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: $$ex ableq cfrac{1}{p}a^p+cfrac{1}{q}b^q. eex$$
(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 $1leq pleq infty$, 定义 $$ex sen{f}_p=sedd{a{ll} dps{sex{int_a^b |f(x)|^p d x}^frac{1}{p}},&1leq p<infty,\ dps{max_{aleq xleq b}|f(x)|},&p=infty. ea} eex$$ 试证: 若 $1leq p,qleq infty$ $dps{frac{1}{p}+frac{1}{q}=1}$, $sen{f}_p<infty$, $sen{g}_q<infty$, 则 $$ex int_a^b |f(x)g(x)| d xleq sen{f}_psen{g}_{q}. eex$$
(4) 对 $1leq pleq infty$, $sen{f}_p<infty$, $sen{g}_p<infty$, 试证: $$ex sen{f+g}_pleq sen{f}_p+sen{g}_p. eex$$
(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 $$ex hsex{frac{1}{b-a}int_a^b f(x) d x} geq frac{1}{b-a} int_a^b h(f(x)) d x. eex$$
(6) 再设 $f$ 恒不为零, 对 $pinbR$, 定义 $$ex A_p(f)=sex{frac{1}{b-a}int_a^b |f(x)|^p d x}^frac{1}{p}. eex$$ 试证: $$eex ea lim_{p o-infty} A_p(f)&=min_{aleq xleq b}|f(x)|,\ lim_{p o 0} A_p(f)&=expsez{frac{1}{b-a} int_a^b ln |f(x)| d x},\ lim_{p o+infty} A_p(f)&=max_{aleq xleq b}|f(x)|. eea eeex$$
(7) 试证: 存在 $xiin (a,b)$, 使得 $dps{int_a^b f(x) d x=f(xi)(b-a)}$.
(8) 若再设 $f$ 非负严格递增, 则由 (7) 知对 $forall p>0$, $$ex exists | x_pin (a,b),st f^p(x_p)=frac{1}{b-a}int_a^b f^p(x) d x. eex$$ 试证: $vlm{p}x_p$ 存在, 并求之.
证明: (1) $$eex ea &quad ableq frac{1}{p}a^p+frac{1}{q}b^q=frac{1}{p}a^p +sex{1-frac{1}{p}}b^frac{p}{p-1}\ &lra frac{a}{b^frac{1}{p-1}}leq frac{1}{p}sex{frac{a}{b^frac{1}{p-1}}}^p+1-frac{1}{p}\ &lra xleq frac{1}{p}x^p+1-frac{1}{p}\ &lra x-1leq frac{1}{p}(x^p-1)\ &la frac{x-1}{frac{1}{p}(x^p-1)}=frac{1}{xi^{p-1}}sedd{ a{ll} >1,&0leq x<1, x<xi<1,\ <1,&1<x<infty, 1<xi<x. ea} eea eeex$$ 也可另证明如下: 设 $$ex f(x)=frac{1}{p}x^p-x+1-frac{1}{p}, eex$$ 则 $f(1)=0$, $$ex f'(x)=x^{p-1}-1sedd{a{ll} <0,&0<x<1,\ >0,&x>1. ea} eex$$ 而 $f(x)geq f(1)=0, xin [0,infty)$.
(2) 由 (1) 知 $$eex ea vsm{n}a_n^{1-frac{1}{n}} &=2vsm{n}a_n^{1-frac{1}{n}}cdot frac{1}{2}\ &leq 2vsm{n} sex{frac{n-1}{n}a_n+frac{1}{n}frac{1}{2^n}}\ &<infty. eea eeex$$
(3) 不妨设 $sen{f}_p eq 0$, $sen{g}_q eq 0$, $p eq 1$, $q eq infty$, 而有 $$eex ea int_a^b sev{frac{f(x)}{sen{f}_p}cdot frac{g(x)}{sen{g}_q}} d x leq int_a^b sez{frac{1}{p}sex{frac{f(x)}{sen{f}_p}}^p +frac{1}{q}sex{frac{g(x)}{sen{g}_q}}^q} d x =frac{1}{p}+frac{1}{q}=1. eea eeex$$
(4) 不妨设 $p eq 1$, $p eq infty$, 而有 $$eex ea sen{f+g}_p^p &=int_a^b |f(x)+g(x)|^p d x\ &leq int_a^b |f(x)+g(x)|cdot |f(x)+g(x)|^{p-1} d x\ &leq int_a^b (|f(x)|+|g(x)|) cdot |f(x)+g(x)|^{p-1} d x\ &leq int_a^b |f(x)| cdot |f(x)+g(x)|^{p-1} d x +int_a^b |g(x)| cdot |f(x)+g(x)|^{p-1} d x\ &leq sex{int_a^b |f(x)|^p d x}^frac{1}{p} cdot sex{int_a^b |f(x)+g(x)|^{(p-1)cdot frac{p}{p-1}} d x}^frac{p-1}{p}\ &quad+sex{int_a^b |g(x)|^p d x}^frac{1}{p} cdot sex{int_a^b |f(x)+g(x)|^{(p-1)cdot frac{p}{p-1}} d x}^frac{p-1}{p}\ &=sen{f}_pcdot sen{f+g}_p^{p-1}+ sen{g}_pcdotsen{f+g}_p^{p-1}. eea eeex$$
