Proving $L^p$ is a uniformly convex Banach space. Let $1<p<\infty$ and consider the $L^p$ space (for any $\sigma$-finite measure space $(X,\mu)$, say $\mathbb{R}$). How can we prove that it is uniformly convex, i.e. that that for any $\epsilon\in (0,1)$ one can find $\eta<1$ such that: If $||f||_p=||g||_p=1$ and $||\frac{f-g}{2}||>\epsilon$ then $||\frac{f+g}{2}||<\eta$? 
Via easy reductions (replacing $f,g$ with $\frac{f}{(|f|^p+|g|^p)^{1/p}}$ and $\frac{g}{(|f|^p+|g|^p)^{1/p}}$ and $d\mu$ with $(|f|^p+|g|^p)^{1/p}d\mu$), one can assume $\mu(X)=2$ and $|f|\le 1, |g|\le 1$ everywhere. My goal is to prove that there exists a constant $c>0$ depending only on $p$ such that
$$\bigg|\frac{f+g}{2}\bigg|^p + c\bigg|\frac{f-g}{2}\bigg|^p \le \frac{|f|^p}{2} + \frac{|g|^p}{2},$$
for $f$ and $g$ any numbers. Integrating we can then obtain our desired result, with $\eta=1-c\epsilon$ I believe. Unfortunately, I can only prove the above inequality for $p\ge 2$, and even then my solution is very messy (and possibly flawed). Does anyone know how to finish the problem?
 A: Carothers has a proof of this, along the lines that you are pursuing. He leaves out quite a few details, so it would make for a nice exercise. My favorite proof appears in Bogachev's Measure Theory book. The key is the following inequality, which makes the proof a simple calculation:
$\epsilon^p(|x|^p+|y|^p)\le 4|x-y|^p\Rightarrow \left |\frac{x+y}{2}\right|\le (1-\delta)\frac{|x|^p+|y|^p}{2}\tag1$.
To prove this, it is enough to consider $1\le x^2+y^2\le 2$ because any pair $(x,y)$ can always be brought into this annulus by taking some $(tx,ty):t>0$. Now it is enough to note that substitution into $(1)$ produces $t^p$ on both sides of the inequlality. Now argue by contradiction, using compactness of the annulus, we find that if $(1)$ is not true then there is a pair $(x,y)$ in the annulus such that the first inequlity is true and yet $\left |\frac{x+y}{2}\right|\ge \frac{|x|^p+|y|^p}{2},$ which, if $p>1,$ is only possible if $x=y$. But then $(1)$ says that $x=y=0$ so neither $x$ nor $y$ lies in the annulus. Contradiction.
Suppose $\|f\|_p=\|g\|_p=1$ such that $\|f − g\|_p\ge \epsilon$. Set 
$S=\{x:\epsilon^p(|f(x)|^p+|g(x)|^p)\le 4|f(x)-f(y)|^p\}.$ Then, $(1)$ implies that for all $x\in S,$
$\left |\frac{f(x)+g(x)}{2}\right|\le (1-\delta)\frac{|f(x)|^p+|g(x)|^p}{2}$ and therefore that $\int_{X\setminus S}|f-g|^p\le \frac{\epsilon^p}{2}$. 
It follows that $\int_S|f-g|^p\ge \frac{\epsilon^p}{2}$. We also have since $p>1,$
$\frac{|f(x)|^p+|g(x)|^p}{2}\ge \left|\frac{f(x)+g(x)}{2}\right|^p\tag2$
To finish, combine $(2)$ and $(1)$ to obtain the result by calcuating:
$\displaystyle\int_X\left(\frac{|f(x)|^p+|g(x)|^p}{2}-\left|\frac{f(x)+g(x)}{2}\right|^p\right )\ge \int_S\left(\frac{|f(x)|^p+|g(x)|^p}{2}-\left|\frac{f(x)+g(x)}{2}\right|^p\right )\ge$
$ \displaystyle\delta\int_S\left(\frac{|f(x)|^p+|g(x)|^p}{2}\right)\ge \frac{\delta}{2^{p+1}}\int_S|f(x)-g(x)|^p\ge \frac{\delta\cdot \epsilon^p}{4\cdot 2^p}\Rightarrow$
$ \int_X\left|\frac{f(x)+g(x)}{2}\right|^p\le 1-\frac{\delta\cdot \epsilon^p}{4\cdot 2^p}.$
