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 $\|f+g\|_p \leq 2 \eta$.
It was written that above is the simplified version of the uniform convexity of Banach space.
1: I cannot see how this is the "simplified" statement of uniform convexity of Banach space.
2: I have no clue of how to prove this, I found to prove the uniform convexity clarkson inequality is used. But I cannot see that it to be similar to this as here the question seem not to requiring to divide $p\in (1,\infty)$ into $p\geq2$ and $1<p<2$.
The 'standard' form of uniform convexity for a Banach space $X$ is:
For $L^p$ spaces in particular we can extend the notion a little: a space is $p$-uniformly convex for $2\leq p< \infty$ if there exists a constant $C > 0$ and $x,y \in S_X$ with $$ \|x+y\| \leq 2-C\| x-y \|^p \tag{2}\label{2}$$
I think that $p$-uniformly convex is the 'unsimplified' version that your notes allude to.
Note that these are both definitions and so not provable as such. It's easy enough to see that $\eqref{1}$ and $\eqref{2}$ are related if we can take $\delta(\varepsilon) = C\varepsilon ^p$ whenever $\|x-y\|<\epsilon$.
You can, however, use the generalised Clarkson inequalities to show that $L^p$ is $p$-uniformly convex for $p\in(2,\infty)$. You can also show that $L^p$ is $2-$uniformly convex for $p\in(1,2]$ but this requires a reasonable amount of extra work.