Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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
2  
"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
show 1 more comment

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.