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I'm trying to solve a problem from an old exam which investigates the construction of functionals on $L^1(\mathbb R)$ and $L^\infty(\mathbb R)$ by extension of 'essential limit'. We say that $ess lim_{x\rightarrow 0} f(x)=\lambda$ if there is a function $g$ such that $lim_{x\rightarrow 0} g(x)=\lambda$ and $g=f$ $a.e.$ I'm trying to prove that there is a functional on $L^\infty (\mathbb R)$ which agrees with the essential limit at $0$ whenever it exists. The question also asks to show that such an extension is not possible for $L^1(\mathbb R)$.

I think the first part can be solved by using the Hahn-Banach theorem. But what subspace and functional should I take? I thought that we should prove that the space of functions for which essential limit exists is a linear subspace and the essential limit is a bounded by the limit. Does that work? Also, why doesn't this hold for $L^1(\mathbb R)$? I'd appreciate some help. Thank you.

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    As a subspace, you can take the subspace of functions from $L^\infty$ for which the essential limit at zero exists. As a functional, take this essential limit. – 2017-02-03
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    Yes, I think that should work. But then what goes wrong if I try the same argument for $L^1$? – 2017-02-03
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    This example should show that $(L^\infty)^*\ne L^1$. So "is not possible for $L_1$" should be interpreted as "no function in $L^1$ cannot represent this functional". – 2017-02-03
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    Sorry, I didn't understand the last comment, could you please clarify? – 2017-02-03
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    Year, I guess I misinterpreted the question about $L^1$, sorry. – 2017-02-03
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    The essential limit isn't bounded on the subspace of $L^1$ with existing essential limit. – 2017-02-03

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