是否存在$f:[0,∞)→\Bbb R$是次加的但不是凹函数

hbghlyj

$f:[0,∞)→\Bbb R$是凹函数,且$f(0) ≥ 0$,则$f$是次加的.
证: $f$是凹函数, 所以$f(a)+f(b)≥f(a+b)+f(0)$, 而$f(0) ≥ 0$, 所以$f(a)+f(b)≥f(a+b)$. $\boxed{\hskip0.15ex}$

又见Wikipedia - Concave function

If a function $f$ is concave, and $f(0) ≥ 0$, then $f$ is subadditive on $[0,\infty)$.
Proof:
• Since $f$ is concave and $1 ≥ t ≥ 0$, letting $1=y= 0$ we have$$f(tx) = f(tx+(1-t)\cdot 0) \ge t f(x)+(1-t)f(0) \ge t f(x) .$$
• For $a,b\in[0,\infty)$:
$$f(a) + f(b) = f \left((a+b) \frac{a}{a+b} \right) + f \left((a+b) \frac{b}{a+b} \right)
\ge \frac{a}{a+b} f(a+b) + \frac{b}{a+b} f(a+b) = f(a+b)$$

问题: 是否存在$f:[0,∞)→\Bbb R$是次加的但不是凹函数

hbghlyj

$f:[0,∞)→\Bbb R$定义为
$$f(x)=\min(1,x)$$
则$f$是次加的但不是凹函数.

hbghlyj

Is a function concave if and only if $f(x)/x$ decreasing?
Suppose that a function $f:[0,1] \to [0,1]$ is non-decreasing and $f(0)=0$. Then $f$ is concave if $f(x)/x$ is non-increasing.
Counter-example:
Consider $$f(x)=\begin{cases}
2x & 0 \leq x < \tfrac13 \\
\tfrac23 & \tfrac13 \leq x < \tfrac23 \\
x & \tfrac23 \leq x \leq 1
\end{cases}
$$

hbghlyj

Wikipedia:
A function $p : X → \mathbb {R}$ is convex and satisfies $p ( 0 ) ≤ 0$ if and only if $p ( a x + b y ) ≤ a p ( x ) + b p ( y ) $ for all vectors $x , y ∈ X $ and all non-negative real $a , b ≥ 0$ such that $a + b ≤ 1$. Every sublinear function is a convex function. On the other hand, if $p : X → \mathbb {R}$ is convex, the function $p_{0}\leq p$ defined by $p_{0}(x):=\inf _{t>0}{\frac {p(tx)}{t}}$ is sublinear, and satisfies $ F\leq p_{0}$ for all linear functional $F ≤ p$. So the extension of the Hahn-Banach theorem to convex functionals is easy, but has not a much larger content than the classical one stated for sublinear functionals.