Let $A \subset \mathbb F\_p \setminus \{0\} $ and let $A+1/A = \{x+1/y:x,y \in A\}$.


**Question:** For fixed $c>0$, if $|A| \geq cp$, is $|A+1/A|$ at least $(1-o(1))p$ when $p \rightarrow \infty$?


**Known:**
This is false for $c<1/4$ according to Seva's comments.


Take $A$ to be $I \cap I^{-1}$ where $I=[1,m]$ and $m=(1/2-\epsilon)p$. It's sure that $A+1/A=A+A$ is not the whole $\mathbb F\_p$. The hard part is that $A$ contains $(m/p)^2p(1-o(1))$ elements.


Let $X$ be the indicator function of $\{(x,1/x):x \in \mathbb F\_p \setminus 0 \}$, and $Y$ the indicator function of $\{(x,y):x,y \in [1,m]\}$. The size of $A$ is the value of the inner product $\langle X,Y\rangle$.


To compute the inner product, we decompose $Y$ into a sum of Fourier coefficients.


Let $e(x)=\exp(2\pi i x/p)$, and let $K\_\omega$ be the $\omega$-th coefficient of the Fourier transform of the indicator function of the interval $[1,m]$, i.e. $K\_\omega = \underset{x=1}{\overset{m}{\sum}} e(\omega x)/p$, where $\omega=0,1,2,...,p-1$.


By the Fourier transform on cartesian products, $\langle X,Y\rangle = \underset{\psi=0}{\overset{p-1}{\sum}} \underset{\omega=0}{\overset{p-1}{\sum}} K\_\psi K\_\omega \langle e(\psi x)e(\omega y),X \rangle $.


Let $I\_0$ be the term where $\psi$ and $\omega$ are both $0$, $I\_1$ the sum of the terms where exactly one of $\psi$ and $\omega$ is $0$, and $I\_2$ the sum of the terms where both $\psi$ and $\omega$ are nonzero. We have $\langle X,Y\rangle = I\_0+I\_1+I\_2$ and $I\_0=m^2(p-1)/p^2$.


To estimate $I\_1$ and $I\_2$, we need to bound the sum of $|K\_\omega|$ over nonzero $\omega$. There is $K\_\omega=\frac{e(\omega(m+1))-e(\omega)}{e(\omega)-1}/p$. As $|e(\omega)-1|=2\sin(\pi\omega/p)>4\omega/p$ when $0<\omega<p/2$, we have $|K\_\omega|<\frac{2p}{4 \omega} /p = 1/2\omega$ when $0<\omega<p/2$. Thus the sum of $|K\_\omega|$ over nonzero $\omega$ is $O(\log p)$.


The terms $K\_\psi K\_\omega \langle e(\psi x)e(\omega y),X \rangle$ appearing in $I\_1$ has norm $K\_\psi m/p$ when $\psi$ is nonzero or $K\_\omega m/p$ when $\omega$ is nonzero, so $I\_1=O(\log p)$.


If both $\psi$ and $\omega$ are nonzero, the term $\langle e(\psi x)e(\omega y),X \rangle$ can be estimated by the [Kloosterman sum](https://en.wikipedia.org/wiki/Kloosterman_sum) and it's $O(\sqrt{p})$. Summing over all nonzero $\psi$ and $\omega$, we have $I\_2=O(\sqrt{p} \log^2 p)$.


Thus $\langle X,Y\rangle = I\_0+I\_1+I\_2 = m^2(p-1)/p^2 + O(\log p)+O(\sqrt{p} \log^2 p) = (m/p)^2p(1-o(1))$.


This shows that the answer is false for any $c<1/4$.



A sum-product phenomenon on reciprocals
The claim is true for $c > 1/4$, although the proof I have either requires the machinery of the arithmetic regularity lemma, or the (morally equivalent) language of additive limits (as discussed in [this blog post](https://terrytao.wordpress.com/2014/10/12/additive-limits/) of mine, or in [this paper of Szegedy](https://mathscinet.ams.org/mathscinet-getitem?mr=3852460)). I will sketch a proof with the latter approach, which is shorter (it avoids a lot of "epsilon management"). One can expand this argument into a lengthier argument using instead the arithmetic regularity lemma (which would most likely give some tower-exponential type quantitative dependencies on constants), analogously to how arguments using graph limits (graphons) can often be converted into lengthier arguments relying instead on the Szemeredi regularity lemma, but I will leave this to the interested reader.


In what follows I assume familiarity with the language of nonstandard analysis and Loeb measure.


