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Does there exist any unital normed algebra $(A,\|\cdot\|)$ enjoying another norm $\|\cdot\|_1$ such that

  1. $(A,\|\cdot\|_1)$ forms a unital normed algebra with the same unit.

  2. Any element contained in the intersection
    $$ \{x\in A : \|x-1\|<1\}\bigcap \{x\in A : \|x-1\|_1<1\} $$ is in the form of $\alpha.1$ where $\alpha$ is a complex number with $0<|\alpha|\leq 1$?

ABB
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    For any non-scalar element $x$, let $a > {\rm min}(|x|, |x|_1)$. Then $1 + \frac{1}{a}x$ belongs to both sets. – Nik Weaver Mar 23 '22 at 10:27
  • Your delightful "enjoying another norm" has made my day – J.J. Green Mar 23 '22 at 10:34
  • @DenisSerre I usually write the scalar before the vector, not sure what your convention is. – Nik Weaver Mar 23 '22 at 11:12
  • @NiK Your argument is really nice and makes one more point which I arrange it as another question. – ABB Mar 23 '22 at 11:57
  • @AliBagheri I don't understand how your edit changes the problem (or what it has to do with my counterexample). – Nik Weaver Mar 23 '22 at 12:06
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    @Nik Your answer is completely right. Based on the point mentioned by Meisam Soleimani (below) I made some changes to remove trivial cases. – ABB Mar 23 '22 at 12:18

1 Answers1

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No, because 1 is in the intersection mentioned in hypothesis (2), but $1\neq\alpha1$ for any $\alpha$ with $|\alpha|<1$.

Meisam Soleimani Malekan
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