$\newcommand{\F}{{\mathbb F}}$
$\newcommand{\R}{{\mathbb R}}$
$\renewcommand{\phi}{\varphi}$


Let $p\ge 5$ be a prime. 


If the functions $\phi\_1,\phi\_2,\phi\_3\colon\F\_p\to\R$ satisfy $\phi\_1(x)+\phi\_2(y)=\phi\_3(x+y)$ for all pairs $(x,y)\in\F\_p^2$, then they are constant functions. The easiest way to see this is by comparing the images $A\_i:=\mathrm{Im}(\phi\_i)$:
 $$ A\_1+A\_2 \subseteq A\_3,\ A\_3-A\_1\subseteq A\_2,\ A\_3-A\_2\subseteq A\_1, $$
whence $|A\_1|=|A\_2|=|A\_3|=1$. 



> 
> Given that $\phi\_1,\phi\_2,\phi\_3\colon\F\_p\to\R$ are non-constant, what is the largest possible number of pairs $(x,y)\in\F\_p^2$ satisfying
>  $$ \phi\_1(x)+\phi\_2(y)=\phi\_3(x+y)? \tag{$\ast$} $$
>  Equivalently, what is the smallest possible number of pairs $(x,y)\in\F\_p^2$, for all choices of the non-constant functions $\phi\_1,\phi\_2,\phi\_3$, such that ($\ast$) fails to hold?
> 
> 
> 


If $u,v,w\in\F\_p$ are pairwise distinct and $w\ne u+v$ then, letting 
 $$ \phi\_1=1\_{\{u\}},\ \phi\_2=1\_{\{v\}},\ \phi\_3=1\_{\{w\}} $$ 
(the indicator functions of the corresponding singletons), we have $3p-5$ pairs $(x,y)$ violating ($\ast$); is this the worst case?



> 
> Is it true that for any non-constant functions $\phi\_1,\phi\_2,\phi\_3\colon\F\_p\to\R$, there are at least $3p-5$ pairs $(x,y)\in\F\_p^2$ with
>  $$ \phi\_1(x)+\phi\_2(y) \ne\phi\_3(x+y)? \tag{$\circ$}$$
> 
> 
> 


What I can show, in two different ways, is that there are at least $p-1$ pairs $(x,y)$ satisfying ($\circ$). 



Near-linear mappings from $\mathbb F_p$ to $\mathbb R$
Yes, for $p$ sufficiently large.


Assume that $\phi\_1,\phi\_2,\phi\_3$ satisfy this condition for at most $3p-5$ values of $x,y$. 


Hence for each $a$, all but at most $6p-10$ pairs $(x,y)$ satisfy


$$\phi\_1(x+a) + \phi\_2(y) - \phi\_3(x+a+y) = 0 = \phi\_1(x) + \phi\_2(y+a)- \phi\_3(x+a+y)$$


in other words


$$\phi\_1(x+a) - \phi\_1(x) = \phi\_2(y+a) - \phi\_2(y)$$


Now when we have the equation $$F(x)=G(y)$$ for all but at most $6p-10$ pairs $(x,y)$, then for some value of $y$ the equation must be satisfied for all but at most $6$ values of $x$.


So for all $a \in \mathbb F\_p$, there is some $c\_a$ with $\phi\_1(x+a) - \phi\_1(x) = c\_a$ for all but at most $6$ values of $x$.


Now given $a,b$ there are at least $p-18$ values of $x$ where $\phi\_1(x+a)-\phi\_1(x) = c\_a$, $\phi\_1(x+a+b)-\phi\_1(x+a)=c\_b$, $\phi\_2(x+a+b)-\phi\_1(x) = c\_{a+b}$, and hence as soon as $p>18$, $c\_{a+b}=c\_a+c\_b$, $c$ is a homomorphism $\mathbb F\_p \to \mathbb R$ and hence vanishes.


Hence for all $a$ in $\mathbb F\_p$, $\phi\_1(x+a)= \phi\_1(x)$ for all but $6$ values of $x$. Hence $\phi\_1(x)=\phi\_1(y)$ for all but $6p$ values of $x,y$. Hence for some $y$, $\phi\_1(x)=\phi\_1(y)$ for all but $6$ values of $x$.


Applying the same logic to $\phi\_1$ and $\phi\_2$, there are constants $A,B,C$ such that $\phi\_1(x)=A, \phi\_2(x)=B,\phi\_3(x)=C$, each with at most $6$ exceptions. So as soon as $p^3- 18 p^2 > 3p-5$, $A+B=C$.


Now let $\alpha,\beta,\gamma$ be the number of exceptions to these equations. There are $(\alpha+\beta+\gamma)p - (\alpha\beta+ \alpha\gamma+\beta\gamma)$ pairs $x,y$, where exactly one of $x,y,x+y$ is an exception, and for none of these pairs is $\phi\_1(x)+\phi\_2(y)=\phi\_3(x+y)$. For $p$ sufficiently large, this is minimized by minimizing $\alpha+\beta+\gamma$, which is minimized when $\alpha=\beta=\gamma=1$, since all are at least one.


But in this case, the best we can do is your example.