Suppose for contradiction that the claim failed, thus there is $c>1/4$ and $\varepsilon>0$ and a sequence $A\_i \subset {\mathbb F}\_{p\_i}^\*$ of sets of density $c + o(1)$ indexed by some primes $p\_i$ going off to infinity such that $|A\_i + 1/A\_i| \leq (1-\varepsilon+o(1)) p\_i$. Taking an ultralimit, one obtains an internal subset $A$ of of a pseudofinite field ${\mathbb F} = {\mathbb F}\_{p}$ (associated to a nonstandard prime $p$) of Loeb measure $c$ such that $A + \phi(A)$ has Loeb measure at most $1-\varepsilon$, where $\phi$ denotes the almost-everywhere defined inversion map $n \mapsto n^{-1}$. In particular the convolution $1\_A \* 1\_{\phi(A)}$ avoids a set of positive measure.


The $\sigma$-algebra ${\mathcal L}$ of Loeb-measurable subsets of ${\mathbb F}$ has two notable factors. One is the Kronecker factor ${\mathcal K}$, generated by the eigenfunctions of $L^\infty({\mathcal L})$ with respect to the translation action (or equivalently, by the Fourier phases $n \mapsto \mathrm{st} e( an/p )$ for some non-standard $a \in {\mathbb F}$); the other is the pullback $\phi^\* {\mathcal K}$ of the Kronecker factor ${\mathcal K}$ by the inversion map $\phi$. Because of Kloosterman sum estimates, the two factors ${\mathcal K}, \phi^\* {\mathcal K}$ are orthogonal. The Kronecker factor is (up to some canonical isomorphisms) a compact abelian group (with Loeb measure restricting to the Haar probability measure on that group); in fact that compact abelian group is nothing more than the Pontryagin dual of the group of Fourier phases. Because this latter group is torsion-free (it is isomorphic as a group to the pseudofinite field ${\mathbb F}$), the Kronecker factor is then a *connected* compact abelian group.


Because the Kronecker factor is characteristic for convolutions (see the previous references), we have the identity
$$ 1\_A \* 1\_{\phi(A)} = {\bf E}(1\_A|{\mathcal K}) \* {\bf E}(1\_{\phi(A)}|{\mathcal K})$$
Loeb-almost everywhere. In particular,
$$ \mathrm{supp}({\bf E}(1\_A|{\mathcal K})) + \mathrm{supp}({\bf E}(1\_\phi(A)|{\mathcal K}))$$
fails to have full measure (here I gloss over a technical but fixable issue involving null sets). Applying [Kemperman's theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=202913)
$$ \mu( A + B ) \geq \min( \mu(A)+\mu(B), 1 )$$
valid for any connected compact group with Haar probability measure (actually for the technical reason mentioned earlier one should actually use [Pollard's theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=354517) here, or the variant of Kemperman's theorem in [this blog post](https://terrytao.wordpress.com/2011/12/26/a-variant-of-kempermans-theorem/) of mine), we conclude that
$$ \mu(\mathrm{supp}({\bf E}(1\_A|{\mathcal K}))) + \mu(\mathrm{supp}({\bf E}(1\_\phi(A)|{\mathcal K}))) < 1,$$
thus ${\bf E}(1\_A|{\mathcal K})$ is supported in a ${\mathcal K}$-measurable set $E$ of measure at most $\alpha$ and ${\bf E}(1\_\phi(A)|{\mathcal K})$ is supported in a set of measure at most $1-\alpha$ for some $0 \leq \alpha \leq 1$. Since $\phi$ is measure-preserving, we conclude that ${\bf E}(1\_A|\phi^\* {\mathcal K})$ is supported on a $\phi^\* {\mathcal K}$-measurable set $F$ of measure at most $1-\alpha$. In particular we have the pointwise bounds
$$ 1\_A \leq 1\_E$$
and
$$ 1\_A \leq 1\_F$$
Loeb-almost everywhere, hence
$$ 1\_A \leq 1\_E 1\_F.$$
Loeb-almost everywhere. Integrating and using the orthogonality of ${\mathcal K}$ and $\phi^\* {\mathcal K}$, we conclude that
$$ c \leq \alpha(1-\alpha)$$
and hence
$$ c \leq 1/4,$$
giving the desired contradiction.


**Remark**: I think that a refinement of this argument in fact gives the lower bound $|A + 1/A| \geq (\min(2\sqrt{c}, 1)-o(1)) p$ for any fixed $0 \leq c \leq 1$, and a variant of Seva's construction should show that this bound is asymptotically best possible.


