# “Elementary” proof that the dual space of $\ell^p$ is $\ell^q$

Let $1<p<\infty$. Given a measure space $X$ one typically proves that $(L^p)^* =L^q$ using (essentially) two components: (i) the Holder inequality shows that $L^q\subset (L^p)^*$ and (ii) the Radon-Nikodym derivative establishes the reverse inclusion.

In particular, I find part (ii) of this proof somewhat inaccessible for first time learners, and so I am wondering if there is a more elementary'' proof to show that $(L^p)^*\subset L^q$ in the case that we have the counting measure. Many questions on this site have established that $(\ell^1)^*=\ell^\infty$, so my question lies specifically with $p>1.$

So far, I have the idea to represent $\phi \in (\ell^p)^*$ through projections: $a_n := (\phi, e_n)$. Then, assuming that the sequence $a_n\not\in \ell^q$ I hope to construct some particular $b_n\in \ell^p$ so that $(a_n,b_n) = \infty$, contradicting the boundedness of $\phi$.

Any reference or direction would be very helpful. Thank you!

• See page 3 here – RideTheWavelet Feb 2 '17 at 3:59
• Ah thank you! Yes this seems to have the construction I was having trouble cooking up myself. – Matt Feb 2 '17 at 4:03