1. 设 $f,g$ 是 $[a,b]$ 上的连续函数.

(1) 对 $1<p<q<\infty$, $\cfrac{1}{p}+\cfrac{1}{q}=1, a,b>0$, 试证: $$\bex ab\leq \cfrac{1}{p}a^p+\cfrac{1}{q}b^q. \eex$$

(2) 设 $\dps{\vsm{n}a_n}$ 为收敛的正项级数, 试证: $\dps{\vsm{n}a_n^{1-\frac{1}{n}}}$ 也收敛.

(3) 对 $1\leq p\leq \infty$, 定义 $$\bex \sen{f}_p=\sedd{\ba{ll} \dps{\sex{\int_a^b |f(x)|^p\rd x}^\frac{1}{p}},&1\leq p<\infty,\\ \dps{\max_{a\leq x\leq b}|f(x)|},&p=\infty. \ea} \eex$$ 试证: 若 $1\leq p,q\leq \infty$ $\dps{\frac{1}{p}+\frac{1}{q}=1}$, $\sen{f}_p<\infty$, $\sen{g}_q<\infty$, 则 $$\bex \int_a^b |f(x)g(x)|\rd x\leq \sen{f}_p\sen{g}_{q}. \eex$$

(4) 对 $1\leq p\leq \infty$, $\sen{f}_p<\infty$, $\sen{g}_p<\infty$, 试证: $$\bex \sen{f+g}_p\leq \sen{f}_p+\sen{g}_p. \eex$$

(5) 再设 $h$ 在 $f$ 的值域 (是一个区间) 上二阶连续可导, 且 $h‘‘\leq 0$, 则 $$\bex h\sex{\frac{1}{b-a}\int_a^b f(x)\rd x} \geq \frac{1}{b-a} \int_a^b h(f(x))\rd x. \eex$$

(6) 再设 $f$ 恒不为零, 对 $p\in\bbR$, 定义 $$\bex A_p(f)=\sex{\frac{1}{b-a}\int_a^b |f(x)|^p\rd x}^\frac{1}{p}. \eex$$ 试证: $$\beex \bea \lim_{p\to-\infty} A_p(f)&=\min_{a\leq x\leq b}|f(x)|,\\ \lim_{p\to 0} A_p(f)&=\exp\sez{\frac{1}{b-a} \int_a^b \ln |f(x)|\rd x},\\ \lim_{p\to+\infty} A_p(f)&=\max_{a\leq x\leq b}|f(x)|. \eea \eeex$$

(7) 试证: 存在 $\xi\in (a,b)$, 使得 $\dps{\int_a^b f(x)\rd x=f(\xi)(b-a)}$.

(8) 若再设 $f$ 非负严格递增, 则由 (7) 知对 $\forall\ p>0$, $$\bex \exists |\ x_p\in (a,b),\st f^p(x_p)=\frac{1}{b-a}\int_a^b f^p(x)\rd x. \eex$$ 试证: $\vlm{p}x_p$ 存在, 并求之.

证明: (1) $$\beex \bea &\quad ab\leq \frac{1}{p}a^p+\frac{1}{q}b^q=\frac{1}{p}a^p +\sex{1-\frac{1}{p}}b^\frac{p}{p-1}\\ &\lra \frac{a}{b^\frac{1}{p-1}}\leq \frac{1}{p}\sex{\frac{a}{b^\frac{1}{p-1}}}^p+1-\frac{1}{p}\\ &\lra x\leq \frac{1}{p}x^p+1-\frac{1}{p}\\ &\lra x-1\leq \frac{1}{p}(x^p-1)\\ &\la \frac{x-1}{\frac{1}{p}(x^p-1)}=\frac{1}{\xi^{p-1}}\sedd{ \ba{ll} >1,&0\leq x<1,\ x<\xi<1,\\ <1,&1<x<\infty,\ 1<\xi<x. \ea} \eea \eeex$$ 也可另证明如下: 设 $$\bex f(x)=\frac{1}{p}x^p-x+1-\frac{1}{p}, \eex$$ 则 $f(1)=0$, $$\bex f‘(x)=x^{p-1}-1\sedd{\ba{ll} <0,&0<x<1,\\ >0,&x>1. \ea} \eex$$ 而 $f(x)\geq f(1)=0,\ x\in [0,\infty)$.

