中值定理证明凸-切线不等式

青青子衿

※）命题1：若函数$\,f(x)\,$在区间$\,(a,b)\,$内可导，在区间$\,[a,b]\,$恒下凸，
则对$\,\forall\,x_0\in(a,b)\,$，恒有$\,\forall\,x\in[a,b]\,$满足
\[f(x)\geq\,\!f'(x_0)(x-x_0)+f(x_0)\]
※）命题1的证明：构造
\[ g(x)=f(x)-\dfrac{f(u)-f(v)}{u-v}(x-v) \]
其中$a\leq\,\!u<v\leq\,\!b$，则$\,g(u)=f(v)=g(v)\,$，
由Rolle中值定理，可知$\,\exists\,x_0\in(u,v)\,$使得
\[ g'(x_0)=0\quad\Rightarrow\quad\,\!f'(x_0)=\dfrac{f(u)-f(v)}{u-v} \]
则$\,g(x)\,$可以改写成$\,g(x)=f(x)-f'(x_0)(x-v)\,$，
从而有$\,g'(x)=f'(x)-f'(x_0)\,$
根据引理1，可知$\,f(x)\,$在$\,[a,b]\,$上是单调递增的.
当$\,a<x<x_0\,$时，$f'(x)<f'(x_0)$，即$\,g'(x)<0\,$；
当$\,x_0<x<b\,$时，$f'(x)>f'(x_0)$，即$\,g'(x)>0\,$.
所以
\begin{alignat*}{4}
&&g(x_0)
&\leq\,\!g(x)\\
\Rightarrow\quad
&&f(x_0)-f'(x_0)(x_0-v)
&\leq\,\!f(x)-f'(x_0)(x-v)\\
\Rightarrow\quad
&&f(x_0)+f'(x_0)(x-x_0)
&\leq\,\!f(x)\\
\end{alignat*}
又因为$u,v$的任意性，所以$x_0$也具有任意性，于是命题成立.



*）引理1：一阶可导凸函数的导函数单调递增.
*）引理1的证明：不妨设$\,x_1,x_2\in[a,b]\,$，且$\,x_1<x_2\,$，
可导函数$\,f(x)\,$为凸函数，以及$\,\lambda\in(0,1)\,$，
记$\,t=\lambda\,\!x_1+(1-\lambda)x_2\,$，则
\begin{alignat*}{4}
&&f\big[\lambda\,\!x_1+(1-\lambda)x_2\big]
&\leq\lambda\,\!f(x_1)+(1-\lambda)f(x_2)\\
\Rightarrow\quad
&&f\big(t\big)
&\leq\lambda\,\!f(x_1)+(1-\lambda)f(x_2)\\
\Rightarrow\quad
&&\lambda\,\!f(t)+(1-\lambda)f(t)
&\leq\lambda\,\!f(x_1)+(1-\lambda)f(x_2)\\
\Rightarrow\quad
&&0
&\geq\lambda[f(t)-f(x_1)]+(1-\lambda)[f(t)-f(x_2)]\\
\end{alignat*}
利用Lagrange中值定理，$\,\exists\,\xi,\eta\>$满足$\,x_1<\xi<t<\eta<x_2\,$，使得
\begin{align*}
&\quad\>\>\lambda[f(t)-f(x_1)]+(1-\lambda)[f(t)-f(x_2)]\\
&=\lambda\,\!f'(\xi)(t-x_1)+(1-\lambda)f'(\eta)(t-x_2)
\end{align*}
又因为$\,t=\lambda\,\!x_1+(1-\lambda)x_2\,$，则
\begin{aligned}
\left\{
\begin{split}
t-x_1&=(1-\lambda)(x_2-x_1)\\
t-x_2&=\lambda(x_1-x_2)\\
\end{split}\right.
\end{aligned}
因此，有
\begin{alignat*}{4}
&&
&\quad\>\>\lambda[f(t)-f(x_1)]+(1-\lambda)[f(t)-f(x_2)]\\
&&
&=\lambda(1-\lambda)(x_2-x_1)[f'(\xi)-f'(\eta)]\leq0\\
\Rightarrow\quad
&&
&\qquad\qquad\qquad\,f'(\xi)\leq\,\!f'(\eta)
\end{alignat*}
所以引理成立.

