Let $f:\mathbb{R}\to\mathbb{R}$ be a twice-differentiable function with a point of inflection at $\ x=0.\ $ Specifically, $\ f''(0)=0,\ $ and there exists $\ b>0, c>0\ $ such that $\ f''(x) < 0\ $ on $\ (-b,0),\ $ and $\ f''(x) > 0\ $ on $\ (0,c).\ $
Does there exist $\ a<0,\ d > 0,\ $ such that $\ f(a), f(a+d), f(a+2d)\ $ forms an arithmetic progression, in particular, $\ f(a+d) - f(a) = f(a+2d) - f(a+d)\ ?$
I think the answer is yes. In particular, I think there exists $\ \mu < 0\ $ such that the result is true for every $\ a\in [\mu,0).\ $ However, I don't know how to prove this. Instinctively I think we can apply the Intermediate Value theorem on $\ f'(x),\ $ but I'm not sure how to implement this.
Yes and yes. The following is a proof.
Since $f(x)$ is continuous and $f''(x)>0$ on $(0, c)$, $f(x)$ is strictly convex on $[0,c]$.
Let $\epsilon=f(0)+f(c)-2f(\frac c2)>0$.
Since $f$ is continuous at $x=0$, there exists $\delta_1<0$ such that $f(x) > f(0)-\frac\epsilon3$ for all $x\in[\delta_1,0]$.
Since $f$ is continuous at $x=\frac c2$, there exists $\delta_2<0$ such that $ f(x) <f(\frac c2) +\frac\epsilon3$ for all $x\in[\delta_2+\frac c2,\frac c2]$.
Let $\mu=\max(\delta_1, \delta_2, -b)$. Let $a\in[\mu,0)$. To fulfill OP's expectation, we will show there exists $d>0$ such that $f(a), f(a+d), f(a+2d)$ is an arithmetic progression.
Since $f(x)$ is continuous and $f''(x)< 0$ on $(a,0)$, $f(x)$ is strictly concave on $[a,0]$. So $$f(a)+f(0)-2f(\frac a2)<0.$$
On the other hand, $$\begin{aligned} &\quad f(a)+f(c)-2f(\frac{a+c}2)\\ &=\left(f(0) + f(c)- 2f(\frac c2)\right) + \left(f(a)-f(0)\right) - 2\left(f(\frac a2+ \frac c2)-f(\frac c2)\right)\\ &>\epsilon - \frac\epsilon3-2\frac\epsilon3\\ &=0 \end{aligned}$$
Consider function $g(x)=f(a)+f(a+2x)-2f(a+x)$.
What we just showed is $$g(-\frac a2) < 0 < g(\frac{c-a}2).$$ Thanks to the Intermediate Value theorem, there exists $d\in [-\frac a2,\frac{c-a}2)$ such that $g(d)=0$, i.e., $f(a), f(a+d), f(a+2d)$ is an arithmetic progression.