(5) 设 $dps{A=frac{1}{b-a}int_a^b f(x) d x}$, 则由 $h''leq 0$ 知 $$eex ea &quad h(y)leq h(A)+h'(A)(y-A)\ & a h(f(x))leq h(A)+h'(A)(f(x)-A)\ & a frac{1}{b-a}int_a^b h(f(x)) d xleq h(A). eea eeex$$
(6) 先证 $dps{lim_{p o+infty} A_p(f)=max_{aleq xleq b}|f(x)|}$. 显然有 $dps{limsup_{p o+infty} A_p(f)leq max_{aleq xleq b}|f(x)|}$, 往证 $$ex liminf_{p o+infty} A_p(f)geqmax_{aleq xleq b}|f(x)|, eex$$ 而有结论. 设 $$ex xi in [a,b],st |f(xi)|=max_{aleq xleq b}|f(x)|. eex$$ 而由连续函数的保号性 (注意: $f(x) eq 0, forall x$), $$ex exists [a,b]supset [c,d] i xi,st xin [c,d] a |f(x)|>|f(x)|-ve. eex$$ 于是 $$eex ea A_p(f)&geq sex{frac{1}{b-a}int_c^d |f(x)|^p d x}^frac{1}{p}\ &geq sez{frac{1}{b-a}int_c^d (|f(xi)|-ve)^p d x}^frac{1}{p}\ &=sex{frac{d-c}{b-a}}^frac{1}{p}(|f(xi)|-ve),\ liminf_{p o+infty} A_p(f)&geq |f(xi)|-ve. eea eeex$$ 再证 $dps{lim_{p o-infty} A_p(f)=min_{aleq xleq b}|f(x)|}$ 如下: $$eex ea lim_{p o-infty} A_p(f) &=lim_{p o-infty} sed{frac{1}{b-a}sez{frac{1}{|f(x)|}}^{-p} d x}^{frac{1}{-p}cdot (-1)}\ &=lim_{q o+infty} sed{frac{1}{b-a}sez{frac{1}{|f(x)|}}^q d x}^{frac{1}{q}cdot(-1)}\ &=sez{lim_{q o+infty} A_qsex{frac{1}{|f(x)|}}}^{-1}\ &=sez{max_{aleq xleq b} frac{1}{|f(x)|}}^{-1}\ &=min_{aleq xleq b}|f(x)|. eea eeex$$ 最后证明 $dps{lim_{p o 0} A_p(f)=expsez{frac{1}{b-a} int_a^b ln |f(x)| d x}}$. 一方面, $$eex ea ln A_p(f)&=frac{1}{p}lnsez{frac{1}{b-a}int_a^b |f(x)|^p d x}\ &geq frac{1}{p}frac{1}{b-a}int_a^b ln |f(x)|^p d xquadsex{(ln x)''<0, mbox{由 }(5)}\ &=frac{1}{b-a}int_a^b ln |f(x)| d x. eea eeex$$ 另一方面, $$eex ea ln A_p(f)&=ln sez{frac{1}{b-a}int_a^b |f(x)|^p d x}^frac{1}{p}\ &leq cfrac{ dps{frac{1}{b-a}int_a^b |f(x)|^p d x-1} }{p}\ &quadsex{ ln xleq x-1 (x>0) a ln x^pleq x^p-1 a ln xleq frac{x^p-1}{p} (p>0) }\ &=frac{1}{b-a} int_a^b frac{|f(x)|^p-1}{p} d x,\ liminf_{p o 0}A_p(f)&leq frac{1}{b-a} int_a^b lim_{p o 0}frac{|f(x)|^p-1}{p} d x =frac{1}{b-a} int_a^b ln |f(x)| d x. eea eeex$$
(7) 设 $dps{F(x)=int_a^x f(t) d t}$, 则由 Lagrange 中值定理, $$ex exists xiin (a,b),st f(xi)=F'(xi)=frac{F(b)-F(a)}{b-a}=frac{1}{b-a}int_a^b f(t) d t. eex$$
(8) 由 (1) 知 $$eex ea f^p(x_p)&=cfrac{1}{b-a}int_a^b f^p(t)cdot 1 d t\ &leq cfrac{1}{b-a}sex{ int_a^b f^{pcdotfrac{p+1}{p}}(t) d t }^frac{p}{p+1} sex{ int_a^b 1^{p+1} d t }^{frac{1}{p+1}}\ &=cfrac{1}{b-a} sex{int_a^b f^{p+1}(t) d t}^{frac{p}{p+1}} (b-a)^{frac{1}{p+1}}\ &=sex{cfrac{1}{b-a}int_a^b f^{p+1}(t) d t}^frac{p}{p+1}\ &=f^p(x_{p+1}). eea eeex$$ 又 $f$ 严格递增, 我们有 $x_pleq x_{p+1}$. 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 $$eex ea f(x_p)&=sez{cfrac{1}{b-a}int_a^b f^p(t) d t}^{frac{1}{p}},\ f(x_infty)&=vlm{p}sez{cfrac{1}{b-a}int_a^b f^p(t) d t}^{frac{1}{p}} =max_{aleq tleq b}f(t)=f(b),\ x_infty&=b, eea eeex$$