1计算.

(1) \dps{\lim_{n\to\infty} \ln\sqrt[n]{\sex{1+\frac{1}{n}} \sex{1+\frac{2}{n}} \cdots \sex{1+\frac{n}{n}} }}<script type="math/tex">\dps{\lim_{n\to\infty} \ln\sqrt[n]{\sex{1+\frac{1}{n}} \sex{1+\frac{2}{n}} \cdots \sex{1+\frac{n}{n}} }}</script>.

解答: \bex \mbox{原式}&=&\lim_{n\to\infty} \sed{ \frac{1}{n} \sez{\ln\sex{1+\frac{1}{n}}+\cdots+\ln \sex{1+\frac{n-1}{n}}} +\frac{\ln 2}{n}}\\ &=&\int_0^1\ln\sex{1+x}\rd x\\ &=&\int_1^2\ln x\rd x\\ &=&\left.x\ln x-x\right|_1^2\\ &=&2\ln2-1. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\lim_{n\to\infty} \sed{ \frac{1}{n} \sez{\ln\sex{1+\frac{1}{n}}+\cdots+\ln \sex{1+\frac{n-1}{n}}} +\frac{\ln 2}{n}}\\ &=&\int_0^1\ln\sex{1+x}\rd x\\ &=&\int_1^2\ln x\rd x\\ &=&\left.x\ln x-x\right|_1^2\\ &=&2\ln2-1. \eex</script>

(2) \dps{\lim_{n\to\infty} \frac{1\cdot 3\cdot \cdots\cdot \sex{2n-1}} {2\cdot 4\cdot \cdots\cdot \sex{2n}}}<script type="math/tex">\dps{\lim_{n\to\infty} \frac{1\cdot 3\cdot \cdots\cdot \sex{2n-1}} {2\cdot 4\cdot \cdots\cdot \sex{2n}}}</script>.

解答: 由 2k=\frac{\sex{2k-1}+\sex{2k+1}}{2} >\sqrt{\sex{2k-1}\sex{2k+1}},\ k=1,2,\cdots,n, <script type="math/tex; mode=display"> 2k=\frac{\sex{2k-1}+\sex{2k+1}}{2} >\sqrt{\sex{2k-1}\sex{2k+1}},\ k=1,2,\cdots,n, </script> 知 0<\frac{1\cdot 3\cdot \cdots\cdot \sex{2n-1}} {2\cdot 4\cdot \cdots\cdot \sex{2n}} <\frac{1}{\sqrt{2n+1}}, <script type="math/tex; mode=display"> 0<\frac{1\cdot 3\cdot \cdots\cdot \sex{2n-1}} {2\cdot 4\cdot \cdots\cdot \sex{2n}} <\frac{1}{\sqrt{2n+1}}, </script> 而 \bex \mbox{原式}&=&0. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&0. \eex</script>

(3) \dps{\lim_{x\to+\infty} \sed{ \sex{x^3+3x}^\frac{1}{3} -\sex{x^2-2x}^\frac{1}{2} }}<script type="math/tex">\dps{\lim_{x\to+\infty} \sed{ \sex{x^3+3x}^\frac{1}{3} -\sex{x^2-2x}^\frac{1}{2} }}</script>.

解答: \bex \mbox{原式}&=&\lim_{x\to +\infty} \frac{ \dps{\sex{1+\frac{3}{x^2}}^\frac{1}{3} -\sex{1-\frac{2}{x}}^\frac{1}{2}}} {\dps{\frac{1}{x}}}\\ &=&\lim_{t\to 0_+} \frac{\sex{1+3t^2}^\frac{1}{3}-\sex{1-2t}^\frac{1}{2}}{t}\\ &=&\lim_{t\to 0_+}\sez{ \frac{1}{3}\sex{1+3t^2}^{-\frac{2}{3}}\cdot 6t -\frac{1}{2}\sex{1-2t}^{-\frac{1}{2}}\cdot \sex{-2}}\\ &=&1. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\lim_{x\to +\infty} \frac{ \dps{\sex{1+\frac{3}{x^2}}^\frac{1}{3} -\sex{1-\frac{2}{x}}^\frac{1}{2}}} {\dps{\frac{1}{x}}}\\ &=&\lim_{t\to 0_+} \frac{\sex{1+3t^2}^\frac{1}{3}-\sex{1-2t}^\frac{1}{2}}{t}\\ &=&\lim_{t\to 0_+}\sez{ \frac{1}{3}\sex{1+3t^2}^{-\frac{2}{3}}\cdot 6t -\frac{1}{2}\sex{1-2t}^{-\frac{1}{2}}\cdot \sex{-2}}\\ &=&1. \eex</script>

