Is the unit ball: $B(0,1)=\{f \in L_p(X,u): \|f\|_p<1\}$ convex? , $0Let $(X,\mathbb{X},u)$  be a measure space 
$L_p(X,u)=\{ f:X\to \mathbb{C}: \|f\|_p<\infty\}$ , $0< p <1$ , $f$: measurable function
$\|f\|_p=\left( \displaystyle \int_X |f|^p  \;\text{d}u\right)^\frac{1}{p}$  , $\|\cdot\|_p: $  quasi-norm
Is the unit ball: $B(0,1)=\{f \in L_p(X,u): \|f\|_p<1\}$ convex?
Any hints would be appreciated.
 A: This is trivially true if $X$ is a singleton.
But when $X$ is not pathological (it suffices that there exist two disjoint subsets of same positive measure), it is not true.
For instance, take $p=1/2$, $X=\mathbb{R}$, with the Lebesgue measure. 
Now if
$$
f=2\cdot 1_{(0,1/2)}\qquad g=2\cdot 1_{(1/2,1)}
$$
then
$$
\|f\|_{1/2}=\|g\|_{1/2}=\frac{1}{2}
$$
but
$$
\|\frac{1}{2}f+\frac{1}{2}g\|_{1/2}=1.
$$
You can construct a similar counterexample in general as soon as there exist two disjoint measurable sets $A,B$ with same positive measure $m>0$.
Indeed, take 
$$
f=k\cdot 1_A\qquad g=k\cdot 1_B
$$
then 
$$
\|f\|_p=km^{1/p}=\|g\|_p
$$
but
$$
\|\frac{1}{2}f+\frac{1}{2}g\|_p=\frac{k}{2}(2m)^{1/p}=2^{1/p-1}km^{1/p}.
$$
So with
$$
k=\frac{1}{2^{\frac{1-p}{2p}}m^{1/p}}
$$
we get $\|f\|_p=\|g\|_p<1$ while $\|(f+g)/2\|_p>1$.
So neither the open unit ball nor the closed unit ball is convex.
See also the pictures here for the unit ball of $\ell^p$ for various values of $p$.
Trying to go in the converse direction to characterize measure spaces for which this unit ball is convex, the same ideas show they must satisfy: for every disjoint measurable sets $A,B$, 
$$
\min(\mu(A),\mu(B))\leq \frac{1}{2^{1-p}}\max(\mu(A),\mu(B)).
$$
I believe these are pathological measure spaces, but I can't figure this out any further. So I'll wait for help. I am sure this is well known to specialists of Banach spaces/measure theory.
