Let $g:[a,+\infty)\rightarrow \Bbb R$ be twice differentiable with $\lim_{x\rightarrow\infty}g(x)=g(a)$. Prove that $g''(x)=0$. Let $g:[a,+\infty)\rightarrow \Bbb R$ be twice differentiable with $\lim_{x\rightarrow\infty}g(x)=g(a)$. Prove that there exists $c\in(a,+\infty)$ such that $g''(x)=0$.
Taking $g '' (x)\neq0$ then g can be strictly concave($x>c$) or strictly convex ($x<c$). Proving that this point c is critical would solve the question?
 A: Suppose $g''$ is never $0.$ Then by Darboux, either $g''>0$ on $[a,\infty)$ or $g''<0$ on $[a,\infty).$ Suppose WLOG the first case holds.
Claim: $g'(x)>0$ for some $x\in [a,\infty).$ Proof: Suppose not. Then $g'\le 0$ everywhere. That implies $g$ is nonincreasing. We can't have $g\equiv g(a),$ because that would imply $g''\equiv 0.$ Thus $g(x)< g(a)$ for some $x>a.$ Because $g$ is nonincreasing, $g(y)\le g(x) <g(a)$ for $y >x.$ Thus $\lim_{x\to \infty} g(x)$ does not equal $g(a),$ contradiction, giving the claim.
But if $g'(x) > 0,$ then $g''>0$ implies $g'(y) \ge g'(x)$ for $y\in [x,\infty).$ This implies $\lim_{x\to \infty} g(x)=\infty,$ contradiction, and we're done.
A: Note that $$\int_a^{\infty} g'(x)\,\mathrm{d}x = 0$$ which implies that $g'(x) = 0$ for at least one $x\in (a, \infty)$. Furthermore, note that if $g'(x_1) = g'(x_2) = 0$ for some $x_1\neq x_2\in [a, \infty)$, then there is some $x\in (x_1, x_2)$ such that $g''(x) = 0$ by the mean value theorem. Therefore, we let $x^*\in (a, \infty)$ such that the sign of $g'(x)$ for $x\in [a, x^*)$ and the sign of $g'(x)$ for $x\in (x^*, \infty)$ are opposite (without loss of generality, we can let $g'(x) < 0$ for $x\in [a, x^*)$ and $g'(x) > 0$ for $x\in (x^*, \infty)$).
Now, consider that $\int_{x^*}^{\infty} g'(x)\,\mathrm{d}x = \int_{x^*}^a g'(x)\,\mathrm{d}x = g(a)-g(x^*)$ is finite. Let $y\in (x^*, \infty)$ such that $g'(y) > 0$. Then, if $g'(x)\geq g'(y)$ for all $x\in (y, \infty)$, we have that $$\int_{x^*}^{\infty} g'(x)\,\mathrm{d}x = \int_{x^*}^y g'(x)\,\mathrm{d}x+\int_y^{\infty} g'(x)\,\mathrm{d}x > \int_y^{\infty} g'(y)\,\mathrm{d}x = \infty$$ which contradicts that $\int_{x^*}^{\infty} g'(x)\,\mathrm{d}x$ is finite. Therefore, there is some $x_2\in (y, \infty)$ such that $g'(x_2) < g'(y)$. As $g'$ is continuous, we have by the intermediate value theorem that there is some $x_1\in (x^*, y)$ such that $g'(x_1) = g'(x_2)$. Applying the mean value theorem again, we get an $x\in (x_1, x_2)$ such that $g''(x) = 0$.
A: If $g$ is constant then we are done. So let us assume that $g$ is not constant and hence there is a value of $g$ different from $g(a)$. Also let's assume that this value say $g(b)$ for some $b > a$ is greater then $g(a)$ (the proof is similar if this value $g(b)$ is less than $g(a)$).
Now $g(x) \to g(a)$ as $x \to \infty$ and hence there is an $M > a$ such that $g(x) < g(b)$ for all $x \geq M$. And therefore the maximum of $g$ in $[a, M]$ is attained at an interior point $c$ and the derivative $g'(c) = 0$.
Next note that $g(a) < g(c) > g(M)$. Hence we have points $d, e$ such that $d\in (a, c), e\in (c, M)$ and $$g'(d) = \frac{g(c) - g(a)}{c - a} > 0, g'(e) = \frac{g(c) - g(M)}{c - M} < 0$$ Next let's assume that $g''$ never vanishes, then by Darboux theorem it maintains a constant sign and clearly this sign is negative because $d < e, g'(d) > g'(e)$. Thus $g'$ is strictly decreasing and hence $g'(x) <g'(e) <0$ for $x>e$.
By mean value theorem we can see that $$g(x) =g(e) +(x-e) g'(\xi) $$ for some $\xi\in(e, x) $ and then $$g(x) <g(e) +(x-e) g'(e) $$ for all $x>e$. We obtain an obvious contradiction when $x\to\infty$ as LHS tends to a constant $g(a) $ and RHS tends to $-\infty$. 

To avoid any nitpick I explicitly state that the symbol $e$ used in above argument is not related to the constant $e$ defined by $e=\lim_{n\to\infty} (1+n^{-1})^{n}$.