(4) \dps{\int_0^\pi \frac{x\sin x}{1+\cos^2x}\rd x}<script type="math/tex">\dps{\int_0^\pi \frac{x\sin x}{1+\cos^2x}\rd x}</script>.

解答: 由变量替换 x\leftrightsquigarrow \pi-x<script type="math/tex">x\leftrightsquigarrow \pi-x</script> 知 \int_0^\pi \frac{x\sin x}{1+\cos^2x}\rd x =\int_0^\pi \frac{\sex{\pi-x}\sin x}{1+\cos^2x}\rd x, <script type="math/tex; mode=display"> \int_0^\pi \frac{x\sin x}{1+\cos^2x}\rd x =\int_0^\pi \frac{\sex{\pi-x}\sin x}{1+\cos^2x}\rd x, </script> 而 \bex \mbox{原式}&=&\frac{\pi}{2}\int_0^\pi \frac{\sin x}{1+\cos^2x}\rd x\\ &=&\frac{\pi}{2}\int_{-1}^1\frac{1}{1+t^2}dt\\ &=&\frac{\pi^2}{4}. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\frac{\pi}{2}\int_0^\pi \frac{\sin x}{1+\cos^2x}\rd x\\ &=&\frac{\pi}{2}\int_{-1}^1\frac{1}{1+t^2}dt\\ &=&\frac{\pi^2}{4}. \eex</script>

(5) \dps{\lim_{x\to \infty,y\to a}\sex{\cos \frac{y}{x}}^\frac{x^3}{x+y^3}}<script type="math/tex">\dps{\lim_{x\to \infty,y\to a}\sex{\cos \frac{y}{x}}^\frac{x^3}{x+y^3}}</script>.

解答: \bex \mbox{原式}&=&\exp\sez{ \lim_{x\to\infty, y\to a} \frac{x^3}{x+y^3} \ln \cos\frac{y}{x}}\\ &=&\exp\sez{\lim_{x\to\infty, y\to a}\frac{x^3}{x+y^3} \cdot\sex{-2}\cdot \frac{y^2}{x^2}}\\ & &\sex{\ln\cos\frac{y}{x}\sim \cos\frac{y}{x}-1 =-2\sin^2\frac{y}{x} \sim -2\frac{y^2}{x^2}}\\ &=&\exp\sez{-2\lim_{x\to\infty, y\to a}\frac{y^2}{1+\frac{y^3}{x}}}\\ &=&e^{-2a^2} \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\exp\sez{ \lim_{x\to\infty, y\to a} \frac{x^3}{x+y^3} \ln \cos\frac{y}{x}}\\ &=&\exp\sez{\lim_{x\to\infty, y\to a}\frac{x^3}{x+y^3} \cdot\sex{-2}\cdot \frac{y^2}{x^2}}\\ & &\sex{\ln\cos\frac{y}{x}\sim \cos\frac{y}{x}-1 =-2\sin^2\frac{y}{x} \sim -2\frac{y^2}{x^2}}\\ &=&\exp\sez{-2\lim_{x\to\infty, y\to a}\frac{y^2}{1+\frac{y^3}{x}}}\\ &=&e^{-2a^2} \eex</script>

(6) \dps{ \oint\limits_L \frac{x\rd y-y\rd x}{x^2+y^2}, }<script type="math/tex">\dps{ \oint\limits_L \frac{x\rd y-y\rd x}{x^2+y^2}, }</script> 其中 L<script type="math/tex">L</script> 是不过原点的简单闭曲线.

