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This is an exercise problem from a course in functional analysis. However, it is not a homework problem. I think I got it figured out, however my teacher said something during the lecture that I didn't understand, and was lagging behind with taking notes, so I thought I'd think about it at home rather than ask during lecture. Now I can't make sense out of what she said, and I'd rather not wait until the next lecture to ask her. Here's where you guys come in :)


Show that there exists a non-zero linear functional $F \in (L^\infty[a, b])^\ast$ such that for any $f \in C[a, b]$, $F(f) = f((a + b)/2)$

My Solution:

In $L^\infty[a, b]$ we identify all functions that are equal almost everywhere. In each such equivalence class there exist a continuous function and we let that function represent the equivalence class. Therefore we can conclude that $C[a,b] \subseteq L^\infty[a, b]$.

Note that $\| \cdot \|_\infty$ is a norm on $L^\infty[a, b]$ (Hahn-Banach requires only semi-norm). Now define $F(f) = f((a + b)/2)$ on C[a,b]. This is linear and bounded functional,

$$F(\lambda f + \mu g) = (\lambda f + \mu g)\left(\frac{a + b}{2}\right) = \lambda f\left(\frac{a + b}{2}\right) + \mu g\left(\frac{a + b}{2}\right) = \lambda F(f) + \mu F(g)$$

$$|F(f)| = \left| f\left(\frac{a + b}{2}\right) \right| \leq \| f \|_\infty $$

By Hahn-Banach Theorem we can extend $F$ to $L^\infty[a, b]$ with the same norm such that $F(f) \leq \| f \|_\infty$, $\forall f \in L^\infty[a, b]$. Hence $F \in (L^\infty[a, b])^\ast$

Is this a complete solution? My teacher pointed something out, that I couldn't really grasp because I was behind on taking notes. But this is what she said basically:

Teacher's comment:

You can try to show that this $F$ does not come from $f \in L^1[a, b]$ i.e. $F \neq \phi_f$ where $$ \phi_f(g) = \int_a^b fg dx.$$ Does this make any sense? Why is this relevant? I can't see how, but at the same time I am afraid of missing some detail here.

Thank you very much in advance!

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It is NOT true that every equivalence class contains a continuous function. For example, consider the function that is $0$ on $[a,a+b/2]$ and $1$ on $(a+b/2,b]$. If there were a continuous function in every equivalence class, then you would have $C[a,b]\cong L_{\infty}[a,b]$ (where the latter is viewed as equivalence classes of functions). There is definitely containment as you have written, though. – Aaron Nov 27 '12 at 0:00
Yes, you are right. I was thinking too fast! Thanks! – DoubleTrouble Nov 27 '12 at 7:59
up vote 3 down vote accepted

Your proof is correct.

The aim of your teacher is to point out that $L_\infty[a,b]$ is not reflexive. Speaking informally this means that you can not describe all linear functionals in $L_\infty[a,b]^*\cong L_1[a,b]^{**}$. Though every element of $L_1[a,b]$ gives rise in a natural way to some element of $L_\infty[a,b]^*$ they are not sufficient to describe whole the $L_\infty[a,b]^*$.

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