Let $A$ be a finite set of real (or complex) numbers. If I consider sets with small sizes, we have that:
If $A$ is the empty set, then $A+A$ is also empty.
If $A$ is a singleton, then $A+A$ is also a singleton.
If $|A|=2,$ then $|A+A|=3.$
If $|A|=3,$ then $|A+A|$ can be at most $6.$
The set $A=\{1, 2, 3\}$ shows we can have that $|A+A|=5,$ But we can not obtain $4.$
Obvious bounds for $A+A$ are $|A|\le |A+A|\le|A|^2.$ But for large $|A|$ values, it looks like that we can find more sharpe bounds for $|A+A|$ than above ones.
MY QUESTION IS: How can I estimate the size of the set $A+A=\{a+b :a,b\in A\}$ using the size of $A$ ?
For general finite subsets of abelian groups we have $$ |A| \leq |A + A| \leq \frac{|A|(|A| + 1)}{2} $$
In the case of $\mathbb R$ and $\mathbb C$, we are dealing with torsion-free abelian groups (i.e. they have no finite subgroups) we have a lower bound of $2|A| - 1$, which is attained if and only if $A$ is an arithmetic progression, i.e. $A = \{a,\, a + d,\, a + 2d, \dots,\, a + (N-1)d\}$ for $N = |A|$ and numbers $a, d$.
The upper bound is sharp and occurs if and only if all pairwise sums in $A$ are distinct ($A$ is then called a Sidon set). This happens often in $\mathbb R$ (and $\mathbb C$): for example, if you choose a set of $N$ numbers uniformly random from $[0, 1]$ you will get a Sidon set with probability 1.