解答: 记 P=\frac{-y}{x^2+y^2},\ Q=\frac{x}{x^2+y^2}, <script type="math/tex; mode=display"> P=\frac{-y}{x^2+y^2},\ Q=\frac{x}{x^2+y^2}, </script> 则 \frac{\p Q}{\p x}-\frac{\p P}{\p y}=0, <script type="math/tex; mode=display"> \frac{\p Q}{\p x}-\frac{\p P}{\p y}=0, </script> 而由 Green<script type="math/tex">Green</script> 公式,

(1)若 0<script type="math/tex">0</script> 在 L<script type="math/tex">L</script> 的内部,则原式 =0<script type="math/tex">=0</script>;

(2)若 0<script type="math/tex">0</script> 在 L<script type="math/tex">L</script> 的外部,则取 \ve>0<script type="math/tex">\ve>0</script> 充分小,使得 B\sex{0,\ve}<script type="math/tex">B\sex{0,\ve}</script> 也包含在 L<script type="math/tex">L</script> 的内部, 而 \bex \mbox{原式}&=&\int_{B\sex{0,\ve}} \frac{x\rd y-y\rd x}{x^2+y^2}\\ &=&\frac{1}{\ve^2} \int_0^{2\pi} \ve\cos \theta\cdot \sex{\ve \cos\theta} -\ve \sin \theta \cdot\sex{-\ve \sin \theta}d\theta\\ &=&\int_0^{2\pi}d\theta\\ &=&2\pi. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\int_{B\sex{0,\ve}} \frac{x\rd y-y\rd x}{x^2+y^2}\\ &=&\frac{1}{\ve^2} \int_0^{2\pi} \ve\cos \theta\cdot \sex{\ve \cos\theta} -\ve \sin \theta \cdot\sex{-\ve \sin \theta}d\theta\\ &=&\int_0^{2\pi}d\theta\\ &=&2\pi. \eex</script>

2 设 f(0)=0<script type="math/tex">f(0)=0</script>, f‘(0)>1<script type="math/tex">f‘(0)>1</script>. 证明 \exists\ \delta>0,\ s.t.\ x\in \sex{0,\delta}\ra f(x)\geq x<script type="math/tex">\exists\ \delta>0,\ s.t.\ x\in \sex{0,\delta}\ra f(x)\geq x</script>.

证明: 我们假设 f<script type="math/tex">f</script> 在 0<script type="math/tex">0</script> 的一个左邻域 [0,\delta_1]<script type="math/tex">[0,\delta_1]</script> 上二次连续可微, 而 \bex f(x)&=&f(0)+f‘(0)x+\frac{f‘‘\sex{\xi_x}}{2}x^2\\ &\geq&x+\sez{f‘(0)-1}x+Ax^2\\ & &\sex{A=\inf_{x\in [0,\delta_1]}f‘‘\sex{x}}\\ &=&x+\sez{f‘(0)-1+Ax}x\\ &>&x, \eex<script type="math/tex; mode=display">\bex f(x)&=&f(0)+f‘(0)x+\frac{f‘‘\sex{\xi_x}}{2}x^2\\ &\geq&x+\sez{f‘(0)-1}x+Ax^2\\ & &\sex{A=\inf_{x\in [0,\delta_1]}f‘‘\sex{x}}\\ &=&x+\sez{f‘(0)-1+Ax}x\\ &>&x, \eex</script> 当 x<\frac{f‘(0)-1}{A}=\delta, <script type="math/tex; mode=display"> x<\frac{f‘(0)-1}{A}=\delta, </script> 时.

注记: 当然, 如果只假设 f<script type="math/tex">f</script> 在 0<script type="math/tex">0</script> 的一个左邻域 [0,\delta_1]<script type="math/tex">[0,\delta_1]</script> 上连续可微也是可以的. 比如 \bex f(x)&=&f(0)+f‘(\xi_x)x\sex{\xi_x\in (0,x),\mbox{中值定理}}\\ &>&x\sex{\mbox{保号性}}. \eex<script type="math/tex; mode=display">\bex f(x)&=&f(0)+f‘(\xi_x)x\sex{\xi_x\in (0,x),\mbox{中值定理}}\\ &>&x\sex{\mbox{保号性}}. \eex</script>

3 试证明 f(x)=x^{-2}<script type="math/tex">f(x)=x^{-2}</script> 在 (0,1)<script type="math/tex">(0,1)</script> 上不一致连续, 但对任何 \delta>0<script type="math/tex">\delta>0</script>, 在 [\delta,1)<script type="math/tex">[\delta,1)</script> 上一致连续.

