Does there exist a function with following properties? Does there exist a differentiable, everywhere concave down function $f(x)$, deﬁned on the whole real line, such that $\lim_{x→−∞}f(x) = 1$ and $\lim_{x→∞}f(x) = −1$? Give an example justifying that it satisﬁes all these properties or explain your reasoning why such a function does not exist.
It is easy to understand intuitively that such a function does not exist. But how do you prove it? 
 A: Hint: a concave function must always lie below a straight line, for instance the tangent line at some point.
A: If such a fonction exists, then $$\forall x,y, \;\;f\left(\frac 12 x + \frac 12 y\right)\geq \frac 12 f(x) + \frac 12 f(y)$$
Let $x$ be fixed for the moment. 
Letting $y\to \infty$ yields $-1\geq \frac 12 f(x) - \frac 12$
Letting $x\to -\infty$ in the last inequality yields $-1\geq \frac 12 - \frac 12=0$, a contradiction.
It seems differentiability isn't even needed.
A: If concave down means $f''(x)<0$.  It is not possible if f(x) and f'(x) are continuous everywhere.
$\lim\limits{x\to\infty} f(x) = 1 \implies \lim\limits{x\to\infty} f'(x) = 0$
Similarly $\lim\limits{x\to-\infty} f(x) = -1 \implies \lim\limits{x\to\infty} f'(x) = 0$
And since $f(\infty) > f(-\infty), f'(x) > 0$ somewhere in between.
This suggest that there exists some $a,b \, a<b$ such that $f'(a)<f'(b)$.
And by the mean value theorem, there exists a $c \in (a,b)$ such that $f''(c) = \frac{f'(b)-f'(a)}{b-a} > 0$
A: Let $f$ be such a function. As $\lim_{x\to-\infty}f(x)=1$ and $\lim_{x\to+\infty}f(x)=-1$ and $f$ is continuous, we conclude that there exists $c\in\Bbb R$ with $f(c)=0$.
As $f$ is differentiable, let $m=-f'(c)$. 
Then $f$ does not ascend above the tangent line through this point, i.e., 
$$\tag1 f(x)\le f(c)+m\cdot(c-x).$$
As $\lim_{x\to-\infty}f(x)=1$, there exists $a<c$ with $f(a)>f(c)$. 
Then from $(1)$ we find $m\ge\frac{f(a)-f(c)}{c-a}>0$.
Then for all $x>c+\frac 3m$, we have
$$f(x)\le f(c)+m(c-x)<f(c)-3<-2$$
contradicting $\lim_{x\to+\infty}f(x)=-1>-2$.
