$E$ a vector space, $C \subset E$ convex set and $f: C \to \mathbb{R}$ a convex functional. Let $x\in C$ and $h \in E$ such that there is a $\alpha > 0$ with $x + \alpha h $, $x - \alpha h \in C$. We define
$$g(\lambda; x,h) = \frac{f(x+ \lambda h ) - f(x)}{\lambda}$$
I think that the geometric intuition of $g(\lambda; x,h)$ is the slope of a secant line passing through $f(x+ \lambda h )$ and $f(x)$. With this geometric intuition in hands, there is a result that I cannot understand:
$$\quad \quad \quad \quad \quad \quad \quad \quad \quad g(\lambda; x,h) + g(\lambda; x,-h)\geq 0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \hbox{(I)}$$
for example, suppose that $f(x) = x^2$ if $x< 0$ and $f(x) = x^4$ if $x \geq 0$:
If $x = 0$, $h = 0.4$ and $\lambda \leq 1$, the slope $g(\lambda; x,h) > 0$ but $g(\lambda; x,-h) < 0$, $|g(\lambda; x,-h)| > g(\lambda; x,h)$ and
$$g(\lambda; x,h) + g(\lambda; x,-h)< 0$$
Some insight? What is the true intuition of this result $\hbox{(I)}$ ?
I think the best way to visualise this is to see $x$ as the midpoint between $x + \lambda h$ and $x - \lambda h$: $$x = \frac{1}{2}(x + \lambda h) + \frac{1}{2}(x - \lambda h).$$ Using the definition of convexity, $$f(x) \le \frac{1}{2}f(x + \lambda h) + \frac{1}{2}f(x - \lambda h).$$ To put this visually, draw a line segment between the points $(x \pm \lambda h, f(x \pm \lambda h))$, and mark the midpoint, which should lie directly above the point $(x, f(x))$.
From this inequality, we get \begin{align*} 0 &\le \frac{1}{2}(f(x + \lambda h) - f(x)) + \frac{1}{2}(f(x - \lambda h) - f(x)) \\ &= \frac{h}{2}g(\lambda; x, h) + \frac{h}{2}g(\lambda; x, -h). \end{align*}