# Extending bounded linear functionals from $L^q$ to $L^p$ on finite measure spaces

I'm trying to give an example to show the following:

Not every bounded linear functional on $L^3([0,1],\mu)$ is the restriction of a bounded linear functional on $L^2([0,1],\mu)$. (Here $\mu$ is Lebesgue measure)

So far for finite measure spaces with $0<p<q\leq\infty$ I have that $L^q\subset L^p$ (from Folland) so the restriction makes sense. Restricting a bounded linear functional from $L^p$ to $L^q$ shouldn't be a problem, it retains boundedness and linearity from the fact that $L^q\subset L^p$.

I found the Hahn-Banach theorem, but this seems to only give conditions for when one can extend a bounded linear functional. (I didn't find anything on if the conditions it gives are also necessary, in addition to being sufficient)

I'd appreciate an example and/or good places to look on general theory. (Perhaps I just didn't google the right terms)

• Consider Riez-Nagy theorem on representing any bounded linear functional on L_p as an integral operator. – Adelafif Jul 11 '15 at 7:18
• Do you have a precise reference/statement? I don't see anything like it in Folland or Bass and the internet is surprisingly unhelpful. – ttt Jul 11 '15 at 13:28
• Ah, this is Theorem 6.15 in Folland. – ttt Jul 24 '15 at 16:02

You know (why?) that $L^2 \neq L^{3/2}$, so choose an $f \in L^{3/2}\setminus L^2$, and consider the map $$L^3 \mapsto \mathbb{C} \text{ given by } g \mapsto \int fgd\mu$$ This is a bounded linear map. Suppose it is the restriction of a bounded linear map on $L^2$, then $\exists h \in L^2$ such that $$\int fgd\mu = \int hg d\mu \quad\forall g\in L^3$$ From this you can conclude that $f\equiv h$ which contradicts the fact that $f\notin L^2$
• Why is product of $L^1$ and $L^3$ function integrable?/ – C_Al Jul 11 '15 at 22:10
• Bleh. That should be $L^{3/2}$. Fixing it. – Prahlad Vaidyanathan Jul 11 '15 at 22:19