(2) 由 (1) 知 $$\beex \bea \vsm{n}a_n^{1-\frac{1}{n}} &=2\vsm{n}a_n^{1-\frac{1}{n}}\cdot \frac{1}{2}\\ &\leq 2\vsm{n} \sex{\frac{n-1}{n}a_n+\frac{1}{n}\frac{1}{2^n}}\\ &<\infty. \eea \eeex$$

(3) 不妨设 $\sen{f}_p\neq 0$, $\sen{g}_q\neq 0$, $p\neq 1$, $q\neq \infty$, 而有 $$\beex \bea \int_a^b \sev{\frac{f(x)}{\sen{f}_p}\cdot \frac{g(x)}{\sen{g}_q}}\rd x \leq \int_a^b \sez{\frac{1}{p}\sex{\frac{f(x)}{\sen{f}_p}}^p +\frac{1}{q}\sex{\frac{g(x)}{\sen{g}_q}}^q}\rd x =\frac{1}{p}+\frac{1}{q}=1. \eea \eeex$$

(4) 不妨设 $p\neq 1$, $p\neq \infty$, 而有 $$\beex \bea \sen{f+g}_p^p &=\int_a^b |f(x)+g(x)|^p\rd x\\ &\leq \int_a^b |f(x)+g(x)|\cdot |f(x)+g(x)|^{p-1}\rd x\\ &\leq \int_a^b (|f(x)|+|g(x)|) \cdot |f(x)+g(x)|^{p-1}\rd x\\ &\leq \int_a^b |f(x)| \cdot |f(x)+g(x)|^{p-1}\rd x +\int_a^b |g(x)| \cdot |f(x)+g(x)|^{p-1}\rd x\\ &\leq \sex{\int_a^b |f(x)|^p\rd x}^\frac{1}{p} \cdot \sex{\int_a^b |f(x)+g(x)|^{(p-1)\cdot \frac{p}{p-1}}\rd x}^\frac{p-1}{p}\\ &\quad+\sex{\int_a^b |g(x)|^p\rd x}^\frac{1}{p} \cdot \sex{\int_a^b |f(x)+g(x)|^{(p-1)\cdot \frac{p}{p-1}}\rd x}^\frac{p-1}{p}\\ &=\sen{f}_p\cdot \sen{f+g}_p^{p-1}+ \sen{g}_p\cdot\sen{f+g}_p^{p-1}. \eea \eeex$$

(5) 设 $\dps{A=\frac{1}{b-a}\int_a^b f(x)\rd x}$, 则由 $h‘‘\leq 0$ 知 $$\beex \bea &\quad h(y)\leq h(A)+h‘(A)(y-A)\\ &\ra h(f(x))\leq h(A)+h‘(A)(f(x)-A)\\ &\ra \frac{1}{b-a}\int_a^b h(f(x))\rd x\leq h(A). \eea \eeex$$

