If there is a constant $K<0$ such that $f''(x)
Let $f:\Bbb R \to \Bbb R$ be twice-differentiable. If there is a constant $K<0$ such that $f''(x)<K$ for all $x \in \mathbb R$, show that $f$ attains a global maximum.
I am not sure how best to prove this in a rigorous fashion, so I would be interested in seeing some solutions. I felt my solution was rather hand-wavy. I just said that $f$ cannot have asymptotes else $f''(x)\to 0$ as $x \to \pm \infty$. So $f'(x) \to -\infty$ as $x \to +\infty$ and $f'(x)\to +\infty$ as $x \to - \infty$. So by intermediate value theorem applied to $f'$ there exists $c\in \Bbb R$ such that $f'(c)=0$ i.e. we have a local maximum. I have already proven (in an earlier part of the exam question which this question is a part of) that if $f''(x)<0$ for all $x \in \Bbb R$ then $f$ can have at most one local maximum, so now we know $f$ has exactly one local maximum. We already "proved" there are no asymptotes so $f(x) \to -\infty$ as $|x|\to \infty$, so we know that this is the global maximum.
 A: Integrating twice, we see that $$f(x) \le \frac K 2 x^2 + f'(0) x + f(0) \le f(0) - \frac{f'(0)^2}{2K}$$ which shows that $\lim_{x\to \pm \infty} f(x) = -\infty$ since $K < 0$ and that $f$ remains bounded on $\mathbb R$. Thus $S = \sup_{x \in \mathbb R}f(x)$ is finite. Since $f(x) \to -\infty$ as $x\to\pm \infty$, there is $N \in \mathbb N$ such that $$f(x) < S-1 \,\,\,\,\, \text{ when } \,\,\,\,\, \lvert x \rvert > N.$$ Since $[-N,N]$ is compact, $T = \max_{x \in [-N,N]} f(x)$ must be achieved at some $x^* \in [-N,N]$. If $T < S$, then we have violated the definition of supremum since $f(x) \le \max\{T,S-1\}$ everywhere and thus one of $T,S-1$ would be a smaller upper bound. Thus $T = S$ and $f(x^*) = S$ shows that $f$ attains its (global) maximum.
A: Since the second derivative is negative it follows that $f'$ is strictly decreasing and hence must tend to $-\infty$ (it can't tend to a finite number as that will contradict $f''<K$ via mean value theorem) as $x\to\infty$. In a similar fashion note that as $x\to - \infty$ the derivative $f'(x) \to \infty$ and hence there is a unique number $c$ such that $f'(c) =0$ and $f'(x) >0 $ for $x<c$ and $f'(x) <0$ for $x>c$. It should now be clear that $f(c) $ is a global maximum value of $f$. 
