Artificial intelligent assistant

What about $\ell^1$ with pointwise multiplication This question occurred to me after reading this thread. I was working on finding an example of a Banach algebra. The example I came up with was $\ell^1 (\mathbb N)$ with pointwise multiplication. I believe I even proved that $\ell^1 (\mathbb N)$ is closed with respect to pointwise multiplication and that the norm is submultiplicative. Is it really possible that $\ell^1 (\mathbb N)$ can be turned into a Banach algebra in two ways, by convolution and by pointwise multiplication, or is there necessarily a mistake in my proof? (I'm happy to post the proof if the answer is yes -- for now I'm just trying to keep this question short)

I don't see any contradiction, at first sight. In fact, any Banach space can be made into a Banach algebra if you define the product as zero.

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