Math Analysis Homework - Week 10

Class 1

\begin{gathered} f(x)\in C^1[a,b] \\ \implies \max_{x\in [a,b]} \vert f(x) \vert \le \dfrac{1}{b-a} \int_a^b \vert f(x) \vert dx+\int_a^b \vert f'(x) \vert dx \end{gathered}

\begin{gathered} \text{let } \max_{x\in [a,b]}\vert f(x) \vert =\vert f(x_0) \vert \\ \dfrac{1}{b-a} \int_a^b \vert f(x) \vert dx \\ \xlongequal{\text{ Fist Mean Value Theorem }} f(\xi),\xi \in [a,b] \\ \vert f(x_0) \vert = \vert f(\xi)+\int_{\xi}^{x_0} f'(x)dx \vert \\ \le \vert f(\xi) \vert +\int_\xi^{x_0} \vert f'(x) \vert dx \\ \text{Q.E.D} \end{gathered}

\begin{gathered} f(x)\in C^1[0,1],f(0)=0 \\ \implies \int_0^1 \vert f(x) \vert ^2dx\le \int_0^1 \vert f'(x) \vert ^2 dx \end{gathered}

\begin{gathered} f(x)=\int_0^x f'(t)dt \\ =\int_0^x f'(t)\cdot 1dt \\ \le (\int_0^x f'(t)^2 dt)^\frac12(\int_0^x dt)^\frac12 \\ \implies f(x)^2 \le x\int_0^x f'^2(t)dt \\ \le x\int_0^1 f'^2(t)dt \\ \implies \int_0^1 f(x)^2 dx\le \dfrac12 \int_0^1 f'^2(t)dt\le \int_0^1 f'^2(t)dt \end{gathered}

\begin{gathered} f(x)\in C[-1,1] \\ \implies \lim_{h \to 0^+} \int_{-1}^1 \dfrac{h}{h^2+x^2} f(x)dx=\pi f(0) \end{gathered}

\begin{gathered} \lim_{h \to 0^+} \int_{-1}^{-\delta} \dfrac{h}{h^2+x^2} +\int_{\delta}^1 \dfrac{h}{h^2+x^2} \\ \le \lim_{h \to 0^+} \int_{-1}^{-\delta}\dfrac{h}{\delta^2}dx +\int_{\delta}^1 \dfrac{h}{\delta^2} dx \\ \le \lim_{h \to 0^+} \dfrac{2h}{\delta^2} \\ =0 \\ \forall \delta>0, Ans=\lim_{h \to 0^+} \int_{-\delta}^\delta \dfrac{h}{h^2+x^2} f(x)dx \\ f(x)\in C[-1,1] \implies \forall p<1,\exists r \ s.t.\ \\ \text{WLOG,assume } f(x)>0 \\ \vert x \vert <r \implies f(x) \in (pf(0),\dfrac{1}{p}f(0)) \\ \text{let } \delta=r \implies \\ I=f(0)(\lim_{h \to 0^+} \int_{-r}^r \dfrac{h}{h^2+x^2} dx) \\ Ans\in (pI,\dfrac{I}{p} ) \\ I=f(0)\lim_{h \to 0^+} 2\arctan \frac rh=\pi f(0) \\ Ans=\lim_{p \to 1} pI=\lim_{p \to 0} \dfrac{I}{p} =\pi f(0) \end{gathered}

\begin{gathered} \begin{cases} f(x)\in D^2[a,b] \\ f(\dfrac{a+b}{2} )=0 \end{cases} \\ \implies \vert \int_a^b f(x)dx \vert \le \dfrac{(b-a)^3}{24} \sup_{x\in[a,b]}\vert f''(x) \vert \end{gathered}

\begin{gathered} \text{let } m=\dfrac{a+b}{2} \\ f(x)=(x-m )f'(m)+\dfrac{f''(\xi)}{2} (x-m)^2 \\ \le (x-m)f'(m)+\dfrac{\vert \sup f''(\xi)\vert}{2} (x-m)^2 \\ \implies \int_a^b f(x)=f'(m)\int_a^b (x-m)dx+\vert \sup f''(x)\vert \int_a^b \dfrac{(x-m)^2}{2}dx \\ =\dfrac{(b-a)^3}{24} \sup \vert f''(x) \vert \end{gathered}

\begin{gathered} \begin{cases} f(x) \in R[0,1] \\ \int_0^1 f(x)dx=1 \\ \int_0^1 xf(x)dx=0 \end{cases} \\ \implies \begin{cases} \text{calculate } I(a)=\int_0^1 \vert ax-1 \vert dx,a\ge 0 \\ \sup_{x\in [0,1]} \vert f(x) \vert \ge \sqrt{2}+1 \end{cases} \end{gathered}

(1)

By geometry, $a<1$ 显然劣于 $a=1$ ,考虑 $a\ge 1$

\begin{gathered} I(a)=\dfrac{1}{2a} +\dfrac{(a-1)(1-\dfrac{1}{a} )}{2} \\ =\dfrac{a}{2} +\dfrac{1}{a} -1 \\ \ge \sqrt 2-1 \end{gathered}

(2)

考虑反证,则 $\vert f(x) \vert <\sqrt 2+1$

\begin{gathered} -1=\int_0^1 f(x)(\sqrt 2x-1)dx \\ 1=\vert \int_0^1 f(x)(\sqrt 2x-1)dx \vert \\ <(\sqrt2+1)\int_0^1 \vert \sqrt2 x-1 \vert dx \\ =(\sqrt 2+1)(\sqrt 2-1) \\ =1 \end{gathered}

矛盾,得证.

\begin{gathered} f(x)\in R[a,b] \\ \implies \forall \epsilon>0,\exists p(x),q(x) \text{ are step functions} ,f(x)\in [p(x),q(x)]\\ s.t.\\ \int_a^b (q(x)-p(x))dx<\epsilon \end{gathered}

\begin{gathered} f(x)\in R[a,b] \\ \implies \forall \epsilon,\exists T \ s.t.\ \sum _{i = 1} ^{n} (M_i-m_i)\Delta x_i<\epsilon \\ \text{where } M_i=\sup_{x\in [t_{i-1},t_i]} f(x),m_i=\inf_{x\in [t_{i-1},t_i]} f(x) \\ \text{let } q(x)=\sum _{i = 1} ^{n} M_i[x\in [t_{i-1},t_i]] \\ p(x)=\sum _{i = 1} ^{n} m_1[x\in [t_{i-1},t_i]] \\ \text{where } [p]=1 \iff p \text{ is true} \\ \int_a^b (q(x)-p(x))dx=\sum _{i = 1} ^{n} (M_i-m_i)\Delta x_i<\epsilon \end{gathered}

\begin{gathered} f(x) \text{ is increasing at } [a,b] \\ \implies \int_a^b xf(x)dx \ge \dfrac{a+b}{2} \int_a^b f(x)dx \end{gathered}

\begin{gathered} \text{let } m=\dfrac{a+b}{2} \\ \int_a^m (x-m )f(x) \\ =\int_m^b (m-x)f(a+b-x) \\ \int_a^b (x-m)f(x)=\int_a^m (x-m)f(x)+\int_m^b (x-m)f(x) \\ =\int_m^b (x-m)f(x)-\int_m^b (x-m)f(a+b-x) \\ =\int_m^b (x-m)(f(x)-f(a+b-x)) \\ \ge 0 \end{gathered}