Consider the domain $D$ of real numbers $(x,y)\in\mathbb{R}^2$ such that $x\in [0,1]$ and $f(x)\leq y\leq g(x)$, where $f(x)$ and $g(x)$ are twice differentiable convex functions over $x\in [0,1]$ such that $0\leq f(x)\leq g(x)$ for all $x\in [0,1]$ and $f(0)=g(0)=0$.
Now, the cone of feasible directions of $D$ at a point $x_0\in D\subset\mathbb{R}^2$ is defined as the set of numbers $d\in\mathbb{R}^2$ such that $x_0+\lambda d\in D$ for all $\lambda\in (0,\delta)$ for some $\delta>0$.
I am theoretically interested in finding a method for finding the feasible direction at $(0,0)$ for the set $D$ given above.
I have tried the following. Let $h_1(x,y)=f(x)-y$ and $h_2(x,y)=y-g(x)$. Note that $h_i(0,0)=0$ for $i=1,2$, and that for all $(x,y)\in D$ we have $h_1(x,y)\leq 0$ and $h_2(x,y)\leq 0$. This means that if $d$ is to be a feasible direction at $(0,0)$ (i.e., an element of the cone of feasible directions at $(0,0)$), then the value of $h_i$, $i=1,2$, must stay the same at $0$, or must decrease.
I have proven a theorem (in an exercise in a book on convex optimization) that says the following: For a convex function $u:\mathbb{R}^2\to\mathbb{R}$ differentiable at $(x,y)$, the existence of a vector $d\in\mathbb{R}^2$ such that $\nabla u(x,y)^t d\leq 0$ is equivalent to the existence of a vector $d\in\mathbb{R}^2$ and a real number $\delta>0$ such that $u((x,y)+\lambda d)\leq u(x,y)$ for all $\lambda\in (0,\delta)$ (note: such a vector $d$ may be called a non-strict descent direction).
Thus, to find a method, I would first note that $h_1(x,y)=f(x)+(-y)$ is a sum of convex functions, and thus convex, and then apply the above result, to say that a feasible direction of $D$ at $(0,0)$ must be such that $\nabla h_1(0,0)^t d\leq 0$ or $$(f'(x),-1)\begin{pmatrix}d_1\\ d_2\end{pmatrix} \leq 0.$$ This ensures us that as we move along the direction $d$ in such a way that we do not increase the value of $h_1$ from $(0,0)$ at which it attains its highest value $0$.
However, the function $h_2(x,y)=y-g(x)$ is not necessarily convex, as $-g(x)$ is concave since $g(x)$ is convex. Though I would want to say something like " $(-g'(x),1)\begin{pmatrix}d_1\\ d_2\end{pmatrix} \leq 0$ must hold for all feasible non-strict directions of descent $d$ of $D$ at $(0,0)$", my theorem does not guarantee this.
So is there some general method or way of thinking to find feasible directions of $D$ at $(0,0)$?
To take a concrete example, let $f(x)=(e-1)x^2$ and $g(x)=e^x-1$. Both are convex, and they are shown below. The question, then, arises: What are the feasible directions of the set $D=\{(x,y)\in\mathbb{R}^2|0\leq x\leq 1,\quad f(x)\leq y\leq g(x)\}$ at $(0,0)$?
The method by which we answer the above problem will, I hope, shed some light on how to find a feasible direction in general.
I assume that $f$ and $g$ are twice differentiable. Then the set consists of all conic combinations of $[1,f'(0)]$ and $[1,g'(0)]$. The set may not be closed (in your example, $[1,0]$ is not a feasible direction), so you may need to exclude the extreme rays. The ray for $f$ should be excluded if $f$ is strictly convex on $(0,\delta)$, while the ray for $g$ should be excluded if $g$ is strictly concave on $(0,\delta)$ (for some $\delta>0$).
Let me provide the proof for the ray for $f$. Take $\delta$ small enough such that $f$ strictly convex on $(0,\delta)$ and let $x \in (0,\delta)$. By Taylor's theorem and the mean value theorem, $f(x) = f(0) + f'(0)(x-0) + f''(a)(x-0)^2$ for some $a \in (0,\delta)$. This simplifies to $f(x) = f'(0)x + f''(a)x^2$. The second term is positive since $f$ is strictly convex.