra is complete, as any normed algebra can be completed. The reasoning (I do not remember the source) is based on relations between spectra of elements. For $xy-yx=e$ we have $$\sigma(yx)\cup \{0\}=\sigma(xy)\cup\{0\}=[\sigma(yx)+1]\cup \{0\}$$ Denoting $A=\sigma(yx)$ we get $A\cup\{0\}=[A+1]\cup \{0\}.$ No bounded nonempty subset may satisfy that property.