证明:

(1)f(x)=x^{-2}<script type="math/tex">f(x)=x^{-2}</script> 在 (0,1)<script type="math/tex">(0,1)</script> 上不一致连续. 因为 \sev{\frac{2}{n}-\frac{1}{n}} =\frac{1}{n}\to 0 \sex{n\to\infty}, <script type="math/tex; mode=display"> \sev{\frac{2}{n}-\frac{1}{n}} =\frac{1}{n}\to 0 \sex{n\to\infty}, </script> 但 \sev{f\sex{\frac{2}{n}}-f\sex{\frac{1}{n}}} =\frac{3}{4}n^2\to\infty \sex{n\to \infty}. <script type="math/tex; mode=display"> \sev{f\sex{\frac{2}{n}}-f\sex{\frac{1}{n}}} =\frac{3}{4}n^2\to\infty \sex{n\to \infty}. </script>

(2)f(x)=x^{-2}<script type="math/tex">f(x)=x^{-2}</script> 在 [\delta,1)<script type="math/tex">[\delta,1)</script> 上一致连续, 其中 \delta\in (0,1)<script type="math/tex">\delta\in (0,1)</script> 任意. 因为 \sev{f‘(x)} =\sev{-2x^{-3}} \leq 2\delta^{-3}<\infty, <script type="math/tex; mode=display"> \sev{f‘(x)} =\sev{-2x^{-3}} \leq 2\delta^{-3}<\infty, </script> 而 \bex \sev{f(x)-f(x‘)} &\leq&2\delta^{-3}\sev{x-x‘}\sex{\mbox{中值定理}}\\ &\to&0\sex{\sev{x-x‘}\to 0}. \eex<script type="math/tex; mode=display">\bex \sev{f(x)-f(x‘)} &\leq&2\delta^{-3}\sev{x-x‘}\sex{\mbox{中值定理}}\\ &\to&0\sex{\sev{x-x‘}\to 0}. \eex</script>

4 设 0<\alpha<\beta<script type="math/tex">0<\alpha<\beta</script>, \lambda<script type="math/tex">\lambda</script> 为实参数. 记 f_\lambda(x)=x^\alpha-x^\beta-\lambda. <script type="math/tex; mode=display"> f_\lambda(x)=x^\alpha-x^\beta-\lambda. </script> 证明存在 \Lambda>0<script type="math/tex">\Lambda>0</script>, 使得对任何 \lambda\in [0,\Lambda)<script type="math/tex">\lambda\in [0,\Lambda)</script>, 都存在 \delta>0,a>0<script type="math/tex">\delta>0,a>0</script>, 满足 f_\lambda\sex{a}\geq \delta<script type="math/tex">f_\lambda\sex{a}\geq \delta</script>.

证明: 其实真是没搞懂这个题目要做什么... \dps{\exists\ \Lambda=\frac{1}{2}\sex{\frac{1}{2^\alpha}-\frac{1}{2^\beta}}>0}<script type="math/tex">\dps{\exists\ \Lambda=\frac{1}{2}\sex{\frac{1}{2^\alpha}-\frac{1}{2^\beta}}>0}</script>, \forall\ \lambda\in [0,\Lambda)<script type="math/tex">\forall\ \lambda\in [0,\Lambda)</script>, \dps{\exists\ a=\frac{1}{2}}<script type="math/tex">\dps{\exists\ a=\frac{1}{2}}</script>, \delta=\Lambda<script type="math/tex">\delta=\Lambda</script>, \bex f_\lambda(a)&=&2\Lambda-\lambda\\ &>&\Lambda\\ &=&\delta. \eex<script type="math/tex; mode=display">\bex f_\lambda(a)&=&2\Lambda-\lambda\\ &>&\Lambda\\ &=&\delta. \eex</script>

5 设 p>0<script type="math/tex">p>0</script>. 讨论级数 \dps{\sum_{n=1}^\infty \frac{x^n}{n^p}}<script type="math/tex">\dps{\sum_{n=1}^\infty \frac{x^n}{n^p}}</script> 的敛散性.

