函数凹凸性的初等定义

guanmo1

如图，函数凹凸性的初等定义，由两个自变量怎么推广的n个自变量的情形？用数学归纳法吗，求过程。

战巡

回复 1# guanmo1


\[f(\sum_{i=1}^na_ix_i)\le \sum_{i=1}^na_if(x_i)\]
其中$a_i\ge 0$且有
\[\sum_{i=1}^na_i=1\]

guanmo1

回复 2# 战巡


    不是这个哟，是从两个元素出发，推n个元素的式子。不要这个一般的定义。保留这种算术平均数的结构。详见1楼，已编辑。

kuing

反向归纳就行了，思路可以参考这贴中的方法 kuing.cjhb.site/forum.php?mod=viewthread&tid=2776

kuing

不得不提的是，$f\bigl((x_1+x_2)/2\bigr)\leqslant \bigl(f(x_1)+f(x_2)\bigr)/2$ 并不是凸函数的定义式，除非加上 $f(x)$ 是连续函数这一条件。

凸函数的定义式是 $f\bigl(\lambda x_1+(1-\lambda)x_2\bigr)\leqslant \lambda f(x_1)+(1-\lambda)f(x_2)$，其中 $\lambda$ 为 $(0,1)$ 内的任意实数。

kuing

$f\bigl((x_1+x_2)/2\bigr)\leqslant \bigl(f(x_1)+f(x_2)\bigr)/2$ 又称之为“中点凸”，如果没有连续性这一条件，那么就不一定是是凸函数，关于这一点可以参考 kuing.cjhb.site/forum.php?mod=viewthread&tid=100 一贴。

而 $f\bigl(\lambda x_1+(1-\lambda)x_2\bigr)\leqslant \lambda f(x_1)+(1-\lambda)f(x_2)$ 的定义虽然表面看上去没有连续的条件，实际上由这个不等式就可以推出连续。

guanmo1

回复 6# kuing


    是的，现在高中题目的表述是不规范的。

羊1234

全取等号的常函数怎么办？

guanmo1

回复 8# 小芳


    不是这个问题要研究的。那样的话不严格凸就是了。

kuing

常数啥的没所谓，因为一般不需要考虑到这种情况。
非要考虑的话就看是否符合定义就好，而且不同地方定义或许不同，就像高中里的单调性好像是没有等的，但到了大学教材好像又能等。

guanmo1

回复 10# kuing


    是的，完全赞同。

羊1234

原来是这么理解的呀。

caijinzhi

熊斌 王伟叶 函数迭代与函数方程 有详尽的论述。

hbghlyj

kuing 发表于 2014-10-16 08:50
而 $f\bigl(\lambda x_1+(1-\lambda)x_2\bigr)\leqslant \lambda f(x_1)+(1-\lambda)f(x_2)$ 的定义虽然表面看上去没有连续的条件，实际上由这个不等式就可以推出连续。

补充一下，这是 Walter Rudin 所著《数学分析原理》一书中的练习 4.23
Exercise 4.23 A real-valued function $f$ defined in $(a, b)$ is said to be convex if
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
whenever $a<x<b, a<y<b, 0<\lambda<1$.

• Prove that every convex function is continuous.
• Prove that every increasing convex function of a convex function is convex. (For example, if $f$ is convex, so is $e^f$.)
• If $f$ is convex in $(a, b)$ and if $a<s<t<u<b$, show that
  $$
  \frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}
  $$

Solution.

• Fix any points $c$, $d$ with $a<c<d<b$, let $\eta>0$ be any fixed positive number with $\eta<\frac{d-c}{2}$ and consider any two points $x, y$ satisfying $c+\eta \leq x<y \leq d-\eta$. The inequality in the definition implies that $f(t)$ is bounded above on $[c, d]$. Indeed, if $c<t<d$, taking $\lambda=\frac{t-c}{d-c}$, we have $t=(1-\lambda) c+\lambda d$, and so, if $M=\max (f(c), f(d))$, we have
  $$
  f(t) \leq(1-\lambda) f(c)+\lambda f(d) \leq(1-\lambda) M+\lambda M=M .
  $$
  It is less obvious that $f$ is also bounded below on $[c, d]$. In fact if $\frac{c+d}{2}<t<d$, we have
  $$
  \frac{c+d}{2}=(1-\lambda) c+\lambda t
  $$
  where $\lambda=\frac{d-c}{2(t-c)}$, so that
  $$
  f\left(\frac{c+d}{2}\right) \leq\left(\frac{2 t-(c+d)}{2(t-c)}\right) f(c)+\left(\frac{d-c}{2(t-c)}\right) f(t)
  $$
  which implies
  $$f(t) \geq\left(\frac{2(t-c)}{d-c}\right) f\left(\frac{c+d}{2}\right)-\left(\frac{2 t-(c+d)}{d-c}\right) f(c) \geq-2\left|f\left(\frac{c+d}{2}\right)\right|-|f(c)|
  $$
  The proof that $f$ is bounded below on $\left[c, \frac{c+d}{2}\right]$ is similar. Hence there exists $M$ such that $|f(t)| \leq M$ for all $t \in[c, d]$.
  
