= Solution
Let $q$ be the <Holder conjugate exponent> of $p$. On $\mathbb R^n$ with Lebesgue measure, the <duality of Lp spaces> is
$$
\boxed{(L^p)^*\cong L^q\qquad(1\leq p<\infty),}
$$
under the pairing $\Lambda_g(f)=\int fg$. Thus $(L^1)^*=L^\infty$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.
A Banach space is <reflexive Banach space>[reflexive] when its canonical map into its bidual is surjective. The spaces $L^p(\mathbb R^n)$ are reflexive exactly for
$$
\boxed{1<p<\infty.}
$$
For 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 $L^1\hookrightarrow(L^\infty)^*$, but it is not onto. On the subspace
$$
V=\mathbb C1\oplus C_0(\mathbb R^n)\subset L^\infty,
$$
define $\Lambda(c+h)=c$, the limit at infinity. Since $|\Lambda(v)|\leq\lVert v\rVert_\infty$, 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 $\Lambda(1)=1$. Therefore
$$
\boxed{(L^\infty)^*\ne L^1.}
$$
This is the <Dual of L infinity is larger than L1>.
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