解答: 考察级数 \dps{\sum_{n=1}^\infty\frac{\sev{x}^n}{n^p}}<script type="math/tex">\dps{\sum_{n=1}^\infty\frac{\sev{x}^n}{n^p}}</script>, 由 \lim_{n\to\infty} \frac{\sev{x}^{n+1}}{\sex{n+1}^p}\cdot \frac{n^p}{\sev{x}^n} =\lim_{n\to\infty} \sex{\frac{n}{n+1}}^p\sev{x}=\sev{x}. <script type="math/tex; mode=display"> \lim_{n\to\infty} \frac{\sev{x}^{n+1}}{\sex{n+1}^p}\cdot \frac{n^p}{\sev{x}^n} =\lim_{n\to\infty} \sex{\frac{n}{n+1}}^p\sev{x}=\sev{x}. </script> 而

(1)当 \sev{x}<1<script type="math/tex">\sev{x}<1</script> 时, 原级数绝对收敛;

(2)当 \sev{x}=1<script type="math/tex">\sev{x}=1</script> 时,由 \dps{\sum_{n=1}^\infty\frac{\sex{-1}^n}{n^p}}<script type="math/tex">\dps{\sum_{n=1}^\infty\frac{\sex{-1}^n}{n^p}}</script> 收敛, 及 \bex \sum_{n=1}^\infty \frac{1}{n^p}\left\{ \ba{ll} \mbox{发散}, &\mbox{当}p\leq 1,\\ \mbox{收敛}, &\mbox{当}p> 1, \ea \right. \eex<script type="math/tex; mode=display">\bex \sum_{n=1}^\infty \frac{1}{n^p}\left\{ \ba{ll} \mbox{发散}, &\mbox{当}p\leq 1,\\ \mbox{收敛}, &\mbox{当}p> 1, \ea \right. \eex</script> 知 (a) 当 x=-1<script type="math/tex">x=-1</script>, p\leq 1<script type="math/tex">p\leq 1</script> 时, 原级数条件收敛;

(b)当 p>1<script type="math/tex">p>1</script> 时, 原级数绝对收敛;

(c)当 x=1<script type="math/tex">x=1</script>, p\leq 1<script type="math/tex">p\leq 1</script> 时, 原级数发散;

(3)当 \sev{x}>1<script type="math/tex">\sev{x}>1</script> 时, 原级数发散.

6 设 f(x)<script type="math/tex">f(x)</script> 在 [a,b]<script type="math/tex">[a,b]</script> 上有界, 记 f^+\sex{x}=\max\sed{f(x),0},\ f^-\sex{x}=\max\sed{-f(x),0}. <script type="math/tex; mode=display"> f^+\sex{x}=\max\sed{f(x),0},\ f^-\sex{x}=\max\sed{-f(x),0}. </script> 证明 f(x)<script type="math/tex">f(x)</script> 在 [a,b]<script type="math/tex">[a,b]</script> 上可积的充分必要条件是 f^+(x)<script type="math/tex">f^+(x)</script>,f^-(x)<script type="math/tex">f^-(x)</script> 在 [a,b]<script type="math/tex">[a,b]</script> 上均可积, 并且 \int_a^bf(x)\rd x =\int_a^bf^+(x)\rd x +\int_a^bf^-(x)\rd x. <script type="math/tex; mode=display"> \int_a^bf(x)\rd x =\int_a^bf^+(x)\rd x +\int_a^bf^-(x)\rd x. </script>

证明: 注意到 f(x)=f^+(x)+f^-(x), <script type="math/tex; mode=display"> f(x)=f^+(x)+f^-(x), </script> 我们仅须证明等价性. 注意到

(1)f<script type="math/tex">f</script> 连续 \dps{\ra f^+=\frac{\sev{f}+f}{2},\ f^-=\frac{\sev{f}-f}{2}}<script type="math/tex">\dps{\ra f^+=\frac{\sev{f}+f}{2},\ f^-=\frac{\sev{f}-f}{2}}</script> 连续;