(6) 先证 $\dps{\lim_{p\to+\infty} A_p(f)=\max_{a\leq x\leq b}|f(x)|}$. 显然有 $\dps{\limsup_{p\to+\infty} A_p(f)\leq \max_{a\leq x\leq b}|f(x)|}$, 往证 $$\bex \liminf_{p\to+\infty} A_p(f)\geq\max_{a\leq x\leq b}|f(x)|, \eex$$ 而有结论. 设 $$\bex \xi \in [a,b],\st |f(\xi)|=\max_{a\leq x\leq b}|f(x)|. \eex$$ 而由连续函数的保号性 (注意: $f(x)\neq 0,\ \forall\ x$), $$\bex \exists\ [a,b]\supset [c,d]\ni \xi,\st x\in [c,d]\ra |f(x)|>|f(x)|-\ve. \eex$$ 于是 $$\beex \bea A_p(f)&\geq \sex{\frac{1}{b-a}\int_c^d |f(x)|^p\rd x}^\frac{1}{p}\\ &\geq \sez{\frac{1}{b-a}\int_c^d (|f(\xi)|-\ve)^p\rd x}^\frac{1}{p}\\ &=\sex{\frac{d-c}{b-a}}^\frac{1}{p}(|f(\xi)|-\ve),\\ \liminf_{p\to+\infty} A_p(f)&\geq |f(\xi)|-\ve. \eea \eeex$$ 再证 $\dps{\lim_{p\to-\infty} A_p(f)=\min_{a\leq x\leq b}|f(x)|}$ 如下: $$\beex \bea \lim_{p\to-\infty} A_p(f) &=\lim_{p\to-\infty} \sed{\frac{1}{b-a}\sez{\frac{1}{|f(x)|}}^{-p}\rd x}^{\frac{1}{-p}\cdot (-1)}\\ &=\lim_{q\to+\infty} \sed{\frac{1}{b-a}\sez{\frac{1}{|f(x)|}}^q\rd x}^{\frac{1}{q}\cdot(-1)}\\ &=\sez{\lim_{q\to+\infty} A_q\sex{\frac{1}{|f(x)|}}}^{-1}\\ &=\sez{\max_{a\leq x\leq b} \frac{1}{|f(x)|}}^{-1}\\ &=\min_{a\leq x\leq b}|f(x)|. \eea \eeex$$ 最后证明 $\dps{\lim_{p\to 0} A_p(f)=\exp\sez{\frac{1}{b-a} \int_a^b \ln |f(x)|\rd x}}$. 一方面, $$\beex \bea \ln A_p(f)&=\frac{1}{p}\ln\sez{\frac{1}{b-a}\int_a^b |f(x)|^p\rd x}\\ &\geq \frac{1}{p}\frac{1}{b-a}\int_a^b \ln |f(x)|^p\rd x\quad\sex{(\ln x)‘‘<0,\ \mbox{由 }(5)}\\ &=\frac{1}{b-a}\int_a^b \ln |f(x)|\rd x. \eea \eeex$$ 另一方面, $$\beex \bea \ln A_p(f)&=\ln \sez{\frac{1}{b-a}\int_a^b |f(x)|^p\rd x}^\frac{1}{p}\\ &\leq \cfrac{ \dps{\frac{1}{b-a}\int_a^b |f(x)|^p\rd x-1} }{p}\\ &\quad\sex{ \ln x\leq x-1\ (x>0)\ra \ln x^p\leq x^p-1\ra \ln x\leq \frac{x^p-1}{p}\ (p>0) }\\ &=\frac{1}{b-a} \int_a^b \frac{|f(x)|^p-1}{p}\rd x,\\ \liminf_{p\to 0}A_p(f)&\leq \frac{1}{b-a} \int_a^b \lim_{p\to 0}\frac{|f(x)|^p-1}{p}\rd x =\frac{1}{b-a} \int_a^b \ln |f(x)|\rd x. \eea \eeex$$

(7) 设 $\dps{F(x)=\int_a^x f(t)\rd t}$, 则由 Lagrange 中值定理, $$\bex \exists\ \xi\in (a,b),\st f(\xi)=F‘(\xi)=\frac{F(b)-F(a)}{b-a}=\frac{1}{b-a}\int_a^b f(t)\rd t. \eex$$

(8) 由 (1) 知 $$\beex \bea f^p(x_p)&=\cfrac{1}{b-a}\int_a^b f^p(t)\cdot 1\rd t\\ &\leq \cfrac{1}{b-a}\sex{ \int_a^b f^{p\cdot\frac{p+1}{p}}(t)\rd t }^\frac{p}{p+1} \sex{ \int_a^b 1^{p+1}\rd t }^{\frac{1}{p+1}}\\ &=\cfrac{1}{b-a} \sex{\int_a^b f^{p+1}(t)\rd t}^{\frac{p}{p+1}} (b-a)^{\frac{1}{p+1}}\\ &=\sex{\cfrac{1}{b-a}\int_a^b f^{p+1}(t)\rd t}^\frac{p}{p+1}\\ &=f^p(x_{p+1}). \eea \eeex$$ 又 $f$ 严格递增, 我们有 $x_p\leq x_{p+1}$. 如此, $x_p$ 递增有上界. 由单调有界定理, $\dps{\vlm{p}x_p=x_\infty}$ 存在. 另外, 由 (6) 知 $$\beex \bea f(x_p)&=\sez{\cfrac{1}{b-a}\int_a^b f^p(t)\rd t}^{\frac{1}{p}},\\ f(x_\infty)&=\vlm{p}\sez{\cfrac{1}{b-a}\int_a^b f^p(t)\rd t}^{\frac{1}{p}} =\max_{a\leq t\leq b}f(t)=f(b),\\ x_\infty&=b, \eea \eeex$$