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I was working on the following problem:

Let $L^\infty(\mathbf{R})$ denote the essentially bounded functions and let $BC(\mathbf{R})$ denote the bounded continuous functions. Show that there is some $\ell \in (\mathbf{L}^\infty(\mathbf{R})^\ast$ such that $\ell(f) = f(0)$ for each $f \in BC(\mathbf{R})$.

There are two things I could see to do here. The first is apply BLT theorem. Since $\ell(f) = f(0)$ is a bounded linear map on $BC(\mathbf{R})$ this map extends to the completion of $BC(\mathbf{R})$. So it would suffice to show that any $L^\infty(\mathbf{R})$ function is the limit of $BC(\mathbf{R})$ functions. The other thing to do is to try applyinh Hahn-Banach theorem: which seems more straightforward. $BC(\mathbf{R})$ is a linear subspace of $L^\infty(\mathbf{R})$ thus, defining a map $f \mapsto f(0)$ on $BC(\mathbf{R})$ which is clearly a bounded linear map extends to $\ell$ with the same norm (1), on $L^\infty(\mathbf{R})$. Which is the correct way to solve the problem?

Drew Brady
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  • I think the statement that any $L^\infty$ function is the limit of $BC(\mathbf{R})$ functions is wrong in the $|\cdot|_{\infty}$ norm. Can someone confirm? – Drew Brady Feb 09 '18 at 16:34
  • look at $f(x)$ defined by $f(0) = 1$ and $f(x) = 0$ if $x\neq 0$. This is clearly bounded, but cannot be uniformly approximated by continuous functions (since such a limit would be continuous). So your suspicion is correct. – Thomas Feb 09 '18 at 16:50
  • @DrewBrady the uniform limit of continuous functions is c*** – user251257 Feb 09 '18 at 16:50
  • Ok, so this means that my approach with Hahn-Banach is the correct one? All I'd have to verify is that $L^\infty, |\cdot|_\infty$ is complete. – Drew Brady Feb 09 '18 at 16:51
  • yes. If you need a hint for completeness have a look here: https://math.stackexchange.com/questions/150306/prove-that-the-normed-space-l-infty-equipped-with-lvert-cdot-rvert-inft – Thomas Feb 09 '18 at 16:55
  • (uhm -- why do you need completeness?) – Thomas Feb 09 '18 at 16:58
  • @Thomas you need $L^\infty$ to be Banach, no? Isn't that a requirement for Hanh-Banach to hold – Drew Brady Feb 09 '18 at 17:20
  • Oh, apparently not. – Drew Brady Feb 09 '18 at 17:22

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