(2)f^+<script type="math/tex">f^+</script>, f^-<script type="math/tex">f^-</script> 连续 \ra f=f^++f^-<script type="math/tex">\ra f=f^++f^-</script> 连续; 而 \bex f\mbox{ 的不连续点集是零测集} \lra f^+,f^-\mbox{ 的不连续点集是零测集}, \eex<script type="math/tex; mode=display">\bex f\mbox{ 的不连续点集是零测集} \lra f^+,f^-\mbox{ 的不连续点集是零测集}, \eex</script> 于是 (<script type="math/tex">(</script>实变理论)<script type="math/tex">)</script> \bex f\mbox{ 可积}\lra f^+,\ f^-\mbox{ 可积}. \eex<script type="math/tex; mode=display">\bex f\mbox{ 可积}\lra f^+,\ f^-\mbox{ 可积}. \eex</script>

注记: 设 f<script type="math/tex">f</script> 是 [a,b]<script type="math/tex">[a,b]</script> 上有界函数,则 f<script type="math/tex">f</script> 是 Riemann<script type="math/tex">Riemann</script> 可积的充分必要条件是: f<script type="math/tex">f</script> 的不连续点集是零测集.

7 f(x,y)<script type="math/tex">f(x,y)</script> 对两个变元均连续, 且关于其中一个变元单调, 则它就是两个变元混合连续的.

证明: 不妨设 f<script type="math/tex">f</script> 关于 y<script type="math/tex">y</script> 单增.对任意固定的 (x_0,y_0)<script type="math/tex">(x_0,y_0)</script>,

(1)由 f<script type="math/tex">f</script> 关于 y<script type="math/tex">y</script> 连续, \forall\ \ve>0,\ \exists\ \delta_1>0,\ s.t.<script type="math/tex">\forall\ \ve>0,\ \exists\ \delta_1>0,\ s.t.</script> \bee\label{lz08sf_7_f:y} \sev{y-y_0}\leq \delta_1\ra \sev{f(x,y_0)-f(x_0,y_0)}<\ve/2; \eee<script type="math/tex; mode=display">\bee\label{lz08sf_7_f:y} \sev{y-y_0}\leq \delta_1\ra \sev{f(x,y_0)-f(x_0,y_0)}<\ve/2; \eee</script>

(2)由 f<script type="math/tex">f</script> 关于 x<script type="math/tex">x</script> 连续, \exists\ 0<\delta<\delta_1,\ s.t.<script type="math/tex">\exists\ 0<\delta<\delta_1,\ s.t.</script> \bee\label{lz08sf_7_f:x} \sev{x-x_0}\leq \delta\ra \left\{ \ba{ll} \dps{\sev{f(x,y_0+\delta)-f(x_0,y_0+\delta)}<\ve/2,}\\ \dps{\sev{f(x,y_0-\delta)-f(x_0,y_0-\delta)}<\ve/2,} \ea \right. \eee<script type="math/tex; mode=display">\bee\label{lz08sf_7_f:x} \sev{x-x_0}\leq \delta\ra \left\{ \ba{ll} \dps{\sev{f(x,y_0+\delta)-f(x_0,y_0+\delta)}<\ve/2,}\\ \dps{\sev{f(x,y_0-\delta)-f(x_0,y_0-\delta)}<\ve/2,} \ea \right. \eee</script> 现对任意的 (x,y)<script type="math/tex">(x,y)</script> 满足 \dps{\sev{x-x_0}\leq \delta,\ \sev{y-y_0}\leq \delta}<script type="math/tex">\dps{\sev{x-x_0}\leq \delta,\ \sev{y-y_0}\leq \delta}</script>, 有

(1)\bex f(x,y)&\leq&f(x,y_0+\delta)\sex{\mbox{由 }f\mbox{ 关于 }y\mbox{单增}}\\ &\leq&f(x_0,y_0+\delta)+\ve/2\sex{\mbox{由 }\eqref{lz08sf_7_f:x}_1}\\ &\leq&f(x_0,y_0)+\ve\sex{\mbox{由 }\eqref{lz08sf_7_f:y}}, \eex<script type="math/tex; mode=display">\bex f(x,y)&\leq&f(x,y_0+\delta)\sex{\mbox{由 }f\mbox{ 关于 }y\mbox{单增}}\\ &\leq&f(x_0,y_0+\delta)+\ve/2\sex{\mbox{由 }\eqref{lz08sf_7_f:x}_1}\\ &\leq&f(x_0,y_0)+\ve\sex{\mbox{由 }\eqref{lz08sf_7_f:y}}, \eex</script>

