Why is $\overline{L^\infty(\Omega)} \subset L^\infty(\Omega) $ where the closure is in norm of $L^1(\Omega)$?

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Let $\Omega$ be a domain which may or may not be unbounded (eg. $\Omega = B_1(0)\times (0,\infty)$). Why is $$ \overline{L^\infty(\Omega)} \subset L^\infty(\Omega)$$ where the closure is in norm of $L^1(\Omega)$?

I would have thought that the inclusion is the other way around. Source is this paper, see page 11, the third displayed equation.

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As written, this is either false or trivial, or both. The closure of any subset of $X$ in $X$ is trivially a subset of $X$; just because of what the word closure means. Closure is taken with respect to a particular ambient space, and cannot produce elements outside of that space.

If you meant completion with respect to $L^1$ norm, then the inclusion is false. The completion of $L^\infty$ with respect to $L^1$ norm can be identified with $L^1$.

So, you misunderstood something. The only conclusion from here is:

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