 Solution

Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.