青青子衿

\begin{aligned}
\left\{
\begin{split}
&f\big[\lambda\,\!x_1+(1-\lambda)x_2\big]\leq\lambda\,\!f(x_1)+(1-\lambda)f(x_2)\\
&\qquad\qquad\qquad\,f(x)\in\mathcal{C}^1(0,1)
\end{split}\right.
\quad\Rightarrow\quad\,\!f'(x)\nearrow
\end{aligned}
※※※※※
\begin{alignat*}{4}  
&& 0\leqslant\,\!x_1<
&\,\xi<x_0\leqslant1\\
\\
\Rightarrow\quad&&f'(\xi)&\leqslant\,\!f'(x_0)\\
\Rightarrow\quad&&f(x_0)-f(x_1)&=f'(\xi)(x_0-x_1)\\
&&&\leqslant\,\!f'(x_0)(x_0-x_1)\\
\Rightarrow\quad
&&f(x_1)&\geqslant\,\!f'(x_0)(x_1-x_0)+f(x_0)\\
\\
&& 0\leqslant\,\!x_0<
&\,\eta<x_2\leqslant1\\
\\
\Rightarrow\quad
&&f'(\eta)&\geqslant\,\!f'(x_0)\\
\Rightarrow\quad
&&f(x_2)-f(x_0)&=f'(\eta)(x_2-x_0)\\
&&&\geqslant\,\!f'(x_0)(x_2-x_0)\\
\Rightarrow\quad
&&f(x_2)&\geqslant\,\!f'(x_0)(x_2-x_0)+f(x_0)
\end{alignat*}

\begin{align*}
\color{black}{
\begin{split}
\int _{a}^{b}f(x)\,\mathrm{d}x&=1\\
\int _{a}^{b}f(x)g(x)\,\mathrm{d}x&\le\ln\left\{\int _{a}^{b}f(x)\exp \left[g(x)\right]\,\mathrm{d}x\right\}\\
\int _{a}^{b}f(x)g(x)\,\mathrm{d}x&\le\int _{a}^{b}f(x)\ln \left[f(x)\right]\,\mathrm{d}x+\ln\left\{\int _{a}^{b}\exp \left[g(x)\right]\,\mathrm{d}x\right\}
\end{split}}
\end{align*}

hbghlyj

math.stackexchange.com/questions/3150163
If $f:\mathbb{R}\to\mathbb{R}$ is a concave function such that $\lim\limits_{x\to\infty}(f(x)-f(x-1))=0$ then $f$ increasing.  
若 ‎$‎‎f$ ‎可导, 使用中值定理, 推出 ‎‎$\lim\limits_{x\to\infty}f'‎(x)=0$, 使用 $f'$ 是减函数, $‎f'‎\geq0$, 故 $‎‎f$‎ 是增函数. ‎但是不需要可导就可以证明:
Hint. Since $f$ is concave it follows that for any $x_0\in \mathbb{R}$ the function
$$x\to R(x,x_0):=\frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $\mathbb{R}\setminus\{x_0\}$. Here it is a reference for the convex function $-f$.

Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)\geq R(x_0+k+1,x_0)\geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.

Can you take it from here?

hbghlyj

en.wikipedia.org/wiki/Subderivative
Lemma 4.3.1. Let $c: I \rightarrow \mathbb{R}$ be convex and let $m$ be a point in the interior of $I$. Then there exist $a, b \in \mathbb{R}$ such $c(x) \geq a x+b$ for all $x$, with equality at $x=m$.
Proof. By convexity, for $m, x, y \in I$ with $x<m<y$, we have
$$
\frac{c(m)-c(x)}{m-x} \leq \frac{c(y)-c(m)}{y-m}
$$
So, fixing an interior point $m$, there exists $a \in \mathbb{R}$ such that, for all $x<m$ and all $y>m$
$$
\frac{c(m)-c(x)}{m-x} \leq a \leq \frac{c(y)-c(m)}{y-m}
$$
Then $c(x) \geq a(x-m)+c(m)$, for all $x \in I$.