 Solution

For $p\ge 2$, integrate the stated scalar inequality to obtain the Clarkson inequalityIf $f,g$ belong to the unit ball and $\Vert f-g{\Vert }_p\ge \epsilon$, thenThus $L^p[0,1]$ is uniformly convex.

Now let $X$ be uniformly convex. It is enough to show that every $\Phi \in S_{X^{\ast \ast }}$ lies in the canonical image of $X$. Given $\epsilon >0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^{\ast }}$ with $\Phi (f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, thenso $\Vert x-y\Vert <\epsilon$. By Goldstine theorem, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_{\alpha })$; completeness gives $x_{\alpha }\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is reflexive.

For $p\ge 2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p{)}^{\ast }=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.