(2)\bex f(x,y)&\geq&f(x,y_0-\delta)\sex{\mbox{由 }f\mbox{ 关于 }y\mbox{单增}}\\ &\geq&f(x_0,y_0-\delta)+\ve/2\sex{\mbox{由 }\eqref{lz08sf_7_f:x}_2}\\ &\geq&f(x_0,y_0)-\ve\sex{\mbox{由 }\eqref{lz08sf_7_f:y}}. \eex<script type="math/tex; mode=display">\bex f(x,y)&\geq&f(x,y_0-\delta)\sex{\mbox{由 }f\mbox{ 关于 }y\mbox{单增}}\\ &\geq&f(x_0,y_0-\delta)+\ve/2\sex{\mbox{由 }\eqref{lz08sf_7_f:x}_2}\\ &\geq&f(x_0,y_0)-\ve\sex{\mbox{由 }\eqref{lz08sf_7_f:y}}. \eex</script> 这就证到了结论.

8 设连续函数 f(x)<script type="math/tex">f(x)</script> 满足 f(1)=1<script type="math/tex">f(1)=1</script>, 记 F(t)=\int_{x^2+y^2+z^2\leq t^2}f\sex{x^2+y^2+z^2}\rd x\rd y\rd z. <script type="math/tex; mode=display"> F(t)=\int_{x^2+y^2+z^2\leq t^2}f\sex{x^2+y^2+z^2}\rd x\rd y\rd z. </script> 证明: F‘(1)=4\pi<script type="math/tex">F‘(1)=4\pi</script>.

证明:

(1)先化简 F<script type="math/tex">F</script>, \bex F(t)=4\pi\int_0^t r^2f(r^2)dr, \eex<script type="math/tex; mode=display">\bex F(t)=4\pi\int_0^t r^2f(r^2)dr, \eex</script>

(2)再计算, \bex F‘(t)=4\pi t^2f(t^2), \eex<script type="math/tex; mode=display">\bex F‘(t)=4\pi t^2f(t^2), \eex</script> \bex F‘(1)=4\pi f(1)=4\pi. \eex<script type="math/tex; mode=display">\bex F‘(1)=4\pi f(1)=4\pi. \eex</script>

9 称 f(x)<script type="math/tex">f(x)</script> 是凸函数, 如果对任意的 \lambda\in (0,1)<script type="math/tex">\lambda\in (0,1)</script>, 均有 f\sex{\lambda x+\sex{1-\lambda}y}\leq \lambda f(x)+\sex{1-\lambda}f(y). <script type="math/tex; mode=display"> f\sex{\lambda x+\sex{1-\lambda}y}\leq \lambda f(x)+\sex{1-\lambda}f(y). </script>

(1) 试给出凸函数的几何解释;

(2)若 f(x)<script type="math/tex">f(x)</script> 是区间 I<script type="math/tex">I</script> 上的凸函数,试讨论 f(x)<script type="math/tex">f(x)</script> 在 I<script type="math/tex">I</script> 上的连续性; (3)若 f(x)<script type="math/tex">f(x)</script> 下有界, 即存在常数 M<script type="math/tex">M</script>, 使得对任何 x<script type="math/tex">x</script>, 都有 f(x)\geq M<script type="math/tex">f(x)\geq M</script>, 问 f(x)<script type="math/tex">f(x)</script> 是否有最小值? 证明你的结论.

证明:

(1)联结f<script type="math/tex">f</script>图像上两点的直线段总在f<script type="math/tex">f</script>图像的上方.

