Prove of Uniform (rotundity) Convexity of Banach Space using the specified instructions I've been required to prove the uniform convexity of Banach Spaces following the below instruction. However,  I have no clue of how to start and also cannot understand the intuition. 
We have that a Banach space is uniformly convex if for any $\epsilon \in (0, 1)$ there exists $\eta < 1$ such that if $\|f\|_p=\|g\|_p=1$, and $\|f-g\|_p \geq 2 \epsilon$   then $\|f+g\|_p \leq 2 \eta$.
the question asks to prove uniform convexity of $L^p[(0, 1); \lambda]$, $p \in (1,\infty)$, $\mu$ is finite.
$\textbf{1)}$  Suppose $\mu$ is any finite measure on $(0, 1)$. Show that for every $\epsilon  \in (0, 1)$ there exists $\eta < 1$ such that for every
pair of functions $f,g \in  L^p([0; 1], \mu)$, where $p\in (1,\infty)$.  If $\|f\|_p=\|g\|_p=1$ , and $\|f\|_\infty ,\|g\|_\infty\leq M <\infty$ and $\|f-g\|_p \geq 2 \epsilon$   then $\|f+g\|_p \leq 2 \eta$.
$\textbf{2)}$ Rescale the general case to the above simplified version. Use
$$\int|f+g|^p d\lambda=\int|\phi+\psi|^p d\mu$$
where $d\mu = (|f|^p +|g|^p)d\lambda$, $\phi = \frac{f}{(|f|^p +|g|^p)^\frac{1}{p}}$, and  $\psi = \frac{g}{(|f|^p +|g|^p)^\frac{1}{p}}$.
 A: The intuition with uniform convexity, is that supposing $x$ and $y$ lie on the unit sphere, no matter how far apart they get, their (vector) sum does not get too large. I always see this by looking at the unit ball in $\mathbb R^2.$ 
I haven't done the proof as outlined in your question, so if my answer is not what you want, please advise and I will delete it. I submit it because it is basically an application of a weakened form of  Clarkson's inequalities and you may find it worth knowing:
Break it up into cases: if $p\ge 2,$ and $\|f\|=\|g\|=1$, consider   
$\tag1 F(x)=\frac{1}{2}+\frac{|x|^p}{2}-\frac{|1+x|^p}{2}\  \text{and}\ G(x)=\frac{|1-x|^p}{2}.$ 
From elementary calculus, we have that $F(1)=0$; otherwise $F$ is positive; $F''(1)>0$ (because $p\ge 2$) so $F$ has one global minimum at $x=1.$ Similarly, $G(1)=0$ and this is the only global minimum. Also, $\underset{x\to \pm \infty}\lim \frac{G(x)}{F(x)}$ is bounded so there is some some real number $c$ such that  $cG(x)\le F(x).$
Now, suppose $g(y)\neq 0$ everywhere and take $x=\frac{f(y)}{g(y)}.$ Substituting these data into $(1)$, dropping the argument $y$ for convenience, and rearranging a bit, we get 
$\tag2 \frac{|f+g|^p}{2}+c\frac{|f-g|^p}{2}\le \frac{|f|^p}{2}+\frac{|g|^p}{2}.$ 
But we see that the condition $g(y)\neq 0$ is superfluous, because $(2)$ is satisfied in this case as well with $c=1$ and so we can replace $c$ by  $c'=\min\{c,1\}$.
Now, integrate this to get 
$\tag3 \frac{\|f+g\|^p}{2}+c\frac{\|f-g\|^p}{2}\le 1.$
Uniform convexity follows easily from $(3).$
If $1<p<2,$ prove that there is a number $c(p)$ such that for all $x,y\in \mathbb R$
$\tag4 |x-y|^p\leq c(p)\left ((|x|^p+|y|^p)^{1-p/2}(|x|^p+|y|^p-2\left|\frac{x+y}{2}\right |^p)\right )^{p/2}.$ 
Do this by assuming without loss of generality,that  $x=1$ and using elementary calculus as above. Then use Holder to conclude. 
