Do $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1} f(x)$ exist if $f$ is concave over $[0, 1]$? Let $f: [0, 1] \rightarrow \mathbb{R}$ be a concave function. We know that $f$ is continuous over $(0, 1)$. Is it true that both $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1} f(x)$ exist? Normal graphs of a concave function suggest these properties. But I have no idea how to prove them. Thanks.
 A: If $f$ is concave on $[0, 1]$ then
$$ \tag{*}
f(b) \ge \frac{c-b}{c-a} f(a) + \frac{b-a}{c-a} f(c)
$$
for $0 \le a < b < c \le 1$. This can be rewritten as
$$
\frac{c-b}{c-a} \bigl(f(b) - f(a)\bigr) \ge \frac{b-a}{c-a}\bigl(f(c) - f(b)\bigr) \, .
$$
In particular we have the following implication:
$$
\tag{**}
\bigl( 0 \le a < b < c \le 1 \text{ and } f(a) > f(b) \bigr)\implies f(b) > f(c) \, .
$$
Now consider two cases:

*

*$f$ is increasing on $(0, 1)$. $(*)$ with $0 = a < b=1/2 < c < 1$ gives an upper bound for $f(c)$, so that $\lim_{x \to 1-} f(x)$ exists as a finite value.


*Otherwise there is $0 < x_1 < x_2  < 1$ with $f(x_1) > f(x_2)$. Then $(**)$ implies that $f$ is decreasing on $[x_2, 1)$. $(*)$ with $a=0 < b < c=1$ gives a lower bound for $f(b)$, so that $\lim_{x \to 1-} f(x)$ exists as a finite value in this case as well.
A similar argument works for $\lim_{x \to 0+} f(x)$.
It follows from the concavity condition $(*)$ that
$$
\lim_{x \to 0+} f(x) \ge f(0) \\
\lim_{x \to 1-} f(x) \ge f(1)
$$
(i.e. $f$ is lower semi-continuous at the boundary points) but strict inequality can hold.