(2)对 \forall\ x_1, x_2, x_3\in I<script type="math/tex">\forall\ x_1, x_2, x_3\in I</script> 满足 x_1<x_2<x_3<script type="math/tex">x_1 \lambda=\frac{x_3-x_2}{x_3-x_1}, <script type="math/tex; mode=display"> \lambda=\frac{x_3-x_2}{x_3-x_1}, </script> 而 x_2=\lambda x_1+\sex{1-\lambda}x_3<script type="math/tex">x_2=\lambda x_1+\sex{1-\lambda}x_3</script>, \bex f(x_2)&=&f\sex{\lambda x_1+\sex{1-\lambda}x_3}\\ &\leq&\lambda f(x_1)+\sex{1-\lambda}f(x_3)\\ &=&\frac{x_3-x_2}{x_3-x_1}f(x_1) +\frac{x_2-x_1}{x_3-x_1}f(x_3), \eex<script type="math/tex; mode=display">\bex f(x_2)&=&f\sex{\lambda x_1+\sex{1-\lambda}x_3}\\ &\leq&\lambda f(x_1)+\sex{1-\lambda}f(x_3)\\ &=&\frac{x_3-x_2}{x_3-x_1}f(x_1) +\frac{x_2-x_1}{x_3-x_1}f(x_3), \eex</script> 即 \frac{f(x_2)-f(x_1)}{x_2-x} \leq \frac{f(x_3)-f(x_1)}{x_3-x_1},\ \ \ \frac{f(x_2)-f(x_1)}{x_2-x} \leq \frac{f(x_3)-f(x_2)}{x_3-x_2}. <script type="math/tex; mode=display"> \frac{f(x_2)-f(x_1)}{x_2-x} \leq \frac{f(x_3)-f(x_1)}{x_3-x_1},\ \ \ \frac{f(x_2)-f(x_1)}{x_2-x} \leq \frac{f(x_3)-f(x_2)}{x_3-x_2}. </script> 于是对任意固定的 x\in I^o<script type="math/tex">x\in I^o</script> (I<script type="math/tex">(I</script> 的内部)<script type="math/tex">)</script>, \bex h>k\ra \frac{f(x+h)-f(x)}{h}>\frac{f(x+k)-f(x)}{k}>\frac{f(x)-f(x-\delta}{\delta}, \eex<script type="math/tex; mode=display">\bex h>k\ra \frac{f(x+h)-f(x)}{h}>\frac{f(x+k)-f(x)}{k}>\frac{f(x)-f(x-\delta}{\delta}, \eex</script> 其中 \delta>0<script type="math/tex">\delta>0</script> 充分小. 而对 \forall\ h_l\to 0_+<script type="math/tex">\forall\ h_l\to 0_+</script>, \bex \sed{\frac{f(x+h_l)-f(x)}{h_l}}_{l=1}^\infty \eex<script type="math/tex; mode=display">\bex \sed{\frac{f(x+h_l)-f(x)}{h_l}}_{l=1}^\infty \eex</script> 单减有下界, 而极限存在. 简单的论述可以说明极限与所选取之特殊子列无关, 而 f‘_+(x)<script type="math/tex">f‘_+(x)</script> 存在. 类似的, f‘_-(x)<script type="math/tex">f‘_-(x)</script> 也存在. 于是 f<script type="math/tex">f</script> 左右连续,而连续. 当 I<script type="math/tex">I</script> 有左或右端点时, f<script type="math/tex">f</script> 在左端点左连续,或在右端点右连续.

(3)若 I<script type="math/tex">I</script> 是闭的,则 f<script type="math/tex">f</script> 一定存在最小值. 实际上, 令 m=\inf f(x)<script type="math/tex">m=\inf f(x)</script>, 则 \bex \exists\ x_n\in I,\ s.t.\ m\leq f(x_n)<m+1/n. \eex<script type="math/tex; mode=display">\bex \exists\ x_n\in I,\ s.t.\ m\leq f(x_n)Weierstrass<script type="math/tex">Weierstrass</script> 聚点定理, \exists\ \sed{n_k}\subset \sed{n},\ s.t.\ x_{n_k}\to x\in I, <script type="math/tex; mode=display"> \exists\ \sed{n_k}\subset \sed{n},\ s.t.\ x_{n_k}\to x\in I, </script> 而 m\leq f(x)=\lim_{k\to\infty}f(x_{n_k})\leq \lim_{k\to\infty}\sex{m+1/{n_k}}=m, <script type="math/tex; mode=display"> m\leq f(x)=\lim_{k\to\infty}f(x_{n_k})\leq \lim_{k\to\infty}\sex{m+1/{n_k}}=m, </script> 即 f(x)=m. <script type="math/tex; mode=display"> f(x)=m. </script> 但是若 I<script type="math/tex">I</script> 不是闭的, 那就未必了. 最简单的例子是 f(x)=x,\ x\in (0,1)<script type="math/tex">f(x)=x,\ x\in (0,1)</script>.