# Invertible products in Banach algebras

I found this interesting challenge: give an example of an unital Banach algebra that contains two elements $x$ and $y$ such that $xy$ is invertible but $yx$ is not invertible.

I thought it would be kind of simple to find such an example. Now, after some time, I'm not able to come up with one. Any hint?

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In the bounded linear operators on $\ell^2$ let $x$ and $y$ be left and right shifts respectively: $x: (s_1, s_2, s_3, \ldots) \mapsto (s_2, s_3, s_4, \ldots)$, $y: (s_1, s_2, s_3, \ldots) \mapsto (0, s_1, s_2, \ldots)$. Then $xy$ is the identity, but $yx (1,0,0,\ldots) = 0$.

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Wow! That was easy. Shifts never crossed my mind. Thanks, Robert. –  ragrigg Feb 5 '13 at 7:26
Let $X=\ell^1$. There are bounded linear maps $S$, $T:X\to X$ such that $S(x_1,\dots)=(x_2,\dots)$ (drop the first item) and $T(x_1,\dots)=(0,x_1,\dots)$ (add a zero in the front). Then $S$, $T$ are in the Banach algebra $B(X)$ of bounded linear maps $X\to X$.
Compute $ST$ and $TS$.