  We can also write
  $$
  x=(1-\lambda) c+\lambda y
  $$
  where $\lambda=\frac{x-c}{y-c} \in(0,1)$. Accordingly we have
  $$
  \begin{aligned}
  f(x)-f(y) \leq(1-\lambda)(f(c)- & f(y))= \\
  & =\frac{y-x}{y-c}(f(c)-f(y)) \leq \frac{y-x}{\eta}|f(c)-f(y)|
  \end{aligned}
  $$
  Thus
  $$
  f(x)-f(y) \leq \frac{2 M}{\eta}(y-x)
  $$
  Similarly, writing $y=\lambda x+(1-\lambda) d$, where $\lambda=\frac{d-y}{d-x} \in(0,1)$, we find
  $$
  \begin{aligned}
  f(y)-f(x) \leq(1-\lambda)(f(d)- & f(x))= \\
  & =\frac{y-x}{d-x}(f(d)-f(x)) \leq \frac{y-x}{\eta}|f(d)-f(x)|
  \end{aligned}
  $$
  Hence we also have
  $$
  f(y)-f(x) \leq \frac{2 M}{\eta}(y-x)
  $$
  Therefore
  $$
  |f(y)-f(x)| \leq \frac{2 M}{\eta}|y-x|
  $$
  for all $x, y \in[c+\eta, d-\eta]$. Since $c, d$, and $\eta$ are arbitrary, it follows that $f$ is continuous on $(a, b)$.
• If $f(x)$ is convex on $(a, b)$, and $g(x)$ is an increasing convex function on $f((a, b))$, we have
  $$
  g(f(\lambda x+(1-\lambda) y)) \leq g(\lambda f(x)+(1-\lambda) f(y)) \leq \lambda g(f(x))+(1-\lambda) g(f(y))
  $$
• The inequality
  $$
  \frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s}
  $$
  can be rewritten as
  $$
  f(t) \leq \frac{t-s}{u-s} f(u)+\left(1-\frac{t-s}{u-s}\right) f(s)
  $$
  which is precisely the definition of convexity if we note that
  $$
  t=\lambda u+(1-\lambda) s
  $$
  when $\lambda=\frac{t-s}{u-s}$.
  The other inequality is proved in exactly the same way.

hbghlyj

kuing 发表于 2014-10-16 08:36
$f\bigl((x_1+x_2)/2\bigr)\leqslant \bigl(f(x_1)+f(x_2)\bigr)/2$ 并不是凸函数的定义式，除非加上 $f(x)$ 是连续函数这一条件。

真是长知识了！
补充一下，这是 Walter Rudin 所著《数学分析原理》一书中的练习 4.24
Exercise 4.24 Assume that $f$ is a continuous real function defined in $(a, b)$ such that
$$
f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}
$$
for all $x, y \in(a, b)$. Prove that $f$ is convex.

Solution. We shall prove that
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
for all dyadic rational numbers, i.e., all numbers of the form $\lambda=\frac{k}{2^n}$, where $k$ is a nonnegative integer not larger than $2^n$. We do this by induction on $n$. The case $n=0$ is trivial (since $\lambda=0$ or $\lambda=1$ ). In the case $n=1$ we have $\lambda=0$ or $\lambda=1$ or $\lambda=\frac{1}{2}$. The first two cases are again trivial, and the third is precisely the hypothesis of the theorem. Suppose the result is proved for $n \leq r$, and consider $\lambda=\frac{k}{2^{r+1}}$. If $k$ is even, say $k=2 l$, then $\frac{k}{2^{r+1}}=\frac{l}{2^r}$, and we can appeal to the induction hypothesis. Now suppose $k$ is odd. Then $1 \leq k \leq 2^{r+1}-1$, and so the numbers $l=\frac{k-1}{2}$ and $m=\frac{k+1}{2}$ are integers with $0 \leq l<m \leq 2^r$. We can now write
$$
\lambda=\frac{s+t}{2}
$$
where $s=\frac{k-1}{2^{r+1}}=\frac{l}{2^r}$ and $t=\frac{k+1}{2^{r+1}}=\frac{m}{2^r}$. We then have
$$
\lambda x+(1-\lambda) y=\frac{[s x+(1-s) y]+[t x+(1-t) y]}{2}
$$
Hence by the hypothesis of the theorem and the induction hypothesis we have
$$
\begin{aligned}
f(\lambda x+(1-\lambda) y) & \leq \frac{f(s x+(1-s) y)+f(t x+(1-t) y)}{2} \\
& \leq \frac{s f(x)+(1-s) f(y)+t f(x)+(1-t) f(y)}{2} \\
& =\left(\frac{s+t}{2}\right) f(x)+\left(1-\frac{s+t}{2}\right) f(y) \\
& =\lambda f(x)+(1-\lambda) f(y) .
\end{aligned}
$$
This completes the induction.
Now for each fixed $x$ and $y$ both sides of the inequality
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
are continuous functions of $\lambda$. Hence the set on which this inequality holds (the inverse image of the closed set $[0, \infty)$ under the mapping $\lambda \mapsto \lambda f(x)+(1-\lambda ) f(y)-f(\lambda x+(1-\lambda) y)$.) is a closed set. Since it contains all the points $\frac{k}{2^n}$, $0 \leq k \leq n, n=1,2, \ldots$, it must contain the closure of this set of points, i.e., it must contain all of $[0,1]$. Thus $f$ is convex.