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There is the following result:

Suppose $X$ is a normed algebra with identity $e$. Then $\|e\| \geq 1$.

I am looking for an example to show that even a Banach algebra $X$ with identity $e$ not necessarily satisfies $\|e\| = 1$. A general example such that $\|e\| = t$, where $t \geq 1$ would be even more interesting.

I am aware of several examples of Banach algebras with identity, but these all have an identity with norm $1$. As algebras have exactly one identity, I think I cannot use the well-known Banach algebras as an example.

Any help or comment is highly appreciated.

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Let $X = (X, \lVert · \rVert)$ be a Banach algebra and $t ≥ 1$. Then $X$ with $t\lVert · \rVert$ is also a Banach algebra.

Thus, for example $ℝ$ with the norm $\lVert · \rVert_t \colon ℝ → [0..∞), x ↦ t|x|$ becomes a Banach algebra with $\lVert 1 \rVert_t = t$. Submultiplicativity of the norm follows from $t ≤ t^2$.

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  • $\begingroup$ Thanks. This seems easy; I did not think of it. This also implies that we can in fact equip any Banach algebra with a norm such that $\|e\| = 1$, doesn't it? $\endgroup$ – user342207 Aug 20 '16 at 9:59
  • $\begingroup$ @jnv I don’t think it does: You need a scalar factor of $t ≥ 1$ to scale a norm into another one, so you can always strech norms, but not necessarily shrink them. I’m actually not proficient in functional analysis, though, so I don’t know of any counterexamples. Wikipedia says that one can at least embed every Banach algebra into a unital one. What do you mean by “equip”, though? If you change the norm, you change the Banach algebra. Do you mean: “… without changing the topology”? $\endgroup$ – k.stm Aug 20 '16 at 10:17
  • $\begingroup$ Yes, I meant without changing the topology, that is, an equivalent norm. In the meantime I have found a proof for this fact. In case you are interested: it is Proposition 1.3 in Takesaki's Theory of Operator Algebras vol. 1. $\endgroup$ – user342207 Aug 20 '16 at 10:22

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