Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 15

Problem

Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a,\infty)$, and $M_0,M_1,M_2$ are the least upper bounds of $|f(x)|,|f'(x)|,|f''(x)|$, respectively, on $(a,\infty)$. Prove that

$$ M_1^2 \leq 4M_0M_2.$$

Does $M_1^2 \leq 4M_0M_2$ hold for vector-valued functions too?

Answer

For $x \in (a,\infty)$ and $h>0$, $f(x+2h)=f(x)+2f'(x)h + \frac{1}{2}f''(\xi)(2h)^2$ for some $\xi$ between $x$ and $x+2h$ by Theorem 5.15. Then $|f'(x)| \leq \frac{|f(x+2h) - f(x)|}{2h} + |f''(\xi)h| \leq \frac{M_0}{h} + M_2h$. This holds for all $x$ and $h$ and

$$ \inf_h(\frac{M_0}{h} + M_2h) = 2\sqrt{M_0M_2}$$

Hence, we have $M_1 \leq 2\sqrt{M_0M_2}$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2.$

This inequality also holds for vector-valued functions $f=(f_1,\cdots,f_k)$. We may assume that $M_1 \neq 0$. For a fixed point $y \in (a,\infty)$, define a function

$$ g(x):=f_1'(y)f_1(x) + \cdots + f_k'(y)f_k(x). $$

Then we have

$$|g'(x)|^2 \leq (\sup_x |g'(x)|)^2 \leq 4(\sup_x|g(x)|)(\sup_x|g''(x)|) \\ \leq 4(\sum_{i=1}^k|f_i'(y)|^2)^{1/2}\sup_x(\sum_{i=1}^k|f_i(x)|^2)^{1/2}(\sum_{i=1}^k|f_i'(y)|^2)^{1/2}\sup_x(\sum_{i=1}^k|f_i''(x)|^2)^{1/2} \leq 4 M_1^2M_0M_2.$$

For $y=x$, we have $(\sum_{i=1}^k|f_i'(x)|^2)^2 \leq 4 M_1^2M_0M_2$. This holds for all $x$, and hence $M_1^4 = (\sup_x\sum_{i=1}^k|f_i'(x)|^2)^2 \leq 4 M_1^2M_0M_2$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2$.

Comments

Post a Comment

Popular posts from this blog

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 4

Problem Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p)=f(p)$ for all $p \in E$, prove that $g(p)=f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Answer For $y \in f(X)$, there exists $x \in X$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $E$ is dense in $X$, there exists $z \in E$ such that $z \in N_\delta(x)$, which implies $f(x) \in N_\epsilon(f(z))$. This means $f(E)$ is dense in $f(X)$. For $p \in X$ and $\epsilon >0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2$. For $q \in E$ satisfying $q \in N_\delta(p)$, $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q)...
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 14

Problem Give an example of an open cover of the segment $(0,1)$ which has no finite subcover. Answer Consider a family of sets $\{(1/n,1)\}_{n=2}^\infty$. For any $x \in (0,1)$, there exists $N \in \mathbb{N}$ such that $1/N < x$, which means $x \in (1/N,1)$. This means that $\{(1/n,1)\}_{n=2}^\infty$ covers $(0,1)$. Consider an arbitrary finite subcover $\{(1/n_1,1),\cdots,(1/n_K)\}$ of $\{(1/n,1)\}_{n=2}^\infty$. Then $\bigcup_{k=1}^K(1/n_k,1)=(1/n_\ast,1)$ where $n_\ast=\max(n_1,\cdots,n_K)$, which cannot cover $(0,1)$.
Read more