**Step 1: Scalar Clarkson inequality for $p \ge 2$.** For $\theta \in [0, 1]$ and $p \ge 2$, convexity of $t \mapsto t^{p/2}$ gives $\theta^{p/2} + (1-\theta)^{p/2} \le 1$. Setting $\theta = a^2/(a^2+b^2)$ yields $a^p + b^p \le (a^2 + b^2)^{p/2}$ for $a, b \ge 0$. Taking $a = |x+y|/2$, $b = |x-y|/2$ and using $(a^2+b^2) = (x^2+y^2)/2$: \begin{align*} \left|\frac{x+y}{2}\right|^p + \left|\frac{x-y}{2}\right|^p \le \left(\frac{x^2+y^2}{2}\right)^{p/2} \le \frac{|x|^p + |y|^p}{2} \end{align*} where the last step uses convexity of $t \mapsto t^{p/2}$ again. **Step 2: Uniform convexity for $p \ge 2$.** Integrating the scalar inequality: for $f, g \in L^p(\mathbb{R}^n)$, \begin{align*} \left\|\frac{f+g}{2}\right\|_p^p + \left\|\frac{f-g}{2}\right\|_p^p \le \frac{\|f\|_p^p + \|g\|_p^p}{2}. \end{align*} If $\|f\|_p, \|g\|_p \le 1$ and $\|f - g\|_p > \varepsilon$, then $\left\|\frac{f+g}{2}\right\|_p \le \left(1 - (\varepsilon/2)^p\right)^{1/p} =: 1 - \delta$, giving uniform convexity. **Step 3: Uniform convexity for $1 < p < 2$.** The second Clarkson inequality states: \begin{align*} \left\|\frac{f+g}{2}\right\|_p^q + \left\|\frac{f-g}{2}\right\|_p^q \le \left(\frac{\|f\|_p^p + \|g\|_p^p}{2}\right)^{q/p} \end{align*} where $q = p/(p-1)$. This is proved by duality: apply the first Clarkson inequality (for $q \ge 2$) on $L^q$ and use the identification $(L^p)^* \cong L^q$. The same argument as in Step 2 then gives uniform convexity. **Step 4: Failure at $p = 1$ and $p = \infty$.** For $L^1$: take $f = \mathbb{1}_{[0,1]}$, $g = \mathbb{1}_{[1,2]}$. Then $\|f\|_1 = \|g\|_1 = 1$, $\|f - g\|_1 = 2$, but $\|(f+g)/2\|_1 = 1$. For $L^\infty$: take $f \equiv 1$, $g(x) = \operatorname{sign}(x)$.