# A continuous linear functional on $L^\infty(R)$ that vanishes on $C(R)$.

Intuitively, I want to construct the functional F in this way:

$$F(f)=\lim_{x\rightarrow 0+}f(x)-\lim_{x\rightarrow 0-}f(x)$$

for $f\in L^\infty$. I know this is not well defined so I'd like to find a way to use this idea. Maybe find an extension using Hahn-Banach, etc.

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How about the following: try averaging f over [0,1/N], averaging it over [-1/N,0], then taking the difference. That gives you a functional $F_N$. Now take a Banach limit or similar as $N\to\infty$ – Yemon Choi Sep 21 '12 at 21:51
Yes, define $F(f)$ for all piecewise continuous functions for which that limit exists, and extend using Hahn-Banach. That works. – George Lowther Sep 21 '12 at 22:08
This looks like homework; voting to close. – John Pardon Sep 21 '12 at 23:15
It does not look like homework to me. It looks like a student who has been told that such functionals exist by the Hahn-Banach theorem and wonders about a specific example. Which, for a student learning functional analysis, is a perfectly reasonable question. That this is not the right forum for it is another matter. – Michael Renardy Sep 21 '12 at 23:32
"This looks like homework; voting to close. – unknown (google)" In all honesty, if you want to vote to close, start with choosing a unique username! – fedja Sep 22 '12 at 3:32

I am not sure what is your question. But you can construct a linear functional in this way. The limit does not exist for some $f$ in $L^\infty$. But you can use the Banach limit. Consider the bounded linear functional on $C(R)$ defined as the limit as $x\to 0$. By Hahn-Banach this has an extension to a bounded linear functioal on $L^\infty(-\infty,0)$. and also has an extension to $L^\infty(0,\infty)$. Now define your functional as you propose, using these Banach limits.
But all this is a kind of trivial nonsense. Hahn Banach says you from the beginning that you have (very many) linear functionals on $L^\infty$ which are zero on $C$. This is a pure existence theorem. But your "construction" really adds nothing to it, because the existence of a Banach limit is again a pure existence theorem.