Proof of Convexity Using Quasiconvexity Suppose $f_{1}, \ldots, f_{n},~n\geq 3$, are continuously differentiable functions defined on $\mathbb{R}$ such that for each $i\in \{1,\ldots,n\}$, we have:
(i) $f_{i}(v_{i}^{0})=0$ and $f_{i}^{'}(v_{i}^{0})=0$, $v_{i}^{0}\in \mathbb{R}$
(ii) the function $F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ defined by
\begin{eqnarray}
F(\textbf{v})=\sum\limits_{i=1}^{n} f_{i}(v_{i}),~\textbf{v}=(v_{1},\ldots,v_{n})^{T},~ v_{i}\in \mathbb{R}
\end{eqnarray}
is nonnegative and strictly quasiconvex on $\mathbb{R}$, i.e., for any $\textbf{v},\textbf{v}^{'}\in \mathbb{R}^{n},~\textbf{v}\neq\textbf{v}^{'}$, and any $0<\alpha<1$, we have 
\begin{eqnarray}
F(\alpha\textbf{v}+(1-\alpha)\textbf{v}^{'})<\max(F(\textbf{v}),F(\textbf{v}^{'})).
\end{eqnarray}
Show that $f_{i}$, $1\leq i\leq n$, have to necessarily be convex.
I started by taking $\textbf{v}=(v,v_{2}^{0},\ldots,v_{n}^{0})^{T}$ and $\textbf{v}^{'}=(v^{'},v_{2}^{0},\ldots,v_{n}^{0})^{T}$, $v\neq v^{'}$, and using these in the quasiconvex relation for $F$, but could not proceed far.
Can someone please provide a proof of this?
 A: I don't think that $f_i,\;1\leq i\leq n$, have to be convex. Maybe this helps:
Let's choose
$$f_1(x)=1-e^{-x^2},\;x\in\mathbb{R},$$
$$f_i\equiv 0,\; \;2\leq i\leq n.$$
All the $f$ are differentiable and clearly satisfies condition (i) in zero. Now, as they are non negative and strictly quasiconvex (in the graph of the function $f_1$ it is pretty clear, but of course it must be proven), the function
$$F(\textbf{v})=\sum\limits_{i=1}^n{f_i(v_i)}=f_1(v_1),$$
satisfies (ii). But $f_1$ is not a convex functions.
$\textbf{Additional}$: I think that for $f_i$ to be convex and with your start in the proof, then it must be truth that if a differentiable function $F$ non negative and strictly quasiconvex have a common zero with its derivative, then $F$ is convex, but it is not.
A: I don't quite have the whole proof but I think I am close.
We know already (using the reasoning you had at the bottom) that each $f_i$ must be quasiconvex and non-negative as well (otherwise you could force $F$ to have a negative value).
So now the only question is are the functions psuedoconvex (strictly quasiconvex and continuously differenentiable) and not convex?  Well the best way I could figure out to do it is proof by contradiction.  Lets assume at least one of the functions $f_k$ is pseudoconvex but not convex.
Then there must be at least an $x$ and $y$ and $\alpha^*$ where $f_k(\alpha^* x + (1-\alpha^*) y) = f(v^*) > \alpha^* f_k(x) + (1-\alpha^*) f_k(y)$.  Now note that without loss of generality we can have $x = v_k^0$ and $y = \alpha^* x + (1-\alpha^*) y + \epsilon$ where $\epsilon$ is really small. Basically I am just squeezing $y$ closer to our special point and then adjusting $\alpha^*$ appropriately.
Now we consider $F(\alpha^* v + (1-\alpha^*) v')$ where $v = (v^-_1,\ldots,v_k^0,\ldots,v_n^-)$ and $v' = (v_1^+,\ldots,y,\ldots,v_n^+)$.  If we pick our $v_i^-$ and $v_i^+$ right then we can get that $F$ isn't strictly quasiconvex which is the contradiction we need to finish the proof.
$F(\alpha^* v + (1-\alpha^*) v') > F(v)$ if we just make each $v_i^-$ close enough to $v_i^0$.  Now I want to say that we can choose $v_i^+$ so that $F(\alpha^* v + (1-\alpha^*) v') > F(v')$ then we would be done.
Note: Since we know the functions are pseudoconvex then the global minimum has to be at each $v^0_i$ for the respective function. That is why we are choosing values around each $v^0_i$ because we know we can get arbitrarily small function values around those points i.e. we can pick $v_i^-$ so that $f(v_i^-)$ is very small.
