Problem. Let $F\colon \mathbb{R}^2 \to \mathbb R$ be a non-negative, $C^2$ function which is also strictly convex, meaning that $$ F(\lambda P + (1-\lambda)Q) < \lambda F(P) + (1-\lambda)F(Q), \qquad \forall P\ne Q \in \mathbb R^2, \,\, \forall \lambda \in (0,1). $$ Assume $F(0,0)=0$.
(1) Prove that the equation $F(x,y)=F(0,1)$ defines implicitly a $C^2$ function $y=\phi(x)$ in a neighbourhood of $(0,1)$ with $\phi(0)=1$.
(2) Show that $\phi^{\prime\prime}(0)\le 0$.
Point (1) is an application of Implicit function theorem. Indeed we have that $$ \frac{\partial F}{\partial y}(0,1) = \lim_{t \to 0} \frac{F(0,1)-F(0,1-t)}{t} = \lim_{t \to 0^+} \frac{F(0,1)-F(0,1-t)}{t}. $$ Now using strictly convexity and $F(0,0)=0$ we have for $t \in (0,1)$ $$ \frac{F(0,1)-F(0,1-t)}{t} = \frac{F(0,1)-F(t(0,0) + (1-t)(0,1))}{t} > \frac{F(0,1)- (1-t)F(0,1)}{t} = F(0,1) \ge 0 $$ which is non negative by assumption. Therefore, $F_y(0,1) \ne 0$ and the claim now follows. Can I ask you what you think about this? Is this correct?
EDIT: I am now doubtful about this first point. I have proved that the difference quotients are strictly bounded from below by a non negative constant. But this is not enough to conclude the derivative is nonzero, right? I have just realized that $F(0,1)$ cannot be $0$ because a strictly convex function has at most one global minimum (which in this case will be $(0,0)$). Indeed, if we had $a\ne b$ global minima then $$ f(ta+(1-t)b)<tf(a)+(1-t)f(b) \le tf(b)+(1-t)f(b) = f(b) $$ hence $b$ would not be a minimum. Is this eventually correct?
Concerning point 2 I am a bit in trouble. If I am not wrong I have computed $$ \phi^{\prime\prime}(0) = - \frac{F_y^2F_{xx}-2F_{xy}F_xF_y+F_{yy}F_x^2}{F_y^3}(0,1) $$ but I do not know how to estimate the sign of this expression... Could you help me, please?
Thanks in advance.
Scrict convexity implies that the Hessian of $F$ in any point is positive definite, so $\det H_f>0$ implies that the discriminant of the quadratic form of $F_x$ and $F_y$ appearing in the numerator